Mahesh

Students get through AP Inter 2nd Year Physics Important Questions 1st Lesson Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 1st Lesson Waves

Very Short Answer Questions

Question 1.
Write the formula for speed of sound in solids and gases.
Answer:
Speed of sound in solids,
V_s = \(\sqrt{Y / \rho}\) [y = Young’s modulus of solid, ρ = density of solid]
Speed of sound in gases,
V_s = \(\sqrt{\gamma P / \rho}\) [γP = Adiabatic Bulk modulus of gas, ρ = density of gas] .

Question 2.
What does a wave represent ?
Answer:
A wave represents the transport of energy through a medium from one point to another without translation of the medium.

Question 3.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare factions are formed alternatively.

Question 4.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = \(\frac{2 \pi}{T}\); k = \(\frac{2 \pi}{\lambda}\)
Parameters:

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 5.
What is the principle of superposition of waves ? .
Answer:
When two or more waves are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.
If y₁, y₂, ……………… y_n be the individual displacements of the particles,then resultant displacement
y = y₁ + y₂ + ……………… + y_n.

Question 6.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 7.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary ?
Answer:
Phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary is radian or 180°.

Question 8.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transeverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 9.
What do you understand by the terms node’ and ‘antinode’ ?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes: The points at which the amplitude is maximum, are called antinodes.

Question 10.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is \(\frac{\lambda}{4}\).

Question 11.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 12.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics..
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 13.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + \(\frac{1}{2}\)) \(\frac{v}{2l}\) where n = 0, 1, 2, 3, ……….

Question 14.
If the air column in a long tube, closed at one end, is set in vibration, what harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n +1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3,

Question 15.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible ?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by no
v_n = \(\frac{\mathrm{nv}}{21}\)
where n = 1, 2, 3, ……………….

Question 16.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.
Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.
It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard, ∆υ = υ₁ ~ υ₂
Importance:
1. It can be used to tune musical instruments.
2. Beats are used to detect dangerous gases.
Explanation for tuning musical instruments with beats:
Musicians use. the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 4.
What is ‘Doppler effect’ ? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between .the observer and the source of sound is called doppler effect.
Examples:

1. The frequency of whistling, engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Sample Problem on Doppler effect:
Two trucks heading in opposite direction with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330m/s). After the two trucks have passed each other, what frequency does the driver of the second truckhear?
Answer:
Speed of first truck = 60 kmph
= 60 × \(\frac{5}{18}\) = 16.66 m/s;
Speed of second truck = 70 kmph 5
= 70 × \(\frac{5}{18}\) = 19.44 m/s
Frequency of horn of first truck = 400 Hz;
Velocity of sound, (V) =330 m/s
Frequency of sound heard by the driver of the second truck when approaching each other,
v¹ = \(\left(\frac{V+V_0}{V-V_s}\right) v=\left(\frac{330+19.44}{330-16.66}\right)\) × 400 = 446 Hz
Frequency of sound heard by the driver of the second truck when approaching each other,
V¹¹ = \(\left(\frac{\mathrm{V}-\mathrm{V}_0}{\mathrm{~V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{v}=\left(\frac{330-19.44}{330+16.66}\right)\) × 400 = 358.5 Hz

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. [IPE]
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length λ and frequency v, travelling in opposite direction be given by

y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω = 2πv and k = \(\frac{2 \pi}{\lambda}\)
The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}, \frac{2 \lambda}{2}, \frac{3 \lambda}{2}\) ……………… etc, the amplitude = zero
These positions are known as “Nodes”.
If x = \(\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4}\) …………… etc., the amplitude = maximum (2a)
These positions are called “Antinodes”.
If the string vibrates in ‘P’ segments and T is its length, then length of each segment = \(\frac{l}{\mathrm{P}}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{P}}=\frac{\lambda}{2} \Rightarrow \lambda=\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v \mathrm{P}}{2 l}\) ………………. (1)
If’ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is
v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) …………… (2)
From the Eqs (1) and (2) :
Harmonic frequency v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\Gamma}{\mu}}\)
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) …………….. (3)

Laws of Transverse Waves Along Stretched String:
Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when ‘T’ and ‘μ’ are constant. .

Second Law: When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ \(\sqrt{T}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.
JT .
Third Law: When the length (l) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and T are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. [A.P. 17; IPE 2015, 2016 (TS)]
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super-imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two anti nodes at two ends of the pipe with a node between them.

∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)
l = \(\frac{\lambda_1}{2}\) ⇒ λ₁ = 2l
fundamental frequency v₁ = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity,of sound in air, v₁ = \(\frac{v}{21}\) = v
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ₂ = \(\frac{21}{2}\)
If ‘v₂’ is frequency of second harmonic then v₂ = \(\frac{v}{\lambda_2}=\frac{v \times 2}{2 l}\) = 2v
v₂ = 2v ……………… (2)
Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)
λ₃ = \(\frac{2l}{3}\)
If ‘v₂‘ is frequency of third harmonic then
v₃ = \(\frac{v}{\lambda_3}=\frac{v \times 3}{2 l}\) = 3V
v₃ = 3v …………… (3)
Similarly we can find the remaining or higher harmonic frequencies i.e., v3, v4 etc., can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : V₁ : v₂ = 1 : 2 : 3

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. [IPE 2015, 2016(A.P.), (T.S) A.P. & T.S. Mar. 15]
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.

∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air
v₁ = \(\frac{v}{4 l}\) = v …………….. (1)
To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.
∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ₃‘ is wave length of third harmonic
λ₃ = \(\frac{4l}{3}\)
where ‘X3’ is wave length of third harmonic.
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = \(\frac{v}{\lambda_3}=\frac{3 v}{41}\)
v₃ = 3v …………….. (2)
Similarly the next overtone in the close pipe is only fifth harmonic, it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ₅)
∴ l = \(\frac{5 \lambda_5}{4}\) where ‘λ₅‘ is wave length of fifth harmonic.
λ₅ = \(\frac{4l}{5}\)
If ‘v₅‘ is frequency of fifth harmonic (second overtone)
v₅ = \(\frac{v}{\lambda_5}=\frac{5 v}{4 l}\)
v₅ = 5v …………….. (3)
∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What is Doppler effect ? Obtain an expression for the apparent frequency of sound heard when the source is in motion with respect to an observer at rest. [Mar. 17, BMP, 2016 (AP) Mar. 14, (TS)]
Answer:
Doppier effect: The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppier effect. When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = Listener
Let ‘S’ be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T = υ_s. T Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength λ’ = λ – υ_sT.
λ’ = λ – \(\frac{v_s}{v}\) [∵ υ = \(\frac{1}{T}\)]
= \(\frac{\lambda v-v_s}{v}=\frac{v-v_s}{v}\) [∵ υ = vλ]
If “v'” is apparent frequency heard by the listener then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is Velocity of sound in air
v’ = \(\frac{v . v}{v-v_S}\)
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency v’ = \(\frac{v . v}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity.

Problems

Question 1.
A stretched wire of length 0.6 m is observed to vibrate with a frequency of 30 Hz in the fundamental mode. If the string has a linear mass of 0.05 kg / m find (a) the velocity of propagation of transverse waves in the string (b) the tension in the string. [IPE 2016 (T.S)
Answer:
v = 30 Hz; I = 0.6 m; μ = 0.05 kg m^-1
υ = ?; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N .

Question 2.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;
μ = \(\frac{\mathrm{M}}{1}=\frac{0.16 \times 10^{-3}}{0.4}\) = 0.4 × 10^-3 kg/m;
T = 70 N; v_n = \(\frac{\mathrm{P}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
v₁ = \(\frac{1}{21} \sqrt{\frac{\mathrm{T}}{\mu}}=\frac{1}{2 \times 0.4} \sqrt{\frac{70}{0.4 \times 10^{-3}}}\) = 523 Hz
v₂ = 2v₁ = 2 × 523 = 1046 Hz
v₃ = 3v₁ = 3 × 523 = 1569 Hz

Question 3.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
l = 70 cm = 70 × 10^-2m; v = 331 m/s ;
v = ?
v = \(\frac{v}{4 l}=\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz.

Question 4.
A steel cable of diameter 3 cm is kept under a tension of lOkN. The density of steel is 7.8 g/ cm3. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = \(\frac{D}{2}=\frac{3}{2}\) cm
= \(\frac{3}{2}\) × 10^-2m;

Question 5.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle. [T.S. Mar. 17]
Solution:
When a whistling train away from rest observer,
v’ = \(\left[\frac{v}{v-v_s}\right] v\) ……………… (1)
When a whistling train away from rest observer,
v” = \(\left[\frac{v}{v+v_s}\right] v\) ……………… (2)
Here v’ = 219 Hz; v” = 184 Hz;
v = 340 m/s

Question 6.
A rocket is moving at a speed of 200 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+v_{\mathrm{S}}}{v}\right]^{-1}\) and not the approximate
v = v₀ [1 – \(\frac{v_s}{v}\)]
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ [1 – \(\frac{v_s}{v}\)]^-1
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v₀, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Textual Examples

Question 1.
Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both.
a) Motion of kink in a longitudinal spring produced by displacing one end of the spring sideways.
b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
c) Waves produced by a motorboat sailing in water.
d) Ultrasonic waves in air produced by a vibrating quartz crystal.
Solution:
a) Transverse and longitudinal
b) Longitudinal
c) Transverse and longitudinal
d) Longitudinal.

Question 2.
A wave travelling along a string is des-cribed by, y(x, t) = 0.005 sin (80.0 x – 3.01), in which the numerical constants are in SI untis (0.005 m, 80.0 rad m^-1, and 3.0 rad s^-1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ?
Solution:
On comparing this displacement equation with Eq. y (x, t)n = a sin(kx – ωt + Φ)
y(x, t) = a sin (kx – ωt).
We find
a) the amplitude of the wave is
0. 005 m = 5 mm.

b) the angular wave number k and angular frequency ω are k = 80.0 m^-1 and ω = 3.0 s^-1
We then relate the wavelength λ to k through Eq.
λ = \(\frac{2 \pi}{K}\)
= \(\frac{2 \pi}{80.0 \mathrm{~m}^{-1}}\) = 7.85 cm

c) Now we relate T to ω by the relation
T = \(\frac{2 \pi}{\omega}\)
= \(\frac{2 \pi}{3.0 \mathrm{~s}^{-1}}\)
= 2.09 s
and frequency, v = \(\frac{1}{T}\) = 0.48 Hz
The displacement y at x = 30.0 cm and time t= 20s is given by
y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)
= (0.005 m) sin (-36 + 12π)
= (0.005 m) sin (1.699)
= (0.005 m) sin (97°) ≃ 5 mm

Question 3.
A steel wire 0.72 m long has a mass of 5.0 × 10^-3 kg. If the .wire is under a tension of 60 N, what is the speed of transverse waves on the wire ? [A.P. Mar. 19]
Solution:
Mass per unit length of the wire,
μ = \(\frac{5.0 \times 10^{-3} \mathrm{~kg}}{0.72 \mathrm{~m}}\) = 6.9 × 10^-3 kg m^-1
Tension, T = 60 N
The speed of wave on the wire is given by
υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{60 \mathrm{~N}}{6.9 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}}}\) = 93 m s^-1

Question 4.
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 × 10^-3 kg.
Solution:
We know that 1 mole of any gas occupies 22.4 litres at STP Therefore, density of air at STP is:
ρ₀ = (mass of one mole of air) / (Volume of one mole of air at STP)
= \(\frac{29.0 \times 10^{-3} \mathrm{~kg}}{22.4 \times 10^{-3} \mathrm{~m}^3}\) = 1.29 kgm^-3
According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,
υ = \(\left[\frac{1.01 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}}{1.29 \mathrm{~kg} \mathrm{~m}^{-3}}\right]^{1 / 2}\) = 280 m s^-1

Question 5.
A pipe, 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source ? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s^-1.
Solution:
The first harmonic frequency is given by
v¹ = \(\frac{v}{\lambda_1}=\frac{v}{2 L}\) (open pipe)
Where L is the length of the pipe. The frequency of its n^th; harmonic is
v_n = \(\frac{n v}{2 L}\) for n = 1, 2, 3, ……………… (open pipe)
First few modes of an open pipe are shown in Fig.
For L = 30.0 cm. υ = 330 m s^-1
v_n = \(\frac{\mathrm{n} \times 330\left(\mathrm{~m} \mathrm{~s}^{-1}\right)}{0.6(\mathrm{~m})}\) = 550 s^-1

Clearly, for a source of frequency 1.1 kHz the air column will resonate at υ₂, i.e. the second harmonic.
Now if one end of the pipe is closed (Fig.), the fundamental frequency is

and only to odd numbered harmonics are present :
v₃ = \(\frac{3 v}{4 L}/latex], v₅ = [latex]\frac{5 v}{4 L}\) and s0 0n.
For L = 30 cm and υ = 330 m s^-1;, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Question 6.
Two sitar strings A and B playing the note ‘Dha’ are slightly put of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Solution:
Increase in the tension of a string increases its frequency. It the original frequency of B (v_B) were greater than that of A(v_A) .further increase in v_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that v_B < v_A. Since v_A – v_B = 5 Hz, and v_a = 427 Hz, we get v_B = 422 Hz.

Question 7.
A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket. [A.P. Mar. 16]
Solution:
1) The observer is at rest and the source is moving with a speed of 200 ms^-1. Since this is comparable with the velocity of sound 330 ms^-1, we must use Eq.
v = v₀ \(\left[\frac{1+\mathrm{v}_{\mathrm{S}}}{\mathrm{v}}\right]^{-1}\) and not the approximate
v = v₀ \(\left[1-\frac{v_s}{v}\right]\)
Since the source is approaching a stationary target, υ₀ = 0, and υ_s must be replaced by -υ_s. Thus, we have
v = v₀ \(\left(1-\frac{v_{\mathrm{S}}}{v}\right)^{-1}\)
v = 1000 Hz × [1 – 200 m s^-1/330 m s^-1]^-1
≃ 2540Hz

2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, υ_s = 0 and υ₀ has a positive value. The frequency of the sound emitted by the source (the target) is v₀, the frequency intercepted by the target and not v0, Therefore, the frequency as registered by the rocket is
v’ = \(v\left(\frac{v+v_0}{v}\right)\)
= 2540 Hz × \(\left(\frac{200 \mathrm{~m} \mathrm{~s}^{-1}+330 \mathrm{~m} \mathrm{~s}^{-1}}{330 \mathrm{~m} \mathrm{~s}^{-1}}\right)\)
≃ 4080Hz

Students get through AP Inter 2nd Year Physics Important Questions 2nd Lesson Ray Optics and Optical Instruments which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 2nd Lesson Ray Optics and Optical Instruments

Very Short Answer Questions

Question 1.
What is optical density and how is it different from mass density ?
Answer:
Optical density: Optical density is defined as the ratio of the speed of light in media. Mass density: Mass per unit volume is defined as mass density.
Mass density of an optically denser medium is less than that of optically rarer medium.

Question 2.
What are the laws of reflection through curved mirrors ?
Answer:

1. “The angle of reflection equals to the angle of incidence”.
2. “The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane”.

Question 3.
Define ‘power’ of a convex lens. What is its unit ? [A.P. Mar. 17, T.S. Mar. 16]
Answer:
Power of a lens : Power of a lens is defined as its bending ability and is measured as reciprocal of focal length in metre.
∴ Power of a lens = \(\frac{1}{f(\text { in metres) }}=\frac{100}{\mathrm{f}(\text { in cms) }}\)
Unit → Dioptre (D)

Problems

Question 1.
A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?
Answer:
f = 10 cm, υ = 35 cm
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{-u}\) (using sign convention)
\(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{35}-\frac{1}{10}\)
\(\frac{1}{u}=\frac{10-35}{35 \times 10}=\frac{-1}{14}\)
U = – 14 cm.
Distance of the object from the wall = 35 – 14 = 21cm.

Question 2.
A concave mirror produces an image of a long vertical pin, placed 40cm from the mirror, at the position of the object. Find the focal length of the mirror. [T.S. Mar. 17, 16]
Answer:
Give u = υ = 40cm
\(\frac{1}{\mathrm{f}}=\frac{1}{v}+\frac{1}{u}\)
\(\frac{1}{f}=\frac{1}{40}+\frac{1}{40}\)
\(\frac{1}{\mathrm{f}}=\frac{2}{40}\)
f = 20 cm.

Question 3.
A small angled prism of 4° deviates a ray through 2.48°. Find the refractive index of the prism.
Answer:
A = 4°, D_m = 2.48°
D_m = A (μ – 1)
μ – 1 = \(\frac{\mathrm{D}_{\mathrm{m}}}{\mathrm{A}}=\frac{2.48}{4}\) = 0.62
μ = 1 + 0.62
μ = 1.62

Question 4.
What is ‘dispersion’? Which colour gets relatively more dispersed ? [Mar. 14]
Answer:
Dispersion : The phenomenon of splitting of white light ipto its constituent colours, on passing through a prism is called dispersion of light.
The deviation is maximum for violet colour.

Question 5.
The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is 1/10 of its size ?
Solution:
f = 30 cm, h₁ = h, h₂ = \(\frac{\mathrm{h}}{10}\)

Question 6.
What is myopia ? How can it be corrected ? [T.S. Mar. 15]
Answer:
Myopia (or) Near sightedness :
The light from a distant object arriving at the eye-lens may get converged at a point infront of the retina. This type of defect is called myopia.
To correct this, we interpose a concave lens between the eye and the object.

Question 7.
What is hypermetropia ? How can it be corrected ? [A.P. Mar. 16]
Answer:
Hypermetropia (or) Farsightedness :
The light from a distant object arriving at the eye-lens may get converged at a point behind the retina. This type of defect is called Hypermetropia.
To correct this, we interpose a convex lens (Convergent lens) between the eye and the object.

Question 8.
Draw neat labelled ray diagram of simple microscope. [PE 2015 (A.P.)]
Answer:

u = object distance
D = distance of near point.

Short Answer Questions

Question 1.
Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal length. [A.P. Mar. 17]
Answer:
Focal length of concave mirror:
The distance between the focus F and the pole P of the mirror is called the focal length of the concave mirror.

Consider a ray AB parallel to principal axis incident on a concave mirror at B and is re-flected along BF. The line CB is normal to the mirror.
Let θ be the angle of incidence, ∠ABC = ∠BCP = θ
Draw BD ⊥ CP
In right angled ∆^le
Tan θ = \(\frac{\mathrm{BD}}{\mathrm{CD}}\) ……………. (1)
From ∆^le BFD, Tan 2θ = \(\frac{\mathrm{BD}}{\mathrm{FD}}\) ……………. (2)
Dividing eq (2) by eq (1), \(\frac{{Tan} 2 \theta}{{Tan} \theta}=\frac{\mathrm{CD}}{\mathrm{FD}}\) ………………….. (3)
If θ is very small, then tan θ ≈ θ and tan 2θ ≈ 2θ since the aperture of the lens is small
∴ The point B lies very close to p.
CD ≈ CP and FD ≈ FP
From eq (3), \(\frac{2 \theta}{\theta}=\frac{\mathrm{CP}}{\mathrm{FP}}=\frac{\mathrm{R}}{\mathrm{f}} \Rightarrow 2=\frac{\mathrm{R}}{\mathrm{f}}\)
R = 2f

Question 2.
Define critical angle. Explain total internal reflection using a neat diagram. [T.S. Mar. 15]
Answer:
Critical angle:
When light ray travelling from denser medium to rarer medium, then the angle of incidence for which angle of refraction in air is 90a is called critical angle.
C = sin^-1 \(\left(\frac{1}{\mu}\right)\)
Total internal reflection:
When a light ray travels from denser to rarer medium, the angle of incidence is greater than the critical angle, then it reflects into the same medium is called total internal reflection.

Explanation:
Consider an object in the denser medium. A ray OA incident on XY bends away from the normal. As the angle of incidence is increased, the angle of refraction goes on increasing. For certain angle of incidence, the refracted ray parallel to XY surface (r = 90°).

When the angle of incidence is further increased, the ray is not refracted but is totally reflected back in the denser medium. This phenomenon is called total internal reflection.

Question 3.
Explain the formation of a mirage. [A.P. Mar. 16]
Answer:
In a desert, the sand becomes very hot during the day time and it rapidly heats the layer of air which is in its contact. So density of air decreases. As a result the successive upward layers are denser than lower layers.

When a beam of light travelling from the top of a tree enters a rarer layer, it is refracted away from the normal. As a result at the surface of layers of air, each time the angle of incidence increases and ultimately a stage is reached, when the angle of incidence becomes greater than the critical angle between the two layers, the incident ray suffers total internal reflection.

So it appears as inverted image of the tree is formed and the same looks like a pool of water to the observer.

Question 4.
Explain the formation of a rainbow. [A.P. Mar. 15]
Answer:

Figure shows how sun light is broken into its segments in the process and a rainbow appears. The dispersion of the violet and the red rays after internal reflection in the drop is shown in figure.

The red rays emerge from the drops of water at one angle (43°) and the violet rays emerge at another angle (41°). The large number of water drops in the sky makes a rain-bow. The rainbow appears semicircular for an observer on earth.

Question 5.
Why does the setting sun appear red ? [T.S. Mar. 17, Mar. 14]
Answer:
As sunlight travels through the earths atmosphere, gets scattered by the large number of molecules present. This scattering of sun light is responsible for the colour of the sky, during sunrise and sunset etc.
The light of shorter wave length is scattered much more than light of larger wave-length. Scattering ∝ \(\frac{1}{\lambda^4}\).
Most of blue light is scattered, hence the bluish colour of sky predominates.

At sunset (or) sunrise, sun rays must pass through a larger atmospheric distance. More of the blue colour is scattered away only red colour which is least scattered ap-pears to come from sun. Hence it appears red.

Question 6.
With a neat labelled diagram explain the formation of image in a simple microscope. [T.S. Mar. 16, A.P. Mar. 15]
Answer:
Simple microscope: It consists a single short focus convex lens. It increases the visual angle to see an object clearly. It is also called magnifying glass (or) reading glass.

Working : The object is adjusted within the principal focus Of the convex lens to form the image at the near point. The image is formed on same side of the object and it is virtual, erect and magnified as shown in fig.

Magnifying power : The ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye is called magnifying power of a simple microscope.
It is denoted by ‘m’.
m = \(\frac{\alpha}{\beta} \simeq \frac{{Tan} \alpha}{{Tan} \beta}\)

Question 7.
What is the position of the object for a simple microscope ? What is the maximum magnifi-cation of a simple microscope for a realistic focal length ?
Answer:
When an object is placed between principal focus and optical centre of a convex lens, a virtual and erect image will be formed on the same side of the object.

Magnifying power: It is defined as the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the eye.
m = \(\frac{\alpha}{\beta} \simeq \frac{{Tan} \alpha}{{Tan} \beta}\)
From figure OJ = IJ’, ∠IO’G = α and ∠IO’J’ = β

The above equation can be written as
m = 1 + \(\frac{\mathrm{D}}{\mathrm{f}}\) …………… (2)
This shows that smaller the focal length of the lens, greater will be the magnifying power of microscope.

Long Answer Questions

Question 1.
Draw a neat labelled diagram of a compound microscope and explain its working. Derive an expression for its magnification.
Answer:
Description : It consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The distance of the object can be adjusted by
means of a rack and pinion arrangement.

Working: The object OJ is placed outside the principal focus of the objective and the real image is formed on the other side of it. The image I₁ G₁ is real, inverted and magnified.
This image acts as the object for the eyepiece. The position of the eyepiece is so adjusted that the image due to the objective is between the optic centre and principal focus to form the final image at the near point. The final image IG is virtual, inverted and magnified.

Magnifying Power: It is defined as the ratio of the angle subtended by the final image at the eye when formed at near point to the angle subtended by the object at the eye when imagined to be at near point.

Imagining that the eye is at the optic centre, the angle subtended by the final image is a. When the object is imagined to be taken at near point it is represented by IJ’ and OJ – IJ’.
The angle made by I J’ at the eye is β. Then by the definition of magnifying power

Dividing and multiplying by I₁ G₁ on the right side, we get
m = \(\left(\frac{\mathrm{IG}}{\mathrm{I}_1 \mathrm{G}_1}\right)\left(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\right)\)
Magnifying power of the objective (m₀) = I₁ G₁ / OJ = Height of the image due to the objective / Height of its object.
Magnifying power of the eye piece (m_e) = IG/I₁G₁ = Height of the final image / Height of the object for the eyepiece.
∴ m = m₀ × m_e ……………… (1)
To find m0: In figure OJ O’ and I₁ G₁ O’ are similar triangles. \(\left(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\right)=\frac{\mathrm{O}^{\prime} \mathrm{I}_1}{\mathrm{O}^{\prime} \mathrm{O}}\)
Using sign convention, we find that O’I₁ = + v₀ and O’O = -u where v₀ is the image distance due to the objective and u is the object distance for the objective or the compound microscope. I₁G₁ is negative and OJ is positive.
∴ m₀ = \(\frac{\mathrm{v}_0}{\mathrm{U}}\) (∵ \(\frac{\mathrm{I}_1 \mathrm{G}_1}{\mathrm{OJ}}\) = m₀)
To find m_e : The eyepiece behaves like a simple microscope. So the magnifying power of the eye piece.
∴ m_e = (1 + \(\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\))
Where f_e is the focal length of the eyepiece.
Substituting m0 and me in equation (1),
m = \(+\frac{\mathrm{v}_0}{\mathrm{u}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)\)
When the object is very close to the principal focus F₀ of the objective, the image due to the objective becomes very close to the eyepiece,
u ≈ -f₀ and v₀ ≈ L
Where L is the length of the microscope. Then
m ≈ \(-\frac{L}{f_0}\left(1+\frac{D}{f_e}\right)\)

Question 2.
a) Define Snell’s Law. Using a neat labelled diagram derive an expression for the refractive index of the material of an equilateral prism.
b) A ray of light, after passing through a medium, meets the surface separating the medium from air at an angle of 45° and is just not refracted. What is the refractive index of the medium ?
Answer:
a) Snell’s law:
The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant, called the refractive index of the medium.
\(\frac{\sin i}{\sin r}\) = μ (constant).
Let ABC be the glass prism. Its angle of prism is A. The refractive index of the material of the prism is p. Let AB and AC be the two refracting surfaces PQ = incident ray, RS = emergent ray.
Let angle of incidence = i₁
angle of emergence = i₂
angle of refraction = r₁
angle of refraction at R = r₂
After travelling through the prism it falls on AC and emerges as RS.

The D angle of deviation.
From the ∆ QRT
r₁ + r₂ + ∠T = 180° ……………….. (1)
From the quadrilateral AQTR
∠A + ∠T = 180°
∠T = 180° – A ………………. (2)
From the equations (1) and (2)
r₁ + r₂ + ∠T = 180° we get
r₁ + r₂ + 180° – A = 180°
r₁ + r₁ = A ………………. (3)
from the ∆ QUR
i₁ – r₁ + i₂ – r₂ + 180° – D = 180°
i₁ + i₂ = (r₁ + r₂) = D
i₁ + i₂ – A = D [∵r₁ + r₂ = A]
i₁ + i₂ = A + D ……………… (4)
Minimum deviation : Experimentally it is found that as the angle of incidence increased the angle of deviation decreases till it reaches a minimum value and then it increases. This least value of deviation is called angle of minimum deviation ‘δ’ as shown in the fig.

When D decreases the two angles i₁ and i₂ become closer to each other at the angle of minimum deviation, the two angles of incidence are same i.e, i ₁ = i₂.
As i₁ = i₂, r₁ = r₂
∴ i₁ = i₂ = i, r₁ = r₂ = r
substituting this in (1) and (2) we get
2r = A ⇒ r = A/2
i + i = A + δ ⇒ i = \(\frac{\mathrm{A}+\delta}{2}\)
According to Snell’s law μ =
μ =
Note :The minimum deviation depends on the refractive index of the prism material and the angle of the prism.

b) Given that i = C = 45°
μ = \(\frac{1}{\sin \mathrm{c}}\) ⇒ μ = \(\frac{1}{\sin 45^{\circ}}\)
μ = \(\frac{1}{1 / \sqrt{2}}=\sqrt{2}\)
μ = 1.414

Textual Examples

Question 1.
Suppose that the lower half of the concave mirror’s reflecting surface in figure is covered with an opaque (non- reflective) material. What effect will this have on the image of an object p’ iced in front of the mirror ?

Solution:
You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the re-fleeting surface has been reduced, the intensity of the image will be low (in this case, half).

Question 2.
A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror ?

Solution:
The ray diagram for the formation of the image of the phone is shown in fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B’C = BC. You can yourself realise why the image is distorted.

Question 3.
An object is placed at (i) 10 cm. (ii) 5 cm in front of a concave mirror of radius of curvature 15cm. Find the position, nature, and magnification of the image in each case.
Solution:
The focal length f = – 15/2 cm = – 7.5 cm
i) The object distance u = – 10 cm. Then
Eq. \(\frac{1}{v}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\) gives
\(\frac{1}{v}+\frac{1}{-10}=\frac{1}{-7.5}\)
or υ = \(\frac{10 \times 7.5}{-2.5}\) = -30 cm
The image is 30 cm from the mirror on the same side as the object.
Also, magnification
m = –\(\frac{v}{u}=-\frac{(-30)}{(-10)}\) = -3
The image is magnified, real and inverted.

ii) The object distance u = -5cm. Then
from Eq. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{v}+\frac{1}{-5}=\frac{1}{-7.5}\)
or υ = \(\frac{5 \times 7.5}{(7.5-5)}\) = 15 cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = – \(\frac{v}{u}=-\frac{15}{(-5)}\) = 3
The image is magnified, virtual and erect.

Question 4.
Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5ms^-1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29m, (c) 19 m and (d) 9 m away.
Solution:
From the mirror equation, Eq. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
we get
υ = \(\frac{\mathrm{fu}}{\mathrm{u}-\mathrm{f}}\)
For convex mirror, since R = 2m, f = 1 m.
Then for u = -39 m. v = \(\frac{(39) \times 1}{-39-1}\)
= \(\frac{39}{40}\)
Since the jogger moves at a constant speed of 5ms”1, after Is the position of the image u (for u = – 39 + 5 = – 34) is (34/35) m.
The shift in the position of image in 1 s is
\(\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}\)
Therefore, the average speed of the im-age when the jogger is between 39 m and 34 m from the mirror, is (1/280) ms^-1. Similarly, it can be seen that for u = – 29 m, -19 m and -9 m, the speed with which the image appears to move is \(\frac{1}{150}\)ms^-1, \(\frac{1}{60}\) ms^-1 and \(\frac{1}{10}\) ms^-1 respectively.

Question 5.
The earth takes 24 h to rotate once about its axis. How much time does the sun take to shift by 1° when viewed from the earth ?
Solution:
Time taken for 360° shift = 24h
Time taken for 1° shift = 24/360
h = 4 min.

Question 6.
Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed ?
Solution:
We use the relation given by Eq.
\(\frac{\mathrm{h}_2}{v}-\frac{\mathrm{h}}{\mathrm{u}}=\frac{\mathrm{h}_2-\mathrm{h}_1}{\mathrm{R}}\) Here
u = – 100 cm, υ = ?. R = + 20 cm, n₁ = 1, and n₂ = 1.5. We then have
\(\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}\)
or υ = + 100 cm

Question 7.
A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid ? Could the liquid be water ?
Answer:
The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n₁ = n₂. This gives 1/f = 0 or f → ∞. The lens in the liquid will act like a plane sheet of glass. No, the liquid is not water. It could be glycerine.

Question 8.
(i) If f = 0.5 m for a glass lens, what is the power of the lens ? (ii) The radii of curvature 6f the faces of a double convex lens are 10 cm and 15cm. Its focal length is 12 cm. What is the refractive index of glass ? (iii) A convex lens has 20 cm focal length in air. What is focal length in water ?
(Refractive index of air-water = 1.33. Refractive index for air – glass = 1.5.)
Solution:
i) Power = + 2 dioptre.

ii) Here, we have f = + 12 cm,
R₁ = + 10 cm, R₂ = – 15 cm.
Refractive index of air is taken as unity.
We use the lens formula of Eq. \(\frac{\mathrm{h}_1}{\mathrm{DB}}+\frac{\mathrm{h}_1}{\mathrm{DB}}=\frac{\mathrm{h}_1}{\mathrm{f}}\). The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac{1}{12}\) = (n – 1) (\(\frac{1}{10}\) – \(\frac{1}{-15}\))
This gives n = 1.5.

iii) For a glass lens in air, n₂ = 1.5, n: = 1, f = + 20cm. Hence, the lens formula gives.
\(\frac{1}{20}=0.5\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\)
For the same glass lens in water. n₂ = 1.5, n₁ = 1.33. Therefore,
\(\frac{1.33}{f}=(1.5-1.33)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\)
Combining the above two equations, we find f = + 78.2 cm.

Question 9.
Find the position of the image formed by the lens combination given in the fig.

Solution:
Image formed by the first lens
\(\frac{1}{v_1}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_1}\)
\(\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}\)
or v₁ = 15 cm
The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens. Though the image is reeil, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.
\(\frac{1}{v_3}-\frac{1}{10}=\frac{1}{-10}\) or v₂ ∞
The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.
\(\frac{1}{v_3}-\frac{1}{\mathrm{u}_3}=\frac{1}{\mathrm{f}_3} \text { or } \frac{1}{v_3}-\frac{1}{\infty}+\frac{1}{30}\)
or v₃ = 30 cm
The final image is formed 30 cm to the right of the third lens.

Question 10.
What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm ?
Solution:
The distance of normal vision is 25cm. So if a book is at u= – 25 cm. Its image should be formed at υ = – 50 cm. There-fore, the desired focal length given by
\(\frac{1}{\mathrm{f}}=\frac{1}{v}-\frac{1}{\mathrm{u}} \text { or } \frac{1}{\mathrm{f}}=\frac{1}{-50}-\frac{1}{-25}=\frac{1}{50}\)
or f = + 50 cm (convex lens).

Question 11.
a) The far point of a myopic person is 80 cm in front of the eye. What is ,. the power of the lens required to enable him to see very distant objects clearly ?
b) In what way does the corrective lens help the above person ? Does the lens magnify very distant objects ? Explain carefully.
c) The above person prefers to remove his spectacles while reading a book. Explain why?
Solution:
a) Solving as in the previous example, we find that the person should use a concave lens of focal length = – 80 cm. i.e:, of power = – 1.25 dioptres.

b) No. The concave lens, in fact, reduces the size of the object, but the angle subtended by the image (at the far point) at the eye. The eye is able to see distant objects not because the corrective lens magnifies the object, but because it brings the (i.e., it produces virtual image of the object) at the far point of the eye which then can be focussed by the eye-lens on the retina.

c) The myopic person may have a normal near point, i.e., about 25 cm (or even less). In order to read a book with the spectacles, such a person must keep the book at a distance greater than 25cm so that the image of the book by the concave lens is produced not closer than 25cm. The angular size of the book (dr its image) at the greater distance is evidently less than the angular size when the book is placed at 25 cm and no spectacles are needed. Hence, the person prefers to remove the spectacles while reading.

Question 12.
a) The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25cm from the eye ?
b) In what way does the corrective lens help the above person ? Does the lens magnify objects held near the eye ?
c) The above person prefers to remove the spectacles while looking at the sky. Explain why? .
Solution:
a) u = – 25 cm, υ = – 75 cm
1/f = 1/25 – 1/75, i.e., f = 37.5 cm.
The corrective lens- needs to have a converging power of +2.67 dioptres.

b) The corrective lens produces a virtual image (at 75 cm) of an object at 25 cm. The angular size of this image is the same as that of the object. In this sense the lens does not magnify the object but merely brings the object to the near point of the hypermetric eye, which then gets focussed on the retina. However, the angular size is greater than that of the same object at the near point (75 cm) viewed without the spectacles.

c) A hypermetropic eye may have normal far point i.e., it may have enough converging power to focus parallel rays from infinity on the retina of the shortened eyeball. Wearing spectacles of converging lenses (used for near vision) will amount to more converging power than needed for parallel rays. Hence the person prefers not to use the spectacles for far objects.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 4th Lesson Electric Charges and Fields

Very Short Answer Questions

Question 1.
What is meant by the statement ‘charge is quantized’ ?
Answer:
The minimum charge that can be transferred from one body to the other is equal to the charge of the electron (e = 1.602 × 10^-19C). A charge always exists an integral multiple of charge of electron (q = ne). Therefore charge is said to be quantized.

Question 2.
Repulsion is.the sure test of charging than attraction. Why ?
Answer:
A charged body may attract a neutral body and also an opposite charged body. But it always repels a like charged body. Hence repulsion is the sure test of electrification.

Question 3.
How many electrons constitute 1 C of charge ?
Answer:
n = \(\frac{q}{e}\) = \(\frac{1}{1.6 \times 10^{-19}}\) = 6.25 × 10¹⁸ electrons

Question 4.
What happens to the weight of a body when it is charged positively ?
Answer:
When a body positively charged it must loose some electrons. Hence weight of the body will decrease.

Question 5.
What happens to the force between two charges if the distance between them is
a) halved
b) doubled ?
Answer:
From Coulombs law, F ∝ \(\frac{1}{\mathrm{~d}^2}\), so
a) When distance is reduced to half, force increases by four times.

b) When distance is doubled, then force is reduced by four times.

Question 6.
The electric lines of force do not intersect. Why ?
Answer:
They do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible.

Question 7.
Consider two charges + q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
Answer:
Charges are scalars, but the .electrical intensities are vectors and add vectorially.

Question 8.
Electrostatic field lines of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force ?
Answer:
It is conservative force.

Question 9.
State Gauss’s law in electrostatics.
Answer:
Gauss’s law : It states that “the total electric flux through any closed surface is equal to – \(\frac{1}{\varepsilon_0}\) times net charge enclosed by the surface”.
\(\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

Question 10.
When is the electric flux negative and when is it positive ?
Answer:
Electric flux ϕ = \(\vec{E} \cdot \vec{A}\). If angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{A}}\) is 180°, then flux will have a ‘-ve’ sign. We consider the flux flowing out of the surface as positive and flux entering into the surface as negative.

Question 11.
Write the expression for electric intensity due to an infinite long charged wire at a distance radial distance r from the wire.
Answer:
The electric intensity due to an infinitely long charged wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) the conductor.
Where λ = Uniform linear charge density
r = Distance of the point from the conductor.

Question 12.
Write the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
The electric intensity due to an infinite plane sheet of charge is E = \(\frac{\sigma}{2 \varepsilon_0}\).

Question 13.
Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.
Answer:
a) Intensity of electric field at any point inside a spherical shell is zero.
b) Intensity of electric field at any point outside a uniformly charged spherical shell is
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity.
Answer:
Coulomb’s law – Statement: Force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the straight line joining the two charges.
Explanation : Let us consider two charges q₁ and q₂ be separated by a distance r.
Then F ∝ q₁q₂ and F ∝ \(\frac{1}{\mathrm{r}^2}\) or F ∝ \(\frac{q_1 q_2}{r^2}\)

∴ F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}\) where \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C^-2
In vector form, in free space \(\overrightarrow{\mathrm{F}}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2} \hat{\mathrm{r}}\). Here \(\hat{\mathrm{r}}\) is a unit vector.
ε₀ is called permittivity of free space.
ε₀ = 8.85 × 10^-12 C²/N-m² or Farad/meter.
Where ε is called permittivity of the medium.

Question 2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
Answer:
Intensity of electric field (E) : Intensity of electric field at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
Expression :
1) Intensity of electric field is a vector. It’s direction is along the direction’ of motion of positive charge.
2) Consider point charge q. Electric field will exist around that charge. Consider any point P in that electric field at a distance r from the given charge. A test charge q₀ is placed at R
3) Force acting on q₀ due to q is F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2}\)
4) Intensity of electric field at that point is equal to the force experienced by a test charge q₀.
Intensity of electric field, E = \(\frac{\mathrm{F}}{\mathrm{q}_0}\)

E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\) N/C or V/m

Question 3.
Derive the equation for the couple acting on a electric dipole in a uniform electric field.
Answer:
1) A pair of opposite charges separated by a small distance is called dipole.
2) Consider the charge of dipole are -q and +q coulomb and the distance between them is 2a.
3) Then the electric dipole moment P is given by P = q × 2a = 2aq. It is a vector. It’s direction is from -q to +q along the axis of dipole.
4) It is placed in a uniform electric field E, making an angle 0 with field direction as shown in fig.
5) Due to electric field force on +q is F = +qE and force on -q is F = -qE.

6) These two equal and opposite charges constitute torque or moment of couple.
i. e., torque, \(\tau\) = ⊥^r distance × magnitude of one of force
∴ \(\tau\) = (2a sin θ)qE = 2aqE sin θ = PE sin θ
In vector form, \(\vec{\tau}\) = \(\overrightarrow{\mathrm{P}}\) × \(\overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.
Answer:
Electric field at a point on the axis of a dipole :
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre ‘O’.

2) We shall calculate electric field E at point P on the axial line of dipole, and at a distance OP = r.
3) Let E₁ and E₂ be the intensities of electric field at P due to charges +q and -q respectively.
4)

The resultant intensity at P is E = E₁ – E₂ [∵ They are opposite and E₁ > E₂]

If r > > a then a² can be neglected in comparision to r².

In vector form, \(\overrightarrow{\mathrm{E}}\) = \(\frac{2 \overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 5.
Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole. (A.P. Mar. ’19, ’15)
Answer:
Electric field intensity on equitorial line of electric dipole:
1) Consider an electric dipole consisting of two charges -q and +q separated by a distance ‘2a’ with centre at ‘O’.
2) We shall calculate electric field E at P on equitorial line of dipole and at a distance OP = r.

3) Let E₁ and E₂ be the electric fields at P due to charges +q and -q respectively.
4) The ⊥^r components (E₁ sin θ and E₂ sin θ) cancel each other because they are equal and opposite. The ||^el components (E₁ cos θ and E₂ cos θ) are in the same direction and hence add up.
5) The resultant field intensity at point P is given by E = E₁ cos θ + E₂ cos θ

6) From figure, cos θ = \(\frac{a}{\left(r^2+a^2\right)^{1 / 2}}\)
∴ E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
7) If r >> a, then a² can be neglected in comparison to r². Then
E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
In vector form \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and explain its importance.
Answer:
Gauss’s law : The total-electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface.
Total electric flux,

Here q is the total charge enclosed by the surface ‘S’, \(\oint\) represents surface integral of the closed surface.

Importance :

1. Gauss’s law is very useful in. calculating the electric field in case of problems where it is possible to construct a closed surface. Such surface is called Gaussian surface.
2. Gauss’s law is true for any closed surface, no matter what its shape or size.
3. Symmetric considerations in many problems make the application of Gauss’s law much easier.

Long Answer Questions

Question 1.
Define electric flux. Applying Gauss’s law and derive the expression for electric intensity due to an infinite long straight charged wise. (Assume that the electric field is everywhere radial and depends only on the radial distance r of the point from the wire.)
Answer:
Electric flux : The number of electric lines of force passing perpendicular to the area is known as electric flux (ϕ). Electric flux ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\). So flux is a scalar.

Expression for E due to an infinite long straight charged wire :

1) Consider an infinitely long thin straight wire with uniform linear charge density ‘λ’.
2) Linear charge density λ = \(\frac{\text { change q }}{\text { length } l}\) ⇒ λl —– (1)
3) Construct a coaxial cylindrical gaussion surface of length T and radius ‘r’. Due to symmetry we will assume that electric field is radial i.e., normal to the conducting wire.
4) The flat surfaces AB and CD are ⊥^r to the wire. Select small area ds₁ and ds₂ on the surface as AB and CD.
They are ⊥^r to \(\overrightarrow{\mathrm{E}}\). So flux coming out through them is zero.
Since flux ϕ = \(\oint \vec{E} \cdot d \vec{s}\) = Eds cos 90° = 0

5) So flux coming out through the cylindrical surface ABCD is taken into account.

6) From Gauss’s law

7) From (2) and (3), E(2πrl) = \(\frac{Q}{\varepsilon_0}\) = \(\frac{\lambda /}{\varepsilon_0}\) (∵ Q = λl)
∴ E = \(\frac{\lambda l}{2 \pi \varepsilon_0 \mathrm{r} l}=\frac{1}{2 \pi \varepsilon_0} \frac{\lambda}{\mathrm{r}}\)

8) Therefore electric intensity due to an infinitely long conducting wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\).

Question 2.
State Gauss’s law in electrostatics. Applying Gauss’s law derive the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
Gauss’s law: The total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface, i.e.,

Expression for E due to an infinite plane sheet of charge :

1. Consider an infinite plane sheet of charge. Let the charge distribution is uniform on this plane.
2. Uniform charge density on this surface σ = \(\frac{\mathrm{dq}}{\mathrm{dS}}\) where dq is the charge over an infinite small area ds.
3. Construct a horizontal cylindrical Gaussian surface ABCD perpendicular to the plane with length 2r.
4. The flat surfaces BC and AD are parallel to the plane sheet and are at equal distance from the plane.
5. Let area of these surfaces are dS₁ and dS₂. They are parallel to \(\overrightarrow{\mathrm{E}}\). So flux through these two surfaces is
   ——– (1)
   Where S is area of plane surface AD or BC. Both are equal in area and intensity.
6. Consider cylindrical surface of AB and CD. Let their areas are say dS₃ and dS₄. These surfaces are ⊥^lr to electric intensity \(\overrightarrow{\mathrm{E}}\).
7. So angle between \(\overrightarrow{\mathrm{E}}\) and d\(\overrightarrow{\mathbf{s}_3}\) or dS₄ is 90°. Total flux through these, surfaces is zero. Since
   
8. From Gauss’s law total flux, ϕ = \(\oint \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = 2ES = \(\frac{\mathrm{q}}{\varepsilon_0}\)
   ∴ 2ES = \(\frac{\sigma S}{\varepsilon_0}\) [∵ \(\text { Q }\) = σ × S]
9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \(\frac{\sigma}{2 \varepsilon_0}\)

Question 3.
Applying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at
(i) a point outside the shell
(ii) a point on the surface of the shell and
(iii) a point inside the shell.
Answer:
Expression for E due to a charged conducting spherical shell:

1. Consider a uniformly charged spherical shell. Let total charge on it is ‘q’ and its radius is R.
   
2. Since the shell is uniformly charged, the intensity of electric field at any point depends on radial distance ‘r’ from centre ‘O’. The direction of E is away from the centre along the radius.

i) E at a point outside the shell:

1) Consider a point at a distance ‘r’ outside the sphere. Construct a Gaussian surface with ‘r’ as radius (where r > R).

2) Total flux coming out of this sphere is



3) Therefore at any point outside the sphere, E = \(\frac{\sigma}{\varepsilon_0} \frac{\mathrm{R}^2}{\mathrm{r}^2}\)

ii) E at a point on the surface of shell:

1) Construct a Gaussian surface with radius r = R.

2)


3) Therefore intensity at any point on surface of the sphere E = \(\frac{\sigma}{\varepsilon_0}\)

iii) E at a point inside the shell :

1) Consider a point P inside the shell. Construct a Gaussian surface with radius r (where r < R). There is no charge inside the shell. So from Gauss’s law \(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)

2) Therefore, intensity of electric field at any point inside a charged shell is zero.

Textual Exercises

Question 1.
Two small identical balls, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal lengths. The balls position themselves at equilibrium such that the angle between the threads is 60°. If the distance between the balls is 0.5 m, find the charge on each ball.
Solution:
Given m = 0.20 g = 0.2 × 10^-3 kg; θ = 60° ⇒ α = \(\frac{\theta}{2}\) = 30°
r = 0.5 m, Let q₁ = q₂ = q

Question 2.
An infinite number of charges-each of magnitude q are placed on x-axis at distance of 1, 2, 4, 8, …….. meter from the origin respectively. Find intensity of the electric field at origin.
Solution:
Let q₁ = q₂ = q₃ = q₄ = ……. = q
r₁ = 1; r₂ = 2; r₃ = 4; r₄ = 8, …….
The resultant electric field at origin ‘O’ is given by
E = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_2}{\mathrm{r}_2^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{q_3}{r_3^2}\) + \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_4}{\mathrm{r}_4^2}\) + ……..

Question 3.
A clock face has negative charges -q, -2q, -3q, ….. -12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to the point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial ?
Solution:
Let distance of each charge from unit charge at centre ‘O’ = r.
Resultant electric field of each charge, E = \(\frac{1}{4 \pi \varepsilon_0} \frac{6 q}{r^2}\) [∵ -6q – (-12q)]
Let OX be the reference axis. The angles of resultant fields with OX-axis are shown.

Resultant field along OX-axis = \(\left(0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\)i = (2 + \(\sqrt{3}\))i
Resultant field along OY-axis = \(\left(1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}-\frac{\sqrt{3}}{2}\right) \hat{\mathrm{j}}\)
= 1\(\hat{\mathrm{i}}\)
∴ Resultant electric field, E_R(OH) = (2 + \(\sqrt{3}\))\(\hat{i}\) + 1\(\hat{j}\)
The direction of resultant field (OH) is given by, tan θ = \(\frac{|\mathrm{OY}|}{|\mathrm{OX}|}\)
⇒ tan θ = \(\frac{1}{2+\sqrt{3}}\) = tan 15°
⇒ θ = 15°, with OX-axis
∴ The hour hand shows at the centre of the dial is at 9.30.

Question 4.
Consider a uniform electric field E = 3 × 10³ N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x – axis ?
Solution:
a) Given E = 3 × 10³ N/C
S = 10² cm² = 10² × (10^-2m)² = 10^-2m²
θ = 0°
ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 0°
∴ ϕ = 30 Nm²C^-1

b) If θ = 60°, ϕ = ES cos θ
= 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 Nm²C^-1

Question 5.
There are four charges, each with a magnitude Q. Two are positive and two are negative. The charges are fixed to the comers of a square of side ‘L’, one to each comer, in such a way that the force on any charge is directed toward the center of the square. Find the magnitude of the net electric force experienced by any charge ?
Solution:

Question 6.
The electric field in a region is given by \(\overrightarrow{\mathbf{E}}\) = a\(\hat{\mathbf{i}}\) + b\(\hat{\mathbf{j}}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = a\(\hat{\mathrm{i}}\) + b\(\hat{\mathrm{j}}\)
\(\vec{S}\) = L²\(\hat{\mathrm{i}}\)

ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (a\(\hat{i}\) + b\(\hat{j}\)) .L²\(\hat{i}\)
∴ ϕ = aL² [∴ \(\hat{i}\). \(\hat{i}\) = 1 and \(\hat{i}\). \(\hat{j}\) = 0]

Question 7.
A hollow spherical shell of radius r has a uniform charge density σ. It is kept in a cube of edge 3r such that the centre of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
Solution:
For spherical shell, charge = q (say)
Radius = r
Charge density = σ = \(\frac{q}{A}\) = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}^2}\)
∴ Charge on spherical shell, q = 4πr²σ
Flux through one of the face of a cube,
ϕ_E = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{1}{6} \times \frac{4 \pi r^2 \sigma}{\varepsilon_0}\) = \(\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}\)

Question 8.
An electric dipole consists of two equal and opposite point charge +Q and -Q, separated by a distance 2l. P is a point collinear with the charges such that its distance from the positive charge is half of its distance from the negative charge. Calculate electric intensity at P.
Solution:
Distance of P from -Q = d (say)
Distance of P from +Q = d/2

Question 9.
Two infinitely long thin straight wires having uniform linear charge densities λ and 2λ are arranged parallel to each other at a distance r apart. Calculate intensity of the electric field at a point midway between them.
Solution:
Distance between two parallel infinite long thin straight wires = r.
Electric field due to infinite long thin straight wire, E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\)

∴ Electric intensity at mid point, E = E₂ – E₁ = 2E₁ – E₁ = E
∴ E = \(\frac{\lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 10.
Two infinitely long thin straight wires having uniform linear charge densities e and 2e are arranged parallel to each other at a distance r apart. Find the intensity of the electric field at a point midway between them.
Solution:
For first infinitely long straight wire, linear charge density λ = e.
For second infinitely long straight wire, linear charge density λ’ = 2e
Distance between two infinite parallel straight wires = r.
Distance of point P from 1^st and 2^nd wire = \(\frac{\mathrm{r}}{2}\)
Electric field intensity at P due 1^st wire, E₁ = \(\frac{\lambda}{2 \pi \varepsilon_0\left(\frac{\mathrm{r}}{2}\right)}=\frac{\mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\) —— (1)

Electric field intensity at P due 2^nd wire, E₂ = \(\frac{\lambda^{\prime}}{2 \pi \varepsilon_0\left(\frac{r}{2}\right)}=\frac{2 \mathrm{e}}{\pi \varepsilon_0 \mathrm{r}}\)
∴ E₂ = 2E₁ [∵ from(1)]
∴ Electric field intensity at middle point due to second infinitely long wire
E₂ = \(\frac{2 \lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Question 11.
An electron of mass m and charge e is fired perpendicular to a uniform electric field of intensity E with an initial velocity u. If the electron tranverses a distance x in the field in the direction of firing, find the transverse displacement y it suffers.
Solution:
Given m_e = m; q = e; d = x; u_x = u; u_y = 0
Electric field between the plates = E
Time taken travel in the field, t = \(\frac{d}{u_x}\) = \(\frac{\mathbf{X}}{\mathbf{u}}\)

Force on electron F = qE = eE
Acceleration of electron, a = \(\frac{F}{m}\) = \(\frac{\mathrm{eE}}{\mathrm{m}}\)
Transverse displacement of electron y = u_yt + \(\frac{1}{2} \mathrm{at}^2\)
⇒ y = 0 + \(\frac{1}{2}\left(\frac{e E}{m}\right)\left(\frac{x}{u}\right)^2\)
∴ y = \(\frac{\mathrm{eEx}^2}{2 \mathrm{mu}^2}\)

Additiona Exercises

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air ?
Solution:
Given, q₁ = 2 × 10^-7 C; q₂ = 3 × 10⁷ C; d = 30 cm = 30 × 10^-2 m = 3 × 10^-1m

As q₁, q₂ are positive charges, the force between them is repulsive.

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres ?
(b) What is the force on the second sphere due to the first ?
Solution:
a) Given q₁ = 0.4 μc ;
= 0.8 × 10^-6C
q₂ = 0.8 μc; F = 0.2 N = 0.4
= 0.4 × 10^-6m
0.2 = \(\frac{9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{\mathrm{r}^2}\)
r² = 16 × 9 × 10^-4
r = 4 × 3 × 10^-2 = 12 × 10^-2 m
∴ Distance between two charges, r = 12 cm

b) Electrostatic force between two charges obeys the Newton’s third law. i.e., force on q₁ due to q₂ = force on q₂ due to q₁
f₁₂ = f₂₁ = 0.2N

Question 3.
Check that the ratio ke²/G m_em_p is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify ?
Solution:
i) In electrostatics, F_e = \(\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}^2}\) ……. (1)
Where q₁ = q₂ = e
In gravitation, F_g = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\) = \(\frac{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}}{\mathrm{r}^2}\) …. (2)
Where m₁ = m_e ; m₂ = m_p

Thus the given ratio is dimensionless.

ii) We know that e = 1.6 × 10^-19 C ; G = 6.67 × 10^-11 N-m²C²

Question 4.
a) Explain the meaning of the statement ‘electric charge of a body is quantized’.
b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e, large scale charges ?
Answer:
a) The electric charge of a body is quantized means that the charge on a body can occur in some particular values only. Charge on any body is the integral multiple of charge on an electron because the charge of an electron is the elementary charge in nature. The charge on any body can be expressed by the formula q = ± ne. Where n = number of electrons transferred and e = charge on one electron. The cause of quantization is that only integral number of electrons can be transferred from one body to other

b) We can ignore the quantization of electric charge when dealing with macroscopic charges because the charge on one electron is 1.6 × 10^-19 C in magnitude, which is very small as compared to the large scale change.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
According to law of conservation of charge, “charge can neither be created nor be destroyed but it can be transferred from one body to another body”. Before rubbing the two bodies they both are neutral i.e., the total charge of the system is zero. When the glass rod is rubbed with a silk cloth, some electrons are transferred from glass rod to silk cloth. Hence glass rod attains positive charge and silk cloth attains same negative charge.

Again the total charge of the system is zero, i.e., the charge before rubbing is same as the charge after rubbing. This is consistent with the law of conservation of charge. Here we can also say that charges can be created only in equal and unlike pairs.

Question 6.
Four point charges q_A = 2 µC, q_B = -5 µC, q_C = 2 µC and q_D = -5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?
Solution:
Let the centre of the square is at O.
The charge placed on the centre is µC
AB = BC = CD = DA = 10 cm; AC = \(\sqrt{2}\) × 10 = 10\(\sqrt{2}\)cm
AO = BO = CO = DO = \(\frac{10 \sqrt{2}}{2}\) = 5\(\sqrt{2}\) cm


Here we observe that, F_A = -F_C and F_D = -F_B
∴ The net resultant force on 1 µC is
F = F_A + F_B + F_C + F_D
= -F_C + F_B + F_C – F_B
= 0.

Question 7.
a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?
b) Explain why two field lines never cross each other at any point ?
Answer:
a) An electrostatic field line represents the actual path travelled by a unit positive charge in an electric field. If the line have sudden breaks it means the unit positive test charge Jumps from one place to another which is not possible. It also means that electric field becomes zero suddenly at the breaks which is not possible. So, the field line cannot have any sudden breaks.

b) If two field lines cross each other, then we can draw two tangents at the point of intersection which indicates that (as tangent drawn at any point on electric line of force gives the direction of electric field at that point) there are two directions of electric field at a particular point, which is not possible at the same instant. Thus, two field lines never cross each other at any point.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vaccum.
a) What is the electric field at the midpoint O of the line AB joining the two charges ?
b) If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge ?
Solution:
a) Given q_A = 3 μC = 3 × 10^-6 C; q_B = -3 μC = -3 × 10^-6C

From fig. AO = OB = 10 cm = 0.1 m
Electric field at midpoint ‘O’ due to q_A

The direction of E_A is A to O.
Electric field at midpoint ‘O’ due to q_B at B is

The direction of E_B is O to B.
Now we see that E_A and E_B are in same direction. So, the resultant electric field at O is E. Hence,
E = E_A + E_B = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C :
The direction of E will be from O to B or toward B.

b) Let test charge q₀ = -1.5 × 10^-9 C is placed at midpoint O’.
Electric field intensity at ‘O’ is E = 5.4 × 10⁶

Force F = E_q = 5.4 × 10⁶ × -1.5 × 10^-9 N
= -8.1 × 10³N
The direction of force is from O to A.

Question 9.
A system has two charges q_A = 2.5 × 10^-7 C, and q_B = -2.5 × 10^-7 C located at points A(0, 0, -15 cm) and B(0, 0, +15 cm). What are the total charge and electric dipole moment of the system ?
Solution:
Given A(0, 0, -15 cm) and B(0, 0, 15 cm)
q_A = 2.5 × 10^-7C
q_B = -2.5 × 10^-7 C
AB = 2a = length of the dipole
= 30 cm = 30 × 10^-2 m

The total charge q on the dipole is
q = q_A + q_B = 2.5 × 10^-7C – 2.5 × 10^-7C = 0
The electric dipolemoment
P = Any charge (q_A) × length of dipole (2a)
= 2.5 × 10^-7 × 10 × 10^-2
∴ P = 7.5 × 10^-8 C-m
The direction of P is from negative charge to positive charge that is along B to A.

Question 10.
An electric dipole with dipole moment 4 × 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Solution:
Given, P = 4 × 10^-9 C-m; E = 5 × 10⁴ N/C; θ = 30°,
Torque, \(\tau\) = PE sin θ

= 4 × 10^-9 × 5 × 10⁴ sin 30° = \(\frac{20 \times 10^{-5}}{2}\) = 10^-4N-m
The direction of torque is ⊥^r to both electric field and dipole moment.

Question 11.
A polythene piece rubbed with wool is found to have a negative charge 3 × 10^-7 C.
a) Estimate the number of electrons transferred (from which to which ?)
b) Is there a transfer of mass from wool to polythene ?
Solution:
a) Given, charge on Polythene, q = -3 × 10^-7 C
e = -1.6 × 10^-19 C
No. of electrons transferred, n = \(\frac{\mathrm{q}}{\mathrm{e}}\) = \(\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}\)
∴ n = 1.875 × 10¹² [∵ q = ± ne]
Electrons are transferred from wool to polythene.
So wool gets positive charge and polythene gets negative charge.

b) The number of electrons transferred = 1.875 × 10¹²
The mass of one electron, m_e = 9.1 × 10^-3 kg
Mass transferred from wool to polythene M = n × m_e
M = 1.875 × 10¹² × 9.1 × 10^-31 = 1.8 × 10^-18 kg

Question 12.
a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each 6.5 × 10^-7 C ? The radii of A and B are negligible compared to the distance of
separation.
b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
Solution:

a) Given, q_A = 6.5 × 10^-7C ; q_B = 6.5 × 10^-7C
r = AB = 50 cm = 50 × 10^-2m

b)

This force is also repulsive in nature because both the charges are similar (positive) in nature.

Question 13.
Suppose the spheres A and B in Exercise – 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with second and finally removed from both. What is the new force of repulsion between A and B?
Solution:
Given q_A = 6.5 × 10^-7C;
q_B = 6.5 × 10^-7 C; q_C = 0

After contact of A and C, the charges will be divided equally on both of them. Then final charge on A, then
\(\mathrm{q}_{\mathrm{A}}^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{A}}+\mathrm{q}_{\mathrm{C}}}{2}\) = \(\frac{6.5 \times 10^{-7}+0}{2}\)
= 3.25 × 10^-7C
Similarly charge on C, \(\mathrm{q}_{\mathrm{c}}^{\prime}\) = 3.25 × 10^-7 C
After contact of B and C, the charges will be divided equally on both of them.

Then final charge on B, \(q_B^{\prime}\) = \(\frac{\mathrm{q}_{\mathrm{B}}+\mathrm{q}_{\mathrm{C}}^{\prime}}{2}\) = \(\frac{6.5 \times 10^{-7}+3.25 \times 10^{-7}}{2}\) = 4.875 × 10^-7 C
Similarly final charge one, \(q_C^{\prime \prime}\) = 4.875 × 10^-7 C

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?

Answer:
We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.

Here, particle 1 and particle 2 are attracted towards positive plate that means particle 1 and particle 2 are negatively charged. Particle 3 is attracted towards negatively charged plate so it is positively charged. As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio.
y ∝ \(\frac{\mathrm{q}}{\mathrm{m}}\)
Here, the deflection in particle 3 is maximum, so the charge to mass ratio of particle 3 is maximum.

Question 15.
Consider a uniform electric field E = 3 × 10³ \(\hat{\mathbf{i}}\) N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis ?
Solution:
Given \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C

a) As the surface is in Y – Z plane, so the area vector (normal to the square) is along X – axis
Area S = 10 × 10 = 100 cm² = 10^-2 m²

Area vector \(\vec{S}\) = 10^-2 \(\hat{\mathbf{i}}\) m²
ϕ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}\) = (3 × 10³ \(\hat{\mathbf{i}}\)). (10^-2i)
∴ ϕ = 3 × 10³ × 10^-2 = 30N-m²/c

b) \(\overrightarrow{\mathrm{E}}\) = 3 × 10³ \(\hat{\mathbf{i}}\) N/C ; \(\vec{S}\) = \(\hat{\mathbf{i}}\) m² ; θ = 60°
ϕ = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{S}}\) = ES cos 60° = 3 × 10³ × 10^-2 × cos 60°
∴ ϕ = 15 N – m²/C

Question 16.
What is the net flux of the uniform electric field of Exercise -15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes ?
Answer:
As we know that the number of lines entering in the cube is the same as that the number of lines leaving the cube. So, no flux is remained on the cube and hence, the net flux over the cube is zero.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ Nm²/C.
(a) What is the net charge inside the box ?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box ? Why or Why not ?
Solution:
a) Given, ϕ = 8.0 × 10³ N – m²/C
ε₀ = 8 × 10³ × 8.854 × 10^-12
∴ q = 0.07 μc
The flux is outward hence the charge is positive in nature

b) Net outward flux = 0
Then, we can conclude that the net charge inside the box is zero. i.e., the box may have either zero charge or have equal amount of positive and negative charges. It means we cannot conclude that there is no charge inside the box.

Question 18.
A point charge +10 μC is a ‘distance 5 cm directly above the centre of a square of side 10 cm, as shown in fig. What is the magnitude of the electric flux through the square ? (Hint: Think of the square as one face of a cube with edge 10 cm).

Solution:
Let the charge q is placed at the centre of cube as shown in fig.
The total flux enclosed through the cube is ϕ = \(\frac{q}{\varepsilon_0}\)
The flux enclosed by one face ϕ = \(\frac{1}{6}\) of total flux.
[∵ Cube has 6 faces]

ϕ = \(\frac{\phi}{6}\) = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\)
Here q = 10 μC = 10 × 10^-6C ; ε₀ = 8.854 × 10^-12C² – N^-1-m^-2
∴ ϕ = \(\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}\)
= 1.88 × 10⁵ N-m²/C

Question 19.
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface ?
Solution:
Given, q = 2.0 μC = 2.0 × 10^-6C
ε₀ = 8.854 × 10^-12 C²-N^-1 – m^-2
The net flux through the surface,
ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\) = \(\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}}\) = 2.26 × 10⁵N-m²/C

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge,
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ?
(b) What is the value of the point charge ?
Solution:
a) From Gauss’s law, ϕ = \(\frac{\mathrm{q}}{\varepsilon_0}\)
Electric flux ϕ depends on charge q.
It is independent of radius of Gaussian surface. Hence the radius of Gaussian surface were doubled, flux does not change.

b) ϕ = – 1.0 × 10³ N-m²/c ; ε₀ = 8.854 × 10^-12 e²-N^-1-m^-2
q = ϕε₀ = -1.0 × 10³ × 8.854 × 10^-12 = -8.85 × 10^-9C.
∴ The value of point charge, q = -8.85 × 10^-9C

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?
Solution:
E = 1.5 × 10³ N/C; r = 20 cm = 20 × 10^-2m.
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)
1.5 × 10³ = \(\frac{9 \times 10^9 \times \mathrm{q}}{\left(20 \times 10^{-2}\right)^2}\)
q = \(\frac{1.5 \times 10^3 \times 20 \times 20 \times 10^{-4}}{9 \times 10^9}\) = 6.67 × 10^-9C.

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density, of 80.0 μC/m².
(a) Find the charge on the sphere,
(b) What is the total electric flux leaving the surface of the sphere ?
Solution:
a) Given D = 2.4 m; r = \(\frac{\mathrm{D}}{2}\) = 1.2 m
σ = 80 µc/m² = 80 × 10^-6 C/m²
σ = \(\frac{\mathrm{q}}{4 \pi r^2}\) ⇒ q = σ 4πr²
⇒ q = 80 × 10^-6 × 4 × 3.14 × 1.2 × 1.2
∴ q = 1.45 × 10^-3C

b) ϕ = \(\frac{Q}{\varepsilon_0}\) = \(\frac{1.4 \times 10^{-3}}{8.854 \times 10^{-12}}\) = 1.6 × 10⁸N-m²/C
Thus, the flux leaving the surface of sphere is 1.6 × 10⁸ N – m²/c

Question 23.
An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
Solution:
Given r = 2 cm = 2 × 10^-2m ; E = 9 × 10⁴ N/C


Thus, the linear charge density is 10^-7 C/m.

Question 24.
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10^-22 C/m². What is E :
(a) in the outer region of the first plate,
(b) in the outer region of the second plate and
(c) between the plates ?
Solution:
Given σ_A = 127.0 × 10^-22 C/m²
σ_B = 17.0 × 10^-22 C/m²

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2; e = 1.60 × 10^-19C).
Solution:
Given n = 12; E = 2.55 × 10⁴ N/C
p = 1.26 g/cm³ = 1.26 × 10³ kg/m³
e = 1.6 × 10^-19C ; g = 9.81 ms^-2

As the oil drop is stationary,
Electrostatic force = Gravitational force
⇒ qE = mg
neE = \(\frac{4}{3} \pi r^3 \mathrm{\rho g}\)
r³ = \(\frac{3 \mathrm{neE}}{4 \pi \rho \mathrm{g}}\) = \(\frac{3 \times 12 \times 1.6 \times 10^{-19} \times 2.55 \times 10^4}{4 \times 3.14 \times 1.26 \times 10^3 \times 9.8}\)
r = 0.94 × 10^-18
r = [0.94 × 10^-18]^{\(\frac{1}{3}\)} = 9.81 × 10^-7m
∴ Radius of the drop = 9.81 × 10^-7 m.

Question 26.
Which among the curves shown in Fig. cannot possibly represent electrostatic field lines ?

Solution:
a) According to the properties of electric lines of force, the lines should be always ⊥r to the surface of a conductor as they starts or they ends. Here, some of the lines are not ⊥r to the surface, thus it cannot represent the electrostatic field lines.

b) According to the property of electrostatic field lines, they never start from negative charge, here some of the lines start from negative charge. So, it cannot represent the electrostatic field lines.

c) As the property of electric field lines that they start outwards from positive charge. Hence, it represents the electrostatic field lines.

d) By the property of electric field lines, two electric field lines never intersect each other. Here, two lines intersect. So it does not represent the electric field lines.

e) By the property of electric field lines that they are not in the form of closed loops. Here, the lines form closed loop. So, it does not represent the electric field lines.

Question 27.
In a certain region of space, electric field is along the Z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive Z-direction, at the rate of 10⁵ NC^-1 per metre. What are the force and torque experienced by a system having a total dipolemoment equal to 10^-7 Cm in the negative Z-direction ?
Solution:
The electric field increases in positive Z – direction. dE
\(\frac{\mathrm{dE}}{\mathrm{dZ}}\) = 10⁵ N/C-m
The direction of dipolemoment is in the negative Z-direction

So the negative charge q is placed at A and positive charge q is placed at B as the direction of dipole moment is from negative charge to positive charge.
P_Z = -10^-7C-m

The negative sign shows its direction in negative Z – axis. According to the basic definition of electric field, F = qdE Now, multiplying and dividing by dz,
F = q\(\frac{\mathrm{dE}}{\mathrm{dz}} \cdot \mathrm{dz}\) .dz = q.dz\(\frac{\mathrm{dE}}{\mathrm{dz}}\)
qdz = dipolement p_z, as the length of the dipole is dz.

∴ F = P_z. \(\frac{\mathrm{dE}}{\mathrm{dz}}\) = -10^-7 × 10⁵ = -10^-2N
Torque, \(\tau\) = PE sin θ (∵ θ = 180° angle between P and E)
\(\tau\) = PE sin 180° = 0
Thus the force is -10^-2 N and the torque is 0.

Question 28.
a) A conductor A with a cavity as shown in Fig. (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor, (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q (Fig. (b)). (c) A sensitive instrument is o he shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Solution:
a) As we know the property of conductor that the net electric field inside a charged conductor is zero, i.e., E = 0.
Now let us choose a Gaussian surface lying completely inside the conductor enclosing the cavity.
So, from Gauss’s theorem \(\oint \text { E. dS }\) = \(\frac{\mathrm{q}}{\varepsilon_0}\)
As E = 0 ⇒ \(\frac{q}{\varepsilon_0}\) = 0 ⇒ q = 0
That means the charge inside the cavity is zero. Thus, the entire charge Q on the conductor must appear on the outer surface of the conductor.

b) As the conductor B carrying a charge +q inserted in the cavity, the charge -q is induced on the metal surface of the cavity and then charge +q induced on the outside surface of the conductor A. Initially the outer surface of A of A has a charge Q and now it has a charge +q induced. So the total charge on the outer surface of A is Q + q.

c) To protect any sensitive instrument from electrostatic field, the sensitive instrument must be put in the metallic cover. This is known as electrostatic shielding.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/εε₀)\(\hat{\mathbf{n}}\), where \(\hat{\mathbf{n}}\) is the unit vector in the outward normal direction and σ is the surface charge density near the hole.
Solution:
Surface charge density near the hole = σ
Unit vector = \(\hat{\mathbf{n}}\) (normal directed outwards)
Let P be the point on the hole.
The electric field at point P closed to the surface of conductor, according to Gauss’s theorem,

\(\oint \mathrm{E} \cdot \mathrm{dS}\) = \(\frac{q}{\varepsilon_0}\)
Where q is the charge near the hole.
E ds cos θ = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\) (∴ σ = \(\frac{\mathrm{q}}{\mathrm{dS}}\) ∴q = σ dS) where dS = area

∴ Angle between electric field and area vector is 0°.
EdS = \(\frac{\sigma \mathrm{dS}}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0}\)
E = \(\frac{\sigma}{\varepsilon_0} \hat{\mathrm{n}}\)

This electric field is due to the filled up hole and the field due to the rest of the charged conductor. The two fields inside the conductor are equal and opposite. So, there is no electric field inside the conductor. Outside the conductor, the electric fields are equal and are in the same direction.
So, the electric field at P due to each part = \(\frac{1}{2} \mathrm{E}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{n}\)

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]
Solution:
Let us consider a long thin wire of linear charge density λ. We have to find the resultant electric field due to this wire at point P.
Now, consider a very small element of length dx at a distance x from C.

The charge on this elementary portion of length dx
q = λ dx ——- (1)

Electric field intensity at point P due to the elementary portion
dE = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{(\mathrm{OP})^2}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d x}{(\mathrm{OP})^2}\) [∵ from (1)]
Now, in ΔPCO (PO)² = (PC)² + (CO)²
(OP)² = r² + x²
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d \mathbf{x}}{\left(x^2+r^2\right)}\) ——- (2)

The components of dE are dE cos θ along PD and dE sin θ along PF.
Here, there are so many elementary portion. So all the dE sin θ components balance each other. The resultant electric field at P is due to only dE cos θ components.
The resultant electric field due to elementary component, dE’ = dE cos θ

dE’ = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\lambda d x}{\left(x^2+r^2\right)} \cos \theta\) —— (3)
In ΔOCP tan θ = \(\frac{x}{r}\) ⇒ x = r tan θ
Differentiating with respect to θ, we get dx = r sec² θ dθ
Putting in equation (3), we get

As the wire is of infinite length, so integrate within the limits –\(\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we get

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge +(2/3)e and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
Solution:
For the protons, the charge on it is +e let the number of up quarks are a, then the number of down quarks are (3 – a) as the total number of quarks are 3.
So, a_x up quark charge + (3 – a) down quark charge = +e
a × \(\frac{2}{3} \mathrm{e}\) + (3 – a)\(\left(\frac{-\mathrm{e}}{3}\right)\) = e
\(\frac{2 \mathrm{ae}}{3}\) – \(\frac{(3-\mathrm{a}) \mathrm{e}}{3}\) = e
2a – 3 + a = 3
3a = 6
a = 2
Thus, in the proton there are two up quarks and one down quark.
∴ Possible quark composition for proton = uud

For the neutron, the charge on neutron is 0.
Let the number of up quarks are b and the number of down quarks are (3 – b)
So, b_x up quark charge + (3 – b) × down quark charge = 0
b\(\left(\frac{2 \mathrm{e}}{3}\right)\) + (3 – b)\(\left(\frac{-\mathrm{e}}{3}\right)\) = 0
2b – 3 + b = 0
3b = 3
∴ b = 1
Thus, in neutron, there are one up quark and two down quarks.
∴ Possible quark composition for neutrons = udd.

Question 32.
a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Solution:
a) Let us consider that initially the test charge is in the stable equilibrium. When the test charge is displaced from the null point (where, E = 0) in any direction, it must experience a restoring force towards the null point.
This means that there is a net inward flux through a closed surface around the null point According to the Gauss’s theorem, the net electric flux through a surface net enclosing any charge must be zero. Hence, the equilibrium is not stable.

b) The middle point of the line joining two like charges is a null point. If we displace a test Charge slightly along the
line, the restoring force try to bring the test charge back to the centre. If we displace the test charge normal to the line, the net force on the test charge takes it further away from the null point. Hence the equilibrium is not stable.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed V_x (as in the fig.). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m \(\mathbf{V}_{\mathbf{x}}^2\)).
Compare this motion with motion of a projectile in gravitational field discussed in section 4.10 of 1^st Year Textbook of Physics.
Solution:
Mass of particle = m
Charge of particle = -q
Speed of particle = V_x
Length of plates = L

Electric field between the plates = E (from positive plate to negative plate).
Let the deflection in the path of the charge – q is Y, because the force acting in +Y axis direction. The direction of force is from negative plate to positive plate because the charge is negative in nature.

Let us discuss the motion in Y axis direction. Initial velocity u = 0
Acceleration a = \(\frac{F}{m}\) = \(\frac{+\mathrm{qE}}{\mathrm{m}}\)
Deflection Y = ?
Time = \(\frac{\text { Distance }}{\text { Velocity }}\) = \(\frac{\mathrm{L}}{\mathrm{V}_{\mathrm{x}}}\)
Using second equation of motion,
S = ut + \(\frac{1}{2} \mathrm{at}^2\)at
Putting the values y = 0 + \(\frac{1}{2} \times\left(+\frac{\mathrm{qE}}{\mathrm{m}}\right) \frac{\mathrm{L}^2}{\mathrm{~V}_{\mathrm{x}}^2}\)
Y = \(\frac{\mathrm{qEL}^2}{2 \mathrm{mV}_{\mathrm{x}}^2}\)

In the case of projectile motion y = \(\frac{1}{2} \mathrm{gt}^2\). Thus, it is exactly similar to the projectile motion in the gravitational field.

Question 34.
Suppose that the particle is an electron projected with velocity V_x = 2.0 × 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate ? (|e| = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg.)
Solution:
Given V_x = 2 × 10⁶ m/s; E = 9.1 × 10² N/C
q = e = 1.6 × 10^-19 C; m_e = 9.1 × 10^-31 kg
d = 0.5 cm = 0.5 × 10^-2 m = 5 × 10^-3 m
The electron will strike the upper plate at its other end at X = L as it get deflected.

Students get through AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 3rd Lesson Wave Optics

Short Answer Questions

Question 1.
Explain Doppler effect in light. Distinguish between red shift and blue shift. [T.S. Mar. 16]
Answer:
Doppler effect in light : The change in the apparent frequency of light, due to relative motion between source of light and observer. This phenomenon is called Doppler effect.

The apparent frequency of light increases when the distance between observer and source of light is decreasing and the apparent frequency of light decreases, if the distance between source of light and observer increasing.
Doppler shift can be expressed as = \(\frac{\Delta v}{v}=\frac{-v_{\text {radial }}}{c}\)
Applications of Doppler effect in light:

1. It is used in measuring the speed of a star and speed of galaxies.
2. Measuring the speed of rotation of the sun.

Red shift: The apparent increase in wave length in the middle of the visible region of the spectrum moves towards the red end of the spectrum is called red shift.

Blue shift: When waves are received from a source moving towards the observer, there is an apparent decrease in wave length, this is called blue shift.

Question 2.
Derive the expression for the intensity at a point where interference of light occurs. Arrive at the conditions for maximum and zero intensity. [A.P. Mar. 16; T.S. Mar. 15]
Answer:
Let y₁ and y₂ be the displacements of the two waves having same amplitude a and <|> is the phase difference between them.

y₁ = a sin ωt …………… (1)
y₂ = a sin (ωt + Φ) …………….. (2)
The resultant displacement y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ)
y = a sin ωt + a sin ωt cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt (a sin Φ)
Let R cos θ = a(1+ cos Φ) ……………….. (4)
R sin θ = a sin Φ ………………. (5)
y = R sin ωt. cos θ + R cos ωt. sin θ
y = R sin (ωt + θ) ………………… (6)
where R is the resultant amplitude at P, squaring equations (4) and (5), then adding
R² [cos² θ + sin2 θ] = a²[l + cos² Φ + 2 cos Φ + sin² Φ]
R² [1] = a² [1 + 1 + 2 cos Φ]
I = R² = 2a² [1 + cos Φ] = 2a² × 2 cos² \(\frac{\phi}{2}\);
I = 4a² cos² \(\frac{\phi}{2}\) …………. (7)

i) Minimum intensity (I_max)
cos² \(\frac{\phi}{2}\) = 1
Φ = 2nπ Where n = 0, 1, 2, 3 ……….
Φ = 0, 2π, 4π, 6π
∴ I_max = 4a².

ii) Minimum intensity (I_min)
cos² \(\frac{\theta}{2}\) = 1
Φ = (2n + 1)π Where n = 0, 1, 2, 3 ……….
Φ = π, 3π, 5π, 7π …………..
∴ I_min = 0

Question 3.
Does the principle of conservation of energy hold for interference and diffraction phenomena? Explain briefly. [Mar. 14]
Answer:
Yes, law of conservation of energy is obeyed. In case of constructive interference, intensity becomes maximum. Hence bright fringes are formed on the screen where as in the case of destructive interference, intensity becomes minimum. Hence dark fringes are formed on the screen.

This establishes that in the interference and diffraction pattern, the intensity of light is simply being redistributed i.e., energy is being transferred from dark fringe to bright fringe. No energy is being created (or) destroyed in the process. Hence energy is redistributed.

Thus the principle of conservation of energy is being obeyed in the process of interference and diffraction.

Question 4.
How do you determine the resolving power of your eye ? [A.P. Mar. 17]
Answer:
Make black strips of equal width separated by white strips. All the black strips having same width, while the width of white strips should increase from left to right.

Now watch the pattern with one eye. By moving away (or) closer to the wall, find the position where you can just see some two black strips as separate strips.

All black strips to the left of this strips would merge into one another and would not be distinguishable on the other hand, the black strips to the right of this would be more and more clearly visible.

Note the width d of the white strips and measure the distance D of the wall from eye.
Then resolution of your eye = \(\frac{\mathrm{d}}{\mathrm{D}}\)

Question 5.
Explain polarisation of light by reflection and arrive at Brewster’s law from it.
Answer:
Polarisation of light by reflection : When unpolarized light is incident on the boundary of a denser medium, at a particular angle of incidence the reflected light is completely plane polarised. This incident angle is called Brewster’s angle (i_B).

Brewster’s law:
When light is incident on a transparent surface at Brewster s angle, then reflected and refracted rays are at right angles to each other.
From snell’s law, refractive index,
∴ i_B + r = \(\frac{\pi}{2}\) ⇒ r = \(\frac{\pi}{2}\) – i_B

n = \(\frac{\sin \mathrm{i}_B}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_B}{\sin \left(\frac{\pi}{2}-\mathrm{i}_B\right)}=\frac{\sin \mathrm{i}_B}{\cos \mathrm{i}_B}\) = tan i_B
n = tani_B, This is known as Brewster’s law.
Brewster’s law – statement: The refractive index of a denser medium is equal to tangent of the polarising angle.

Question 6.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. [T.S. Mar. 17; Mar. 12]
Answer:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ.
Where θ is the angle between pass axes P₁ and P₂. Since P₁ and P₂ are crossed the angle between the pass axes of P₂ and P₃ will be (\(\frac{\pi}{2}\) – θ)
Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ . cos²(\(\frac{\pi}{2}\) – θ)
= I₀cos²θ . sin²θ
I = \(\frac{\mathrm{I}_0}{4}\) sin² 2θ
∴ The transmitted intensity will be maximum when θ = \(\frac{\pi}{4}\).

Long Answer Questions

Question 1.
Distinguish between Coherent and Incoherent addition of waves. Develop the theory of constructive interferences.
Answer:
Coherent sources : The two sources which maintain zero (or) any constant phase relation between themselves are known as Coherent sources.

Incoherent sources : If the phase difference changes with time, the two sources are known as incoherent sources.
Theory of constructive and destructive interference :
Let the waves of two coherent sources be
y₁ = a sin ωt ………………….. (1)
y₂ = a sin (ωt + Φ) …………….. (2)
where a is amplitude and Φ is the phase difference between two displacements.
According to superposition principle, y = y₁ + y₂
y = a sin ωt + a sin (ωt + Φ) = a sin ωt + a sin ω cos Φ + a cos ωt sin Φ
y = a sin ωt [1 + cos Φ] + cos ωt [a sin Φ] ………………….. (3)
Let A cos θ = a(1 + cos Φ], ………………. (4)
A sin θ = a sin Φ ……………… (5)
Substituting equations (4) and (5) in equation (3)
y = A sin ωt. cos θ + A cos ωt sin θ
y = A sin (ωt + θ) …………………… (6)
Where A is resultant amplitude. Squaring equations (4) and (5), then adding
A²[cos² θ + sin² θ] = a²[1 + cos² Φ + 2 cos Φ + sin² Φ]
A² [1] = a² [1 + 1 + 2 cos Φ]
I = A² = 2a² [1 + cos Φ]
I = 2a² × 2 cos² \(\frac{\phi}{2}\)
I = 4a² cos² \(\frac{\phi}{2}\)
I = 4I₀ cos² \(\frac{\phi}{2}\) …………….. (7) [∵ I₀ = a²]

Case (i) For constructive interference: Intensity should be maximum.
cos \(\frac{\phi}{2}\) = 1 ⇒ Φ = 2nπ
Where n = 0, 1, 2, 3 ⇒ Φ = 0, 2π, 4π, 6π ………….. I_max = 4I₀
Case (ii) For destructive interference: Intensity should be minimum
i.e., cos Φ = 0 ⇒ Φ = (2n + 1) π ; where π = 0, 1, 2, 3 ; Φ = π, 3π, 5π ⇒ I_min = 0.

Question 2.
Describe Young’s experiment for observing interference and hence arrive at the expression for ‘fringe width’.
Answer:
Interference: The modification of intensity obtained by the super position of two (or) more light waves is called interference.
Description :

1. Thomas Young experimentally observed the phenomenon of interference of light using two coherent sources.
2. A small pin hole ‘S’ illuminated by monochromatic source of light which produces a spherical wave.
3. S₁ and S₂ are two narrow pin holes equidistant from S.
4. Screen is placed at a distance D.
5. The points at which any two crests (or) any two troughs are superimposed, constructive interference takes place bright fringe will be observed on the screen.
6. The points at which crest of one wave and trough of another wave are superimposed, destructive interference takes place dark fringe will be observed on the screen.
7. Thus on the screen alternately bright and dark frings are observed.

Expression for fringe width :
1. It is the distance between two successive bright (or) dark fringes, denoted by β.

The path difference (δ) = d sin θ
2. If θ is very small then from figure sin θ ≈ tan θ = \(\frac{x}{D}\)

3. For bright fringes path difference S₂P – S₁P = nλ
∴ d sin θ = nλ
d × \(\frac{x}{D}\) = nλ
x = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\) ……………. (1) where n = 0, 1, 2, 3 …………….

This equation represents the position of bright fringe.
When n = 0, x₀ = 0.
n = 1, x₁ = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) and n = 2, x₂ = \(\frac{2\lambda \mathrm{D}}{\mathrm{d}}\)
The distance between any two consecutive bright fringes is
x₂ – x₁ = \(\frac{2 \lambda D}{d}-\frac{\lambda D}{d} \Rightarrow \beta=\frac{\lambda D}{d}\) ………… (2)

4. For dark fringes path difference S₂P – S₁P = (2n + 1) \(\frac{\lambda}{2}\) ∴ d sin θ = (2n + 1) \(\frac{\lambda}{2}\)
d × \(\frac{x}{D}\) = (2n + 1) \(\frac{\lambda}{2}\) ⇒ x = \(\frac{(2 n+1) \lambda D}{2 d}\) ………….. (3) where n = 0, 1, 2, 3…………
This equation (3) represents, position of dark fringe.
When n = 0, x₀ = \(\frac{\lambda D}{2 d}\) ⇒ n = 1, x₁ = \(\frac{3 \lambda D}{2 d}\); n = 2, x₂ = \(\frac{5 \lambda D}{2 d}\) ………….
The distance between any two consecutive dark fringes is x₂ – x₁ = \(\frac{5 \lambda \mathrm{D}}{2 \mathrm{~d}}-\frac{3 \lambda \mathrm{D}}{2 \mathrm{~d}}=\frac{5 \lambda \mathrm{D}-3 \lambda \mathrm{D}}{2 \mathrm{~d}}\)
β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\) …………………….. (4)
Hence fringe width is same for bright and dark fringes.

Problems

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ? .
Solution:
Since vλ = c, \(\frac{\Delta v}{v}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Question 3.
What is the Brewster angle for air to glass transition ? (Refractive index of glass = 1.5.)
Solution:
Here, i_p = ? μ = 1.5; As tan i_p = μ = 1.5
∴ i_p = tan^-1 (1.5); i_p = 56.3

Question 4.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3 ?
Solution:
Let I₁ = I₂ = I. If Φ is phase difference between the two light waves, then resultant intensity,
I_R = I₁ + I₂ + \(2 \sqrt{\mathrm{I}_1 \mathrm{I}_2}\) . cos Φ
When path difference = λ, Phase difference Φ = 0°
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) . cos 0° = 4I = k
When path difference = \(\frac{\lambda}{3}\), phase difference Φ = \(\frac{2 \pi}{3}\) rad
∴ I_R = I + I + 2\(\sqrt{\text { II }}\) cos \(\frac{2 \pi}{3}\) ⇒ I’_R = 2I + 2I(\(\frac{-1}{2}\)) = I = \(\frac{\mathrm{k}}{4}\)

Question 5.
Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then ∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
= 2.9 × 10^-7 radians.

Question 6.
In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m, β = 1.2 × 10^-2 m, n = 4
Since, β = \(\frac{\mathrm{D}}{\mathrm{d}}\) nλ ⇒ \(\frac{\beta}{n} \cdot \frac{d}{D}\) ⇒ λ = \(\frac{1.2 \times 10^{-2} \times 2.8 \times 10^{-2}}{4 \times 1.4}\)
⇒ λ = 600 × 10^-9 m
⇒ λ = 600 nm.

Textual Examples

Question 1.
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm ?
Solution:
Since vλ = c, \(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{\Delta \lambda}{\lambda}\)
(for small changes in v and λ). For
∆λ = 589.6 – 589.0 = + 0.6 nm
we get [usmg Equation \(\frac{\Delta v}{v}=\frac{v_{\text {radial }}}{c}\)]
\(\frac{\Delta v}{v}=-\frac{\Delta \lambda}{\lambda}=-\frac{v_{\text {radial }}}{\dot{c}}\)
or, υ_radial ≅ + c(\(\frac{0.6}{589.0}\)) = +3.06 × 10⁵ m s^-1
= 306 km/s
Therefore, the galaxy is moving away from us.

Question 2.
(a) When monochromatic light is incident on a surface separating, two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why ?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave ?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Solution:
a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) and undergo forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.

b) No. energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing ah unit area per unit time.

Question 3.
Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation when blue*: green light of wavelength 500 nm is used ?
Solution:
Fringe spacing = \(\frac{\mathrm{D} \lambda}{\mathrm{d}}=\frac{1 \times 5 \times 10^{-7}}{1 \times 10^{-3}}\) m
= 5 × 10^-4 m = 0.5 mm

Question 4.
What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations :
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochro-matic) source of shorter wavelength;
(c) the separation between the two slits is increased;
(d) the source slit is moved Closer to the double-slit plane;
(e) the width of the source slit is increased;
(f) the monochromatic source is replaced by a source of white light ?
(In each operation, take all parameters, other than the one specified, to remain unchanged.)
Solution:
a) Angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits.

b) The separation of the fringes (and also angular separation) decreases. See, however, the condition mentioned in (d) below.

c) The separation of the fringes (and also angular separation) decreases. See however, the condition mentioned in (d) below.

d) Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition s/S < λ/d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.

e) Same as in (d). As the source slit which increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s/S ≤ λ/d is not satisfied, the interference pattern disappears.

f) The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are at the same position. Therefore, the central fringe is white. For a point P for which S₂P – S₁P = λ_b/2, where λ_b (≈ 4000 Å) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away where S₂Q – S₁Q = λ_b = λ_r/2 where λ_r (≈ 8000 Å) is the wavelength for the red colour, the fringe will be predominantly blue.
Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 5.
In Textual Example 3, what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern ?
Solution:
We want aθ = λ, θ = \(\frac{\lambda}{\mathrm{a}}\)
10 \(\frac{\lambda}{\mathrm{d}}\) = 2\(\frac{\lambda}{\mathrm{a}}\), a = \(\frac{\mathrm{d}}{5}\) = 0.2 mm

Question 6.
Assume that light of wavelength 6000 Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch ?
Solution:
A 100 inch telescope implies that 2a =100 inch = 254 cm.
Thus if,
λ = 6000 Å = 6 × 10^-5 cm
then
∆θ ≈ \(\frac{0.61 \times 6 \times 10^{-5}}{127}\)
≈ 2.9 × 10^-7 radians.

Question 7.
For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm ?
Solution:
Z_F = \(\frac{\mathrm{a}^2}{\lambda}=\frac{\left(3 \times 10^{-3}\right)^2}{5 \times 10^{-7}}\) = 18 m

Question 8.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ?
Solution:
Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be I = I₀cos²θ,
where θ is the angle between pass axes of P₁ and P₂. Since P₂ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be (π/2 – θ). Hence the intensity of light emerging from P₃ will be
I = I₀cos²θ cos² (\(\frac{\pi}{2}\) – θ)
= I₀ cos² θ sin² θ = (I₀/4) sin² 2θ
Therefore, the ‘transmitted intensity will be maximum when θ = π/4.

Question 9.
Unpolarised light is incident on a plane glass surface. What should be the angle of the incidence so that the reflected and refracted rays are perpendicular to each other ?
Solution:
For i + r to be equal to π/2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.

Students get through AP Inter 2nd Year Physics Important Questions 4th Lesson Electric Charges and Fields which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 4th Lesson Electric Charges and Fields

Very Short Answer Questions

Question 1.
What is meant by the statement ‘charge is quantized’? [IPE 2015 (TS)]
Answer:
The minimum charge that can be transferred from one body to the other is equal to the charge of the electron (e = 1.602 × 10^-19C). A charge always exists an integral multiple of charge of electron (q = ne). Therefore charge is said to be quantized.

Question 2.
Repulsion is the sure test of charging than attraction. Why ?
Answer:
A charged body may attract a neutral body and also an opposite charged body. But it always repels like a charged body. Hence repulsion is the sure test of electrification.

Question 3.
How many electrons constitute 1 C of charge ?
Answer:
n = \(\frac{\mathrm{q}}{\mathrm{e}}=\frac{1}{1.6 \times 10^{-19}}\) = 6.25 × 10¹⁸ electrons

Question 4.
What happens to the weight of a body when it is charged positively ?
Answer:
When a body is positively charged it must loose some electrons. Hence, weight of the body will decrease.

Question 5.
What happens to the force between two charges if the distance between them is [Board Model Paper]
a) halved
b) doubled ?
Answer:
From Coulombs law, F ∝ \(\frac{1}{\mathrm{~d}^2}\). So
a) When distance is reduced to half, force increases by four times.
[∵ F₂ = \(\frac{F_1 d_1^2}{\left(\frac{d_1}{2}\right)^2}\) = 4 F₁]

b) When distance is doubled to half, force increases by four times.
[∵ F₂ = \(\frac{F_1 d^2}{\left|2 d_1\right|^2}\) = \(\frac{1}{4}\) F₁]

Question 6.
The electric lines of force do not intersect. Why ?
Answer:
They do not intersect because if they intersect, at the point of intersection, intensity of electric field must act in two different directions, which is impossible.

Question 7.
Consider two charges + q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
Answer:
Charges are scalars, but the electrical intensities are vectors and add vectorially.

Question 8.
Electrostatic field lines of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements can you guess the nature of electrostatic force ?
Answer:
It is conservative force.

Question 9.
State Gauss’s law in electrostatics. [IPE 2015 (TS)]
Answer:
Gauss’s law: It states that “the total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times net charge enclosed by the surface”.
\(\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}}{\varepsilon_0}\)

Question 10.
When is the electric flux negative and when is it positive ?
Answer:
Electric flux Φ = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\). If angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{A}}\) is 180°, then flux will have a -ve’ sign. We consider the flux flowing out of the surface as positive and flux entering into the surface as negative.

Question 11.
Write the expression for electric intensity due to an infinite long charged wire at a distance radial distance r from the wire.
Answer:
The electric intensity due to an infinitely long charged wire E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\) perpendicular to the conductor.
Where λ = Uniform linear charge density
r = Distance of the point from the conductor.

Question 12.
Write the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
The electric intensity due to an infinite plane sheet of charge is E = \(\frac{\sigma}{2 \varepsilon_0}\).

Question 13.
Write the expression for electric intensity due to a charged conducting spherical shell at points outside and inside the shell.
Answer:
a) Intensity of electric field at any point inside a spherical shell is zero.
b) Intensity of electric field at any point- outside a uniformly charged spherical shell is
E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Question 14.
A proton and an α-particle are released in a uniform electric field. Find the ratio of (a) forces experienced by them (b) accelerations gained by each.
Answer:
a) As F = Eq, F ∝ q, ⇒ \(\frac{F_p}{F_\alpha}=\frac{Q_p}{Q_\alpha}=\frac{1}{2}\)
∴ \(\frac{F_p}{F_\alpha}=\frac{1}{2}\)

b) As a = \(\frac{E Q}{m}\) ⇒ a ∝ \(\frac{Q}{m}\) ⇒ \(\frac{a_p}{a_\alpha}=\frac{e_p}{Q_\alpha} \times \frac{m_\alpha}{m_p}=\frac{1}{2} \times \frac{4}{1}=\frac{2}{1}\)
∴ \(\frac{a_p}{a_\alpha}=\frac{2}{1}\)

Question 15.
The electric field in a region is given by \(\bar{E}=a \bar{i}+b \bar{j}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y-z plane.
Answer:
Electric field, \(\bar{E}=a \bar{i}+b \bar{j}\)
Flux passing through square area, Φ = \(\overline{\mathrm{E}} \cdot \overline{\mathrm{A}}=(\mathrm{a} \overline{\mathrm{i}}+\mathrm{b} \overline{\mathrm{i}}) \cdot\left(\mathrm{L}^2 \overline{\mathrm{i}}\right)\) = aL² Wb

Question 16.
A hollow sphere of radius ‘r’ has a unifrom charge density ‘σ’. It is kept in a cube of edge 3r such that the center of the cube coincides with the center of the shell. Calculate the electric flux that comes out of a face of the cube.
Answer:
Charge on the hallow sphere, q = σ × 4πr².
The flux through a single face of the cube, Φ¹ = \(\frac{f}{6}=\frac{1}{6} \cdot \frac{Q}{\varepsilon_0}=\frac{\sigma \times 4 \pi \mathrm{r}^2}{6 \varepsilon_0}=\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}\)

Question 17.
Consider a uniform electric field . What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the YZ plane ?
Answer:
Given, (field is along positive x-axis)
Surface area of square, S = (10 × 10^-2)(10 × 10^-2) = 10^-2m².
When plane of the square is parallel to yz-plane its area vector points towards 4-ve x-axis.
So θ = 0°.
∴ Flux through square, Φ = EScosθ = 3 × 10³ × 10^-2 × cos0° => <(> = 30 NC^-1m².

Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity. [T.S. Mar. 17; Mar. 14]
Answer:
Coulomb’s law – Statement: Force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The force acts along the straight line joining the two charges.

Explanation : Let us consider two charges q₁ and q₂ be separated by a distance r.

Then F ∝ q₁q₂ and F ∝ \(\frac{1}{\mathrm{r}^2}\) or F ∝ \(\frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\)
∴ F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\) where \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C^-2.
In vector form, in free space \(\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2} \hat{\mathrm{r}}\). Here \(\hat{\mathrm{r}}\) is a unit vector.
ε₀ is called permittivity of free space.
ε₀ = 8.85 × 10^-12 C²/N-m² or Farad/meter.
In a medium, F_m = \(\frac{1}{4 \pi \varepsilon} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{1}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}}} \times \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\) [∵ ε = ε₀ε_r]
Where ε is called permittivity of the medium.

Question 2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge. [A.P. Mar. 16]
Answer:
Intensity of electric field (E) : Intensity of electric field at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point.
Expression :

1. Intensity of electric field is a vector. It’s direction is along the direction of motion of positive charge.
2. Consider point charge q. Electric field will exist around that charge. Consider any point P in that electric field at a distance r from the given charge. A test charge q₀ is placed at P.
3. Force acting on q₀ due to q is F = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qq}_0}{\mathrm{r}^2}\)
4. Intensity of electric field at that point is equal to the force experienced by a test charge q₀.
   
   Intensity of electric field, E = \(\frac{\mathrm{F}}{\mathrm{q}_0}\)
   E = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}\)N/C or V/m

Question 3.
Derive the equation for the couple acting on a electric dipole in a uniform electric field.
Answer:

1. A pair of opposite charges separated by a small distance is called dipole.
2. Consider the charge of dipole are -q and +q coulomb and the distance between them is 2a.
3. Then the electric dipole moment P is given by P = q × 2a = 2aq. It is a vector. It’s direction is from -q to +q along the axis of dipole.
4. It is placed in a uniform electric field E, making an angle θ with, field direction as shown in fig.
5. Due to electric field force on +q is F = +.qE and force on -q is F = -qE.
6. These two equal and opposite charges constitute torque or moment of couple.
   
   i.e., torque, τ = ⊥^r distance × magnitude of one of force
   ∴ τ = (2a sin θ)qE = 2aqE sin θ = PE sin θ
7. In vector form, \(\vec{\tau}=\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole. [A.P. Mar. 17; T.S. Mar. 16]
Answer:
Electric field at a point on the axis of a dipole :

1. Consider an electric dipole consisting of two charges -q and + q separated by a distance ‘2a’ with centre ‘O’.
   
2. We shall calculate electric field E at point P on the axial line of dipole, and at a distance OP = r.
3. Let E₁ and E₂ be the intensities of electric field at P due to charges + q and -q respectively.
   

Question 5.
Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole. [A.P. Mar. 15]
Answer:
Electric field intensity on equatorial line of electric dipole :

1. Consider an electric dipole consisting of two charges-q and +q separated by a distance ‘2a’ with centre at ’O’.
2. We shall calculate electric field E at P on equatorial line of dipole and at a distance OP = r.
   
3. Let E₁ and E₂ be the electric fields at P due to charges +q and -q respectively.
4. The ⊥^r components (E₁ sin θ and E₂ sin θ) cancel each other because they are equal and opposite. The I lel components (E₁ cos θ and E₂ cos θ) are in the same direction and hence add up.
5. The resultant field intensity at point P is given by E = E₁ cos θ + E₂ cos θ
   
6. From figure, cos θ = \(\frac{a}{\left(r^2+a^2\right)^{1 / 2}}\)
   ∴ E = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{2 \mathrm{aq}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{3 / 2}}\)
7. If r >> a, then a² can be neglected in comparison to r². Then E
   E = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0} \times \frac{1}{\mathrm{r}^3}\)
   In vector form E = \(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{P}}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and explain its importance. [T.S. Mar. 15]
Answer:
Gauss’s law : The total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface.
Total electric flux, Φ = \(\oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{\mathrm{q}}{\varepsilon_0}\)
Here q is the total charge enclosed by the surface ‘S’, \(\oint\) represents surface integral of the closed surface.
Importance :

1. Gauss’s law is very useful in calculating the electric field in case of problems where it is possible to construct a closed surface. Such surface is called Gaussian surface.
2. Gauss’s law is true for any closed surface, no matter what its shape or size.
3. Symmetric considerations in many problems make the application of Gauss’s law much easier.

Long Answer Questions

Question 1.
State Gauss’s law in electrostatics. Applying Gauss’s law derive the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
Gauss’s law : The total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge enclosed by the surface. i.e., Φ = \(\oint_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{s}}=\frac{\mathrm{q}}{\varepsilon_0}\)

Expression for E due to an infinite plane sheet of charge:

1. Consider an infinite plane sheet of charge Let the charge distribution is uniform on this plane.
2. Uniform charge density on this surface σ = \(\frac{\mathrm{dq}}{\mathrm{dS}}\) where dq is the charge over an infinite small area ds.
3. Construct a horizontal cylindrical Gaussian surface ABÇD perpendicular to the plane with length 2r.
4. The flat surfaces BC and AD are parallel to the plane sheet and are at equal distance from the plane.
5. Let area of these surfaces are dS₁ and dS₂. They are parallel to \(\overrightarrow{\mathrm{E}}\). So flux through
   these two surfaces is \(\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}=\int\) Eds = E(S + S) = 2ES ……………….. (1)
6. Consider cylindrical surface of AB and CD. Let their areas are say dS₃ and dS₄. These surfaces are ⊥^r to electric intensity \(\overrightarrow{\mathrm{E}}\).
7. So angle between \(\overrightarrow{\mathrm{E}}\) and d\(\overrightarrow{S_3}\) or dS₄ is 90°. Total flux through these surfaces is zero.
   Since \(\oint_S\) E.dS = 0.
8. From Gauss’s law total flux, Φ = \(\oint \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{S}}\) = 2ES = \(\frac{\mathrm{q}}{\varepsilon_0}\)
   ∴ 2ES = \(\frac{\sigma S}{\varepsilon_0}\) [∵Q = σ × S]
9. Therefore intensity of electric field due to an infinite plane sheet of charge E = \(\frac{\sigma}{2 \varepsilon_0}\).

Textual Examples

Question 1.
How can you charge a metal sphere positively without touching it ?
Solution:
Figure (a) shows an uncharged metallic sphere on an insulating metal stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c). Disconnect the sphere from the ground. The positive charge continues to be held at the near end [Fig. 4.5 (d)]. Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

Question 2.
If 10⁹ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? .
Solution:
In one second electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10^-19 × 10⁹C = 1.6 × 10^-10 C. The time required o accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.6 × 10^-10 C/s) = 6.25 × 10⁹ s = 6.25 × 10⁹ ÷ (365 × 24 × 3600) years = 198 years. Thus to collect a charge of one coulomb, from a body from which 10⁹ electrons move, out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.

It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about 2.5 × 10²⁴ electrons.

Question 3.
How much positive and negative charge is there in a cup of.water ?
Solution:
Let us assume that the mass of one cup of water is 250 g. The molecular mass of. water is 18g. Thus, one mole (= 6.02 × 10²³ molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 10²³.

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10^-19 C = 1.34 × 10⁷C.

Question 4.
Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square depen-dence on the distance between the charges/masses, (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a protron and (ii) for two protons (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= 10^-10 m) apart ?
(m_p = 1.67 × 10^-27 kg, m³ = 9.11 × 10^-31 kg).
Solution:
a) i) The electric force between an electron and a proton at a distance r apart is :
F_e = –\(\frac{\mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\)
Where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is :
F_G = -G\(\frac{\mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\)
Where m_p and m_e are the masses of a proton and an electron respectively.
\(\left|\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{G}}}\right|=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}\) = 2.4 × 10³⁹

ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is :
\(\left|\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{G}}}\right|=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{G} \mathrm{m}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}\) = 1.3 × 10³⁶
However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive. The actual values of these forces between two protons inside a nucleus (distance between two protons
is ~ 10^-15m inside a nucleus) are F_e ~ 230 N whereas F_G ~ 1.9 × 10^-34 N.
The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however the masses of an electron and a proton are different. Thus, the magnitude of force is
|F| = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{e}^2}{\mathrm{r}^2}\) = 8.987 × 10⁹ Nm²/C² × (1.6 × 10^-19C)² / (10^-10 m)²
= 2.3 × 10^-8N
Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is a = 2.3 × 10^-8 N/9.11 × 10^-31 kg = 2.5 × 10²² m/s²
Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.
The value for acceleration of the proton is
a = 2.3 × 10^-8 N/1.67 × 10^-27 kg = 1.4 × 10⁹ m/s².

Question 5.
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in Fig. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in Fig. (c). What is the expected repulsion of A on the basis of Coulomb’s law ? Spheres A and C and spheres B and D have identical sizes. Ignore the size of A and B in comparison to the separation between their centres.

Solution:
Let the original charge on sphere A be q and that on B be q’. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}\)
Neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
F’ = \(\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}\) = F
Thus the electrostatic force on A, due to B, remains unaltered.

Question 6.
Consider three charges q₁, q₂, q₃ each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. ?

Solution:
In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC,
AD = AC cos 30° = \(\left(\frac{\sqrt{3}}{2}\right) l\) and the distance AO of the centroid O from A is
(2/3) AD = \(\left(\frac{1}{\sqrt{3}}\right) l\).
By symmetry AO = BO = CO.
Thus,
Force F₁ on Q due to charge q at A = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along AO
Force F₂ on Q due to charge q at B = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along BO
Force F₃ on Q due to charge q at C = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along CO
The resultant of forces F₂ and F₃ is \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}\) along OA, by the parallelogram law. Therefore, the total force on Q = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{l^2}(\hat{\mathrm{r}}-\hat{\mathrm{r}})\) = 0, where \(\hat{\mathrm{r}}\) is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60° about O.

Question 7.
Consider the charges q, q and -q placed at the vertices of an equilateral triangle, as shown in Fig. What is the force on each charge ?

Solution:
The forces acting on charge q at A due to charges q at B and -q at C are F₁₂ along BA and F₁₃ along AC respectively, as shown in Fig. By the parallelogram law, the total force F₁ on the charge q at A is given by
F₁ = F \(\hat{\mathrm{r}}_1\) where \(\hat{\mathrm{r}}_1\) is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude
F = \(\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 l^2}\)
The total force F₂ on charge q at B is thus F₂ = F \(\hat{\mathrm{r}}_2\), where \(\hat{\mathrm{r}}_2\) is a unit vector along AC.
Similarly the total force on charge -q at C is F₃ = \(\sqrt{3} \mathrm{~F} \hat{\mathrm{n}}\), where \(\hat{\mathrm{n}}\) is the unit vector along the direction bisecting the ∠BCA.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
F₁ + F₂ + F₃ = 0
The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise.

Question 8.
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N C^-1 (Fig. a). The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance (Fig. b) Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.

Solution:
In Fig. (a) the field is upward, so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude f the electric field. The acceleration of the electron is a_e = eE/m_e.
Where m_e is the mass of the electron.
Starting from rest, the time required by the electron to fall through a distance h is given by
t_e = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{a}_{\mathrm{e}}}}=\sqrt{\frac{2 \mathrm{hm_{ \textrm {e } }}}{\mathrm{eE}}}\)
For e = 1.6 × 10^-19C, m_e = 9.11 × 10^-31 kg.
E = 2.0 × 10⁴ NC^-1, h 1.5 × 10^-2 m.
t_e = 2.9 × 10^-9 s
In Fig. (b), the field is downward and the positively charged proton experiences a downward force of magnitude eE. The acceleration of the proton is
a_p = eE/m_p
Where mp is the mass of the proton ; m_p = 1.67 × 10^-27 kg. The time of fall for the proton is
t_p = \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{a}_{\mathrm{p}}}}=\sqrt{\frac{2 h \mathrm{~m}_{\mathrm{p}}}{\mathrm{eE}}}\) = 1.3 × 10^-7 s
Thus, the heavier particle (proton) takes a greater time to fall through the same distance. This is in basic contrast to the situation of, free fall under gravity’ where the time of fall is independent of the mass of the body. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. To see if this is justified, let us calculate the acceleration of the proton in the given electric field
a_p = \(\frac{\mathrm{eE}}{\mathrm{m}_{\mathrm{p}}}=\frac{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(2.0 \times 10^4 \mathrm{NC}^{-1}\right)}{1.67 \times 10^{-27} \mathrm{~kg}}\)
= 1.9 × 10¹² ms^-2
Which is enormous compared to the value of g (9.8 ms^-2), the acceleration due to gravity. The acceleration of the electron is even greater. Thus, the effect of acceleration due to gravity can be ignored in this example.

Question 9.
Two point charges q₁ and q₂, of magnitude +10^-8 C and -10^-8 C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig.

Solution:
The electric field vector E_1A at A due to the positive charge qx points towards the right and has a magnitude
E_1A = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.05 \mathrm{~m})^2}\) = 3.6 × 10⁴ NC^-1
The electric field vector E_2A at A due to the negative charge q₂ points towards the right and has the same magnitude. Hence the magnitude of the total electric field E_A at A is
E_A = E_1A + E_2A = 7.2 × 10⁴ NC^-1
E_A is directed toward the right.
The electric field vector E_1B at B due to the positive charge q₁ points towards the left and has a magnitude.
E_1B = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.05 \mathrm{~m})^2}\) = 3.6 × 10⁴ NC^-1
The electric field vector E_2B at B due to the negative charge q₂ points towards the right and has a magnitude.
E_2B = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.15 \mathrm{~m})^2}\) = 4 × 10⁴ NC^-1
The magnitude of the total electric field at B is E_B = E_1B – E_2B = 3.2 × 10⁴ NC^-1
E_B is directed towards the left.
The magnitude of each electric field vector at point C, due to charge q₁ and q₂ is
E_1C = E_2C = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right) \times\left(10^{-8} \mathrm{C}\right)}{(0.10 \mathrm{~m})^2}\) = 9 × 10³ NC^-1
The directions in which these two vectors point are indicated in Fig. The resultant of these two vectors is
E_C = E₁ cos \(\frac{\pi}{3}\) + E₂ cos \(\frac{\pi}{3}\) = 9 × 10³ NC^-1
E_C points towards the right.

Question 10.
Two charges 10 μC are placed 5.0 mm apart. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig. (a) and (b) a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig. (b).

Solution:
a) Field at P due to charge +10 μC
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15-0.25)^2 \times 10^{-4} \mathrm{~m}^2}\)
= 4.13 × 10⁶ NC^-1 along BP
Field at P due to charge -10 μC
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15+0.25)^2 \times 10^{-4} \mathrm{~m}^2}\)
= 3.86 × 10⁶ NC^-1 along PA
The resultant electric field at P due to the two charges at A and B is 2.7 × 10⁵ NC^-1 along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can’expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ±q,.2a distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude.
E = \(\frac{2 \mathrm{p}}{4 \pi \varepsilon_0 \mathrm{r}^3}\) (r/a > > 1)
Where p = 2aq is the magnitude of the dipole moment.
The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from -q to q). Here, p = 10^-5 × C ; 5 × 10^-3 m = 5 × 10^-8 C m
Therefore,
E = \(\frac{2 \times 5 \times 10^{-8} \mathrm{Cm}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15)^3 \times 10^{-6} \mathrm{~m}^3}\)
= 2.6 × 10⁵ N C^-1
Along the dipolemoment direction AB, which is close to the result obtained earlier.

b) Field at Q due to charge +10 μC at B
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{\left(15^2+(0.25)^2\right] \times 10^{-4} \mathrm{~m}^2} \mathrm{x}\)
= 3.99 × 10⁶ N C^-1 along BQ
Field at Q due to charge – 10 μC at A
= \(\frac{10^{-5} \mathrm{C}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{\left[15^2+(0.25)^2\right] \times 10^{-4} \mathrm{~m}^2}\)
= 3.99 × 10⁶ × N C^-1 along QA ‘
Clearly, the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. Therefore, the resultant electric field at Q due to the two charges at A and B is
= 2 × \(\frac{0.25}{\sqrt{15^2+(0.25)^2}}\) × 3.99 × 10⁶NC^-1 along BA
= 1.33 × 10⁵ N C^-1 along BA.
As in (a), we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole :
E = \(\frac{\mathrm{p}}{4 \pi \varepsilon_0 \mathrm{r}^3}\) (r/a > > 1)
= \(\frac{5 \times 10^{-8} \mathrm{Cm}}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)} \times \frac{1}{(15)^3 \times 10^{-6} \mathrm{~m}^3}\)
= 1.33 × 10⁵ N C^-1
The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.

Question 11.
The electric field components in Fig. are E_x = ax^1/2, E_y = E_z = 0, in which a = 800 N/C m^1/2. Calculate (a) the flux through the cube and (b) the charge within the cube. Assume that a = 0.1 m.

Solution:
a) Since the electric field has only an x component, for faces perpendicular to x direction, the angle between E and ∆S is ± π/2. Therefore, the flux Φ = E. ∆S is separately zero for each face of the cube except the two shaded ones. Now the magnitude of the electric field at the left face is E_L = x^1/2 = αa^1/2
(x = a at the left face).
The magnitude of electric field at the right face is E_R = αx^1/2 = α(2a)^1/2
(x = 2a at the right face).
The corresponding fluxes are
Φ_L = E_L . ∆S = ∆SE_L . \(\hat{n}_L\) = E_L ∆S cos θ = -E_L ∆S, since θ = 180°
= -E_La²
Φ_R = E_R . ∆S = E_R ∆S cos θ = E_R ∆S, since θ = 0°
= E_Ra²
Net flux through the cube.
= Φ_R + Φ_L = E_Ra² – E_La² = a² (E_R – E_L) = αa² [(2a)^1/2 – a^1/2]
= αa^5/2 (\(\sqrt{2}\) – 1)
= 800 (0.1)^5/2 (\(\sqrt{2}\) – 1)
= 1.05 N m² C^-1

b) We can use Gauss’s law to find the total charge q inside the cube.
We have f = \(\frac{\mathrm{q}}{\varepsilon_0}\) or q = Φε₀. Therefore,
q = 1.05 × 8.854 × 10^-12 C = 9.27 × 10^-27 C.

Question 12.
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 \(\hat{\mathrm{i}}\) N/C for x > 0 and E = -200 \(\hat{\mathrm{i}}\) N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = – 10 cm (Fig.),
(a) What is the net outward flux through each flat face ?
(b) What is the flux through the side of the cylinder ?
(c) What is the net outward flux through the cylinder ?
(d) What is the net charge inside the cylinder ?
Solution:
a) We can see from the figure that on the left face E and ∆S are parallel. Therefore, the outward flux is
Φ_L = E. ∆S = -200 \(\hat{\mathrm{i}}\) . ∆S
= +200 ∆S, since \(\hat{\mathrm{i}}\) . ∆S = – ∆S
= +200 × π(0.05)² = + 1.57 Nm²C^-1
On the right face, E and AS are parallel and therefore
Φ_R = E. ∆S = +1.57 Nm²C^-1.

b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E. ∆S = 0. Therefore, the flux out of the side of the cylinder is zero.

c) Net outward flux through the cylinder Φ = 1.57 + 1.57 + 0 = 3.14 Nm²C^-1.

d) The net charge within the cylinder can be found by using Gauss’s law which gives
q = ε₀Φ
= 3.14 × 8.854 × 10^-12 C
= 2.78 × 10^-11 C

Question 13.
An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus ?
Solution:
The charge distribution for this model of the atom is as shown in Fig. The total negative charge in the uniform spherical charge distribution of radius R must be -Ze, since the atom (nucleus of charge Z e + negative charge) is neutral.

This immediately gives us the negative charge density p, since we must have
\(\frac{4 \pi \mathrm{R}^3}{3}\) ρ = 0 – Ze or ρ = – \(\frac{3 \mathrm{Ze}}{4 \pi \mathrm{R}^3}\)
To find the electric field E(r) at a point P which is a distance r away from the nucleus, we use Gauss’s law. Because of the spherical symmetry of the charge distribution, the magnitude of the electric field E(r) depends only on the radial distance, no matter what the direction of r. Its direction is along (or opposite to) the radius vector r from the origin to the point P. The obviouis Gaussian surface is a spherical surface centred at the nucleus. We consider two situations, namely r < R and r > R.

i) r < R : The electric flux Φ enclosed by the spherical surface is Φ = E(r) × 4πr²
Where E(r) is the magnitude of the electric field at r. This is because the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there, and has the same magnitude at all points on the surface.
The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r,
i.e., q = Ze + \(\frac{4 \pi \mathrm{r}^3}{3}\) ρ
Substituting for the charge density p obtained earlier, we have
q = Ze – Ze\(\frac{\mathrm{r}^3}{\mathrm{R}^3}\)
Gauss’s law then gives,
The electric field is directed radially outward.

ii) r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. Thus, from Gauss’s law,
E(r) × 4 π r² = 0 or E(r) = 0 ; r > R
At r = R, both cases give the same result: E = 0.

Students get through AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 5th Lesson Electrostatic Potential and Capacitance

Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge. [T.S. Mar. 16]
Answer:
Expression for the electric potential due to a point charge:

1. Electric potential at a point is defined as the amount of workdone in moving a unit + ve charge from infinity to that point.
   
2. Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
3. Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, whereas the potential difference is +ve in the direction A to B.
4. Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P
   

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Expression for the electrostatic potential energy of a system of two point charges :

1. Let two point charges q₁ and q₂ are separated by distance ‘r’ in space.
2. An electric field will develop around the charge q₁.
3. To bring a charge q₂ from infinity to the point B some work must be done.
   
   Workdone = q₂ v_B
   But v_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
4. This amount of workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.
   ∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
5. If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
6. Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Expression for potential energy of an electric dipole placed in a uniform electric field :

1. Consider a electric dipole of length 2a having charges + q and -q.
2. The electric dipole is placed in uniform electric field E and it’s axis makes an angle θ with E.
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   
   Torque τ = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}} \) ⇒ BC = 2a sinθ
   ∴ Torque τ = qE × 2a sinθ = PE sin θ [∴ p = 2aq]
4. Suppose the dipole is rotated through an angle dθ, the workdone dw is given by
   dw = τdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂
   workdone W = \(\int_{\theta_1}^{\theta_2}\) PE sinθdθ = PE(cosθ₁ – cosθ₂)
6. This workdone (W) is then stored as potential energy(U) in the dipole.
   ∴ U = PE(cosθ₁ – cosθ₂)
7. If θ₁ = 90° and θ₂ = 0°, U = – PE cos₁.
   In vector form U = –\(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{E}}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. [A.P. Mar. 16; T.S. Mar. 14]
Answer:
Expression for the capacitance of a parallel plate capacitor:

1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is + q and surface charge density of charge = σ
   ∴ q = Aσ
   
4. The electric intensity at point x , E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   \(\int {\mathrm{dV}}=\int_{\mathrm{d}}^0-\mathrm{Edx}=\int_{\mathrm{d}}^0 \frac{-\sigma}{\varepsilon_0} \mathrm{dx}=\frac{\sigma \mathrm{d}}{\varepsilon_0}\)
6. Capacitance of the capacitor C = \(\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{A} \sigma}{\frac{\sigma \mathrm{d}}{\varepsilon_0}}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\) Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field.
Answer:

1. When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts
   
   opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.
2. Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.
3. If E₀ is the external field strength and E_i is the electric field strength induced, then the net field, \(\overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_0+\overrightarrow{\mathrm{E}}_{\mathrm{i}}\)
   (E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Question 6.
Define electric potential. Derive an expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.
Answer:
Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.

Expression for the potential at a point due to a dipole:

1. Consider A and B having -q and + q charges separated by a distance 2a.
2. The electric dipole moment P = q × 2a along AB
3. The electric potential at the point’P’is to be calculated.
4. P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
5. BN and AM are perpendicular to OP.
6. Potential at ‘P’ due to charge +q at B,
   V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{BP}}\right]\)
   ∴ V₁ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}\right]\) [∵ BP = NP]
7. Potential at P due to charge -q at A, V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{-q}}{\mathrm{AP}}\right]\)
   ∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{MP}}\right]\) [∵ AP = MP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{\mathrm{q}}{\mathrm{NP}}-\frac{\mathrm{q}}{\mathrm{MP}}\right]\) ………… (1)
9. In △le ONB, ON = OB cos0 = a cosθ; ∴NP = OP – ON = r – a cosθ ………………… (2)
10. In △le AMO, OM = AO cos0 = a cosθ; ∴ MP = MO + OP = r + a cosθ ………………. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
13. (a) Electric potential on the axial line of dipole:
    
    • When θ = 0°, point p lies on the side of + q. ____
      ∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]
    • When θ = 180°, point p lies on the side of — q.
      ∴ V = \(\frac{\mathrm{-P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

Qeustion 7.
What is series combination of capacitors. Derive the formula for equivalent capacitance in series combination. [A.P.& T.S. Mar.15]
Answer:
Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.
In this combination

1. Charge on each capacitor is equal.
2. P.D’s across the capacitors is not equal.

Consider three capacitors of capacitances C₁, C₂ and C₃ are connected in series across a battery of P.D V as shown in figure.

Let ‘Q’ be the charge on each capacitor.
Let V₁, V₂ and V₃ be the P.D’s of three
V = V₁ + V₂ + V₃ ……………… (1)
RD across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
RD across III^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across IIII^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1),V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}=\mathrm{Q}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)
\(\frac{\mathrm{V}}{\mathrm{Q}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\) [∵ \(\frac{1}{C}=\frac{V}{Q}\)
For ‘n’ number of capacitors, the effective capacitance
\(\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}+\ldots+\frac{1}{\mathrm{C}_{\mathrm{n}}}\)

Question 8.
What is parallel combination of capacitors. Derive the formula for equivalent capacitance in parallel combination. [T.S. Mar. 17; A.P.& T.S. Mar. 15]
Answer:
Parallel Combination: The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,
1. The P.D’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal.
Consider three capacitors of capacitance C₁, C₂ and C₃ are connected in parallel across a P.D V as shown in fig.
The charge on Ist capacitor Q₁ = C₁ V
The charge on IInd capacitor Q₂ = C₂ V
The charge on IIIrd capacitor Q₃ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{\mathrm{Q}}{\mathrm{V}}\) = C₁ + C₂ + C₃
C = C₁ + C₂ + C₃ [∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]
for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as
C = C₁ + C₂ + C₃ + ……………. + C_n

Question 9.
Derive an expression for the energy stored in a capacitor.
Answer:
Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’
∴ Average potential difference V_A = \(\frac{\mathrm{O}+\mathrm{V}}{2}=\frac{\mathrm{V}}{2}\)
Hence workdone to move the charge Q = W = V_A × Q = \(\frac{\mathrm{VQ}}{2}\)
This is stored as electrostatic potential energy ‘U’
∴ U = \(\frac{\mathrm{QV}}{2}\)
We know Q = CV then ‘U’ can be written as given below.
U = \(\frac{\mathrm{QV}}{2}=\frac{1}{2}\) CV² = \(\frac{Q^2}{2 C}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}=\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Qeustion 10.
What is the energy stored when the space between the plates is filled with dielectric.
a) With charging battery disconnected ?
b) With charging battery connected in the circuit ?
Answer:
Effect of Dielectric on energy stored :
Case (a) : When the charging battery is disconnected from the circuit:
Let the capacitor is charged by a battery and disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge remains constant.
Capacity increases by’K’ times.
New capacity C’ = \(\frac{Q}{V}\) = \(\frac{\frac{Q}{V}}{K}\) = K\(\frac{Q}{V}\) = KC [V’ = \(\frac{V}{K}\); C = \(\frac{Q}{V}\)]
∴ C’ = KC
Energy stored U’ = \(\frac{1}{2}\) QV = \(\frac{1}{2}\) Q \(\frac{V}{K}\) = \(\frac{U}{K}\)
U’ = \(\frac{U}{K}\)
∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b) : When the charging battery is connected in the circuit:
Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V
New charge on the plates Q’= KQ
Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC
Energy stored in the capacitor U’ = \(\frac{1}{2}\) QV; = \(\frac{1}{2}\) (KQ) V = KU
U = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7 C located 9 cm away,
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Qeustion 2.
Two charges 3 × 10^-8 C and -2 × 10^-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Thke the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Question 3.
A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when die slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 4.
Four charges are arranged at the corners of a square ABCD of side d. as shown in fig. Find the work required to put together this arrangement, (b) A charge q₀ is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A and then the charges -q, +q, and -q are brought to B, C and D, respectively. The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
2. Work needed to bring -q to B when + q is at A. This is given by (charge at B) × (electrostatic potential at B due to change +q at A) = -q × \(\left(\frac{q}{4 \pi \varepsilon_0 d}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q₀ to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7μC and -2μC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each ‘ other?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

Question 6.
There is a uniform electric field in the XOY plane represented by (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) Vm^-1. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is
Answer:
Given, uniform Electric field intensity,
\(\overrightarrow{\mathrm{E}}\) = (\(40 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\)) Vm^-1
Electric potential at the origin = 200V
Position vector \(\mathrm{d} \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}+1 \hat{\mathrm{j}})\) m
We know that,

dV = \(-\vec{E} \cdot d \vec{r}=-(40 \hat{i}+30 \hat{j}) \cdot(2 \hat{i}+\hat{j})\)
V_p – V₀ = -(80 + 30) = -110Volt. ’
V_p = V₀ – 110 = (200 – 110) Volt = 90 Volt
∴ Potential at point P, V_p = 90Volt.

Question 7.
An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is
Answer:
Charge at the centroid of an equilateral triangle = +q
The charge + q divides the line segment in ratio 2 : 1.
That means r_max = 2 and r_min = 1

Question 8.
A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged ?
Solution:
(For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V₁² = \(\frac{1}{2}\) CV² …………………. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}=\frac{1}{4}\)CV²; Let potential difference across the capacitor = V₂

Then, U₂ = \(\frac{1}{2}\) C₂V₂²
⇒ \(\frac{1}{4}\) CV² = \(\frac{1}{2}\) × 2C × V₂²
⇒ V₂² = \(\frac{\mathrm{V}^2}{4}\)
∴ V₂ = \(\frac{V}{2}\)

Question 9.
Three Capacitors each of capaitance 9 pF are connected in series.
a) What is the total capacitance of the combination ?
b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Solution:
a) Resultant capacitance in series combination \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \Rightarrow \frac{1}{C_S}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\)
C_S = 3pF

b) Rd across each capacitor = \(\frac{\mathrm{V}}{3}=\frac{120}{3}\) = 40V

Question 10.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. [A.P. Mar. 17]
a) What is the total capacitance of the combination ?
b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
a) C_p = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₂ = C₃V = 4 × 100 = 400pC

Textual Examples

Question 1.
(a) Calculate the potential at a point P due to a charge of 4 × 10^-7C located 9 cm away.
(b) Hence obtain the work done in bringing a charge of 2 × 10^-9C from infinity to the point P. Does the answer depend on the path along which the charge is brought ?
Solution:
(a) V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
= 9 × 10⁹ Nm² C^-2 \(\frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 × 10⁴ V

(b) W = qV = 2 × 10^-9 C × 4 × 10⁴V
= 8 × 10^-5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements. One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Question 2.
Two charges 3 × 10^-8 C and -2 × 10^-8C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be x-axis; the negative charge is taken to be on the right side of the origin fig.

Let P be the required point on the x-axis where the potential is zero. If x is the x – coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_0}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-\mathrm{x}) \times 10^{-2}}\right]\) = 0
where x is in cm. That is \(\frac{3}{x}-\frac{2}{15-x}\) = 0
which gives x = 9 cm.
If x lies on the extended line OA, the required condition is \(\frac{3}{x}-\frac{2}{x-15}\) = 0
Which gives x. = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the claculation required choosing potential to be zero at infinity.

Question 3.
Figure (a) and (b) shows the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference V_p – V_Q; V_B – V_A.
(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.
(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.
(d) Give the sign of the work done by the external agency in moving a small negative charge from B And A.
(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A ?
Solution:
(a) As V ∝ \(\frac{1}{\mathrm{r}}\), V_p > V_Q. Thus, (V_p – V_Q) is positive. Also VB is less negative than V_A. Thus, V_B > V_A or (V_B – V_A) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly. (PE.)_A > (P.E.)_B and hence sign of potential energy differences is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the eletric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going form B to A.

Question 4.
Four charges are arranged at the comets of a square ABCD of side d. as shown in fig. 5.15.(a) Find the work required to put together this arrangement, (b) A charge q0 is brought to the centre of the square, the four charges being held fixed at its comers. How much extra work is needed to do this ?

Solution:
(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges -q, +q, and -q are brought to B, C and D, respectively The total work needed can be calculated in steps.

1. Work needed to bring charge +q to A when no charge is present elesewhere: this is zero.
2. Work needed to bring -q to B when +q is at A. This is given by (charge at B) x (electrostatic .potential at B due to charge +q at A) = -q × \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{~d}}\right)\)
3. Work needed to bring charge +q to C when +q is at A and -q is at B. This is given by – (charge at Q × (potential at C due to charges at A and B) .
   
4. Work needed to bring -q to D when +q at A, -q at B, and +q at C.
   This is given by (charge at D) × (potential at D due to charges at A, B and C)
   
   Add the work done is steps (i), (ii), (iii) and (iv). The total work required is
   
   The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
   (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D.Hence no work is required to bring any charge to point E.

Question 5.
a) Determine the electrostatic potential energy of a system consisting of two charges 7µC and -2µC (and with no external field) placed at (-9 cm, 0,0) and (9cm, 0, 0) respectively.
b) How much work is required to separate the two charges infinitely away from each other ?
c) Suppose that the same system of charges is now placed in an external electric field E = A(1/r²); A = 9 × 10⁵ C m^-2. What would the electrostatic energy of the configuration be ?
Solution:
(a) U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
= 9 × 10⁹ × \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\) = -0.7 J.

(b)W = U₂ – U₁ = 0 – U = 0 – (-0.7)
= 0.7J.

c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,
q₁V(r₁) + q₂V(r₂) = A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
q₁V(r₁) + q₂V(r₂) + \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r_{12}}\)
= A \(\frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}\) + A \(\frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
= 70 – 20 – 0.7 = 49.3J

Question 6.
A molecule of a substance has a perma-nent electric dipole moment of magnitude 10^-29C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10⁶ V m^-1. The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.
Solution:
Here, dipole moment of each molecules = 10^-29C m
As 1 mole of the substance contains 6 × 10²³ molecules,
total dipole moment of all the molecules, p = 6 × 10²³ × 10^-29 Cm
= 6 × 10^-6 C m
Initial potential energy, U_t = -pE cos θ
=-6 × 10^-6 × 10⁶ cos 0° = -6 J
Final potential energy (when θ = 60°),
U_f = -6 × 10^-6 × 10⁶ × cos 60° = – 3J
Change in potential energy
= -3 J – (-6j) = 3 J
So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.

Question 7.
(a) A comb run through one’s dry hair attracts small bits of paper. Why ? What happens if the hair is wet or if it is a rainy day ? (Remember, a paper does not conduct electricity.)
(b) Ordinary rubber is an insulator. But special rubber types of aircraft are made slightly conducting. Why is this necessary ?
(c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?
(d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why ?
Solution:
(a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction, if the hair is wet, or if it is a rainy day. friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper.

(b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire.

(c) Reason similar to (b).

(d) Current passes only when there is difference in potential.

Question 8.
A slab of material of dielectric constant K has the same area as the plates of a parallelplate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates ?
Solution:
Let E₀ = V_g/d be the electric field between the plates when there is no dielectric and the potential difference is V₀. If the dielectric is now inserted, the electric field in the dielectric will be E = E₀/K. The potential difference will then be
V = \(E_0\left(\frac{1}{4} d\right)+\frac{E_0}{K}\left(\frac{3}{4} d\right)\)
= \(\mathrm{E}_0 \mathrm{~d}\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)=\mathrm{V}_0 \frac{\mathrm{K}+3}{4 \mathrm{~K}}\)
The potential difference decreases by the factor (K + 3)/K while the free charge Q₀ on the plates remains unchanged. The capacitance thus increases,
C = \(\frac{\mathrm{Q}_0}{\mathrm{~V}}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \frac{\mathrm{Q}_0}{\mathrm{~V}_0}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_0\)

Question 9.
A network of four 10μF capacitors is connected to a 500V supply, as shown in Fig. Determine
(a) the equivalent capacitance of the network and
(b) the charge on each capacitor, (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Solution:
(a) In the given network, C₁, C₂ and C₃ are connected in series. The effective capacitance C’ of these three capacitors isgivgftby
\(\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\)
For C₁ = C₂ = C₃ = 10 μF. C = (10/3) μF. The network has C and C₄ connected in parallel. Thus, the equivalent capacitance C of the network is
C = C’ + C₄ = (\(\frac{10}{3}\) + 10) μF = 13.3 μF

(b) Clearly, from the figure, the charge on each of the capacitors, C₁, C₂ and C₃ is the same, say Q. Let the charge on C₄ be Q. Now, since the potential difference c across AB is Q/C₁ across BC is Q/C₂. across CD is Q/C₃, we have
\(\frac{\mathrm{Q}}{\mathrm{C}_1}+\frac{\mathrm{Q}}{\mathrm{C}_2}+\frac{\mathrm{Q}}{\mathrm{C}_3}\) = 500 V
Also Q’/C₄ = 500V.
This gives for the given value of the capacitances,
Q = 500 V × \(\frac{10}{3}\) μF = 1.7 × 10^-3C and
Q’ = 500 V × 10μF = 5.0 × 10^-3C

Question 10.
(a) A 900pF capacitor is charged by 100 V battery [Fig.a]. How much electrostatic energy is stored by the capacitor ?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic y energy stored by the system ?

Solution:
The charge on the capacitor is
Q = CV = 900 × 10^-12F × 100 V
= 9 × 10^-8C
The energy stored by the capacitor is
= (1/2) CV² = (1/2) QV
= (1/2) × 9 × 10^-8C × 100 V
= 4.5 × 10^-6 J

(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V. The charge on each capacitor is then Q’ = CV’. By charge conservation, Q’ = Q/2. This implies V’ = V/2. The total energy of the system is
= 2 × \(\frac{1}{2}\)Q’V’ = \(\frac{1}{4}\)QV = 2.25 × 10⁶J
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone ?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b)

I.

Question 1.
In the experiment of tossing a coin n times, if the variable X denotes the number of heads and P(X = 4), P(X = 5), P(X = 6) are in arithmetic progression then find n.
Solution:
X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\) (∵ a coin is tossed)
Hint: a, b, c are in A.P.
⇒ 2b = a + c (or) b – a = c – a
Given, P(X = 4), P(X = 5), P(X = 6) are in A.P.

⇒ 2 × 30(n – 4) = 5[30 + n² – 9n + 20]
⇒ 12n – 48 = n² – 9n – 50
⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0
⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0
⇒ n = 7 or 14

Question 2.
Find the minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8.
Solution:
Let n be the number of times a fair coin tossed
X denotes the number of heads getting
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Given P(X ≥ 1) ≥ 0.8
⇒ 1 – P(X = 0) ≥ 0.8
⇒ P(X = 0) ≤ 0.2
⇒ \({ }^n C_o\left(\frac{1}{2}\right)^n \leq 0.2\)
⇒ \(\left(\frac{1}{2}\right)^n \leq \frac{1}{5}\)
The Maximum value of n is 3

Question 3.
The probability of a bomb hitting a bridge is \(\frac{1}{2}\) and three direct hits (not necessarily consecutive) are needed to destroy it. Find the minimum number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
Solution:
Let n be the minimum number of bombs required and X be the number of bombs that hit the bridge, then
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Now P(X ≥ 3) > 0.9
⇒ 1 – P(X < 3) > 0.9
⇒ P(X < 3) < 0.1
⇒ P(X = 0) + P(X = 1) + P (X = 2) < 0.1

By trial and error, we get n ≥ 9
∴ The least value of n is 9
∴ n = 9

Question 4.
If the difference between the mean and the variance of a binomial variate is \(\frac{5}{9}\) then, find the probability for the event of 2 successes, when the experiment is conducted 5 times.
Solution:
Given n = 5
Let p be the parameters of the Binomial distribution
Mean – Variance = \(\frac{5}{9}\)

∴ Probability of the event of 2 success = \(\frac{80}{243}\)

Question 5.
One in 9 ships is likely to be wrecked when they are set on the sail, when 6 ships are on the sail, find the probability for
(i) Atleast one will arrive safely
(ii) Exactly, 3 will arrive safely
Solution:
p = probability of ship to be wrecked = \(\frac{1}{9}\)
q = 1 – p
= 1 – \(\frac{1}{9}\)
= \(\frac{8}{9}\)
Number of ships = n = 6

Question 6.
If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4).
Solution:
Mean = np = 2.4 ………(1)
Variance = npq = 1.44 ……….(2)
Dividing (2) by (1)
\(\frac{n p q}{n p}=\frac{1.44}{2.4}\)
⇒ q = o.6 = \(\frac{3}{5}\)
p = 1 – q
= 1 – 0.6
= 0.4
= \(\frac{2}{5}\)
Substituting in (1)
n(0.4) = 2.4
⇒ n = 6
P(1 < X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)
= \({ }^6 C_2 q^4 \cdot p^2+{ }^6 C_3 q^3 \cdot p^3+{ }^6 C_4 q^2 \cdot p^4\)

Question 7.
It is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.
Solution:
p = probability of defective bulb = \(\frac{1}{10}\)
q = 1 – p
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)
n = Number of bulbs in the sample = 20
P(X > 2) = 1 – P(X ≤ 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

Question 8.
On average, rain falls on 12 days every 30 days, find the probability that, the rain will fall on just 3 days of a given week.
Solution:
Given p = \(\frac{12}{30}=\frac{2}{5}\)
q = 1 – p
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)
n = 7, r = 3
P(X = 3) = ⁿC_r . q^n-r . p^r
= \({ }^7 C_3\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\text { 35. }\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Let n, p be the parameters of a binomial distribution
Mean (np) = 6 ……..(1)
and variance (npq) = 2 ……..(2)
then \(\frac{n p q}{n p}=\frac{2}{6}\)
⇒ q = \(\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
From (1) np = 6
n(\(\frac{2}{3}\)) = 6
∴ n = 9
The first two terms of the distribution are
P(X = 0) = \({ }^9 C_0\left(\frac{1}{3}\right)^9=\frac{1}{3^9}\)
and P(X = 1) = \({ }^9 C_1\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)=\frac{2}{3^7}\)

Question 10.
In a city, 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.
Solution:
Average number of accidents per day
λ = \(\frac{10}{50}=\frac{1}{5}\) = 0.2
The probability that there win be 3 or more accidents in a day
P(X ≥ 3) = \(\sum_{\mathrm{K}=3}^{\infty} \mathrm{e}^{-\lambda} \cdot \frac{\lambda^{\mathrm{K}}}{\mathrm{K} !}, \lambda=0.2\)

II.

Question 1.
Five coins are tossed 320 times. Find the frequencies of the distribution of a number of heads and tabulate the result.
Solution:
5 coins are tossed 320 times
Probability of getting a head on a coin
p = \(\frac{1}{2}\), n = 5
Probability of having x heads

Question 2.
Find the probability of guessing at least 6 out of 10 answers in (i) True or false type examination (ii) multiple choice with 4 possible answers.
Solution:
(i) Since the answers are in True or false type
probability of success p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10-6} \cdot\left(\frac{1}{2}\right)^6\)
= \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10}\)

(ii) Since the answers are multiple-choice with 4 possible answers
Probability of success p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)
Probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} C_6\left(\frac{3}{4}\right)^{10-6}\left(\frac{1}{4}\right)^6\)
= \({ }^{10} C_6 \cdot \frac{3^4}{4^{10}}\)

Question 3.
The number of persons joining a cinema ticket counter in a minute has Poisson distribution with parameter 6. Find the probability that
(i) no one joins the queue in a particular minute
(ii) two or more persons join the queue in a minute.
Solution:
Here λ = 6
(i) Probability that no one joins the queune in a particular minute
P(X = 0) = \(\frac{\mathrm{e}^{-\lambda} \lambda^0}{0 !}=\mathrm{e}^{-6}\)

(ii) Probability that two or more persons join the queue in a minute

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a)

I.

Question 1.
A p.d.f of a discrete random variable is zero except at the points x = 0, 1, 2. At these points it has the value P(0) = 3c³, P(1) = 4c – 10c², P(2) = 5c – 1 for some c > 0. Find the value of c.
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c³ + 4c – 10c² + 5c – 1 = 1
3c³ – 10c² + 9c – 2 = 0
Put c = 1, then 3 – 10 + 9 – 2 = 12 – 12 = 0
c = 1 satisfies the above equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c² – 7c + 2 = 0
(c – 2) (3c – 1) = 0
∴ c = 2 or c = \(\frac{1}{3}\)
c = 2
⇒ P(X = 0) = 3 . 2³ = 24 which is not possible
∴ c = \(\frac{1}{3}\)

Question 2.
Find the constant C, so that F(x) = \(C\left(\frac{2}{3}\right)^x\), x = 1, 2, 3,……… is the p.d.f of a discrete random variable X.
Solution:

Question 3.

is the probability distribution of a random variable X. Find the value of K and the variance of X.
Solution:
Sum of the probabilities = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 – 0.6 = 0.4
⇒ k = \(\frac{0.4}{4}\) = 0.1
Mean = (-2) (0.1) + (-1) (k) + 0 (0.2) + 1 (2k) + 2(0.3) + 3k
= – 0.2 – k + 0 + 2k + 0.6 + 3k
= 4k + 0.4
= 4(0.1) + 0.4
= 0.4 + 0.4
= 0.8
μ = 0.8
Variance (σ²) = \(\sum_{i=1}^n x_i^2 P\left(x=x_i\right)-\mu^2\)
∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1 (2k) + 4 (0.3) + 9k – μ²
= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ²
= 12k + 0.4 + 1.2 – (0.8)²
= 12(0.1) + 1.6 – 0.64
= 1.2 + 1.6 – 0.64
= 2.8 – 0.64
= 2.16
∴ σ² = 2.16

Question 4.

is the probability distribution of a random variable X. Find the variance of X.
Solution:

Question 5.
A random variable X has the following probability distribution.

Find (i) k (ii) the mean and (iii) P(0 < X < 5)
Solution:
Sum of the probabilities = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\), -1 Since k > 0
∴ k = \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\)

(ii) Mean = \(\sum_{i=1}^n x_i P\left(x=x_i\right)\)
Mean (μ) = 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k2) + 6(2k²) + 7 (7k² + k)
= 0 + k + 4k + 6k + 12k + 5k² + 12k² + 49k² + 7k
= 66k² + 30k
= 66(\(\frac{1}{100}\)) + 30(\(\frac{1}{10}\))
= 0.66 + 3
= 3.66

(iii) P(0 < x < 5)
P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k
= 8k
= 8(\(\frac{1}{10}\))
= \(\frac{4}{5}\)

II.

Question 1.
The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c³, P(X = 1) = 4c – 10c², P(X = 2) = 5c – 1
(i) Find the value of c
(ii) P(X < 1), P(1 ≤ X < 2) and P(0 < X ≤ 3)
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c³ + 4c – 10c² + 5c – 1 = 1
3c³ – 10c² + 9c – 2 = 0
c = 1 satisfy this equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c² – 7c + 2 = 0
(c – 2) (3c – 1) = 0
c = 2 or c = \(\frac{1}{3}\)
c = 2 ⇒ P(X = 0) = 3 . 2³ = 24 which is not possible
∴ c = \(\frac{1}{3}\)

(i) P(X < 1) = P(X = 0)
= 3 . c³
= 3 . \(\left(\frac{1}{3}\right)^3\)
= 3 . \(\frac{1}{2}\)
= \(\frac{1}{9}\)

(ii) P(1 < X ≤ 2) = P(X = 2)
= 5c – 1
= \(\frac{5}{3}\) – 1
= \(\frac{2}{3}\)

(iii) P(0 < X ≤ 3) = P(X = 1) + P(X = 2)
= 4c – 10c² + 5c – 1
= 9c – 10c² – 1
= 9 . \(\frac{1}{3}\) – 10 . \(\frac{1}{9}\) – 1
= 3 – \(\frac{10}{9}\) – 1
= \(\frac{8}{9}\)

Question 2.
The range of a random variable X is {1, 2, 3, …..} and P(X = k) = \(\frac{C^K}{K !}\), (k = 1, 2, 3, ……), Find the value of C and P(0 < X < 3)
Solution:
Sum of the probabilities = 1

Students get through AP Inter 2nd Year Physics Important Questions 6th Lesson Current Electricity which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 6th Lesson Current Electricity

Short Answer Questions

Question 1.
Derive an expression for the effective resistance when three resistors are connected in
(i) series
(ii) parallel.
Answer:
Effective resistance when three resistors are connected :
(i) In series :

1. Three resistors R₁, R₂ and R₃ are connected in series as shown in fig. V₁, V₂, V₃ are the potential differences across R₁, R₂ and R₃. I is the current flowing through them.
2. Applying Ohm’s law to R₁, R₂ and R₃, Then V₁ = IR₁, V₂ = IR₂, V₃ = IR₃
3. In series, V = V₁ + V₂ + V₃
   IR_s = IR₁ + IR₂ + IR₃ [∵V = IR_s]
   ∴ R_s = R₁ + R₂ + R₃

ii) In parallel :

1. Three resistors, R₁, R₂ and R₃ are connected in parallel as shown in fig. Potential differences across each resistor is V.I₁, I₂, I₃ are the currents flowing through them.
2. Applying Ohm’s law to R₁, R₂ and R₃, then
   V = I₁R₁ = I₂R₂ = I₃R₃
   ⇒ I₁ = \(\frac{\mathrm{V}}{\mathrm{R}_1}\); I₂ = \(\frac{\mathrm{V}}{\mathrm{R}_2}\); I₃ = \(\frac{\mathrm{V}}{\mathrm{R}_3}\)
3. In parallel, I = I₁ + I₂ + I₃
   ⇒ \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{V}}{\mathrm{R}_1}+\frac{\mathrm{V}}{\mathrm{R}_2}+\frac{\mathrm{V}}{\mathrm{R}_3}\) [∵ I = \(\frac{V}{R_P}\)]
   ∴ \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}\)

Question 2.
On what factors does the resistance of a conductor depend ? Define electric resistance and write its S.I unit. How does the resistance of a conductor vary if (a) Conductor is stretched to 4 times of its length (b) Temperature of conductor is increased ? [Board Model Paper]
Answer:
Dependence of Resistance on geometry :

1. Resistance is directly proportional to length of the conductor (R ∝ l)
2. Resistance is inversely proportional to area of cross-section (R ∝ 1/A)
   i.e., R ∝ l/A ⇒ R = ρl / A; ρ is called specific resistance or resistivity of the material.

Dependence of Resistance on temperature : R₁ = R₁[1 + α(∆T)] & α = \(\frac{\Delta R / R}{\Delta T}\)
α is called temperature coefficient of resistance and its SI unit: K^-1.

Definition of electrical resistance (R) : The ratio between the potential difference (V) across the ends of the conductor and the current (i) flowing through the conductor is called its electrical resistance, R = V/i; S.I unit of electrical resistance is ohm (Ω)
a) Resistance of the conductor, R = ρ \(\frac{l}{\mathrm{~A}}\) = ρ \(\frac{l^2}{V}\) ⇒ R ∝ l² (∵ ρ and V constant)
⇒ R₁/R₂ = (l₁/l₁)² ∴ R₁/R₂ = (l/4l)
⇒ R₂ = 16R
Hence resistance of the conductor increases by 16 times.
b) The value of the electrical resistance of a conductor increases with increase of temperature. Because, temperature coefficient of resistance is positive for metals.

Long Answer Questions

Question 1.
State Kirchhoffs law for an electrical network. Using th&se laws deduce the condition for balance in a Wheatstone bridge. [T.S. Mar. 16; 15; Mar. 14]
Answer:

1. Kirchhoffs first law (Junction rule or KCL) : The algebraic sum of the currents at any junction is zero.
   ∴ ΣI = 0
   (or)
   The sum of the currents flowing towards a junction is equal to the sum of currents away from the junction.
2. Kirchhoffs second law (Loop rule or KVL) : The algebraic sum of potential around any closed loop is zero.
   ∴ Σ(IR) + ΣE = 0

Wheatstone bridge : Wheatstone’s bridge circuit consists of four resistances R₁, R₂, R₃ and R₄ are connected to form a closed path. A cell of emf is connected between the point A and C and a galvanometer is connected between the points B and D as shown in fig. The current through the various branches are indicated in the figure. The current through the galvanometer is I_g and the resistance of the galvanometer is G.

Applying Kirchhoffs first law at the junction D, I₁ – I₃ – I_g = 0 ………………. (1)
at the junction B, I₂ + I_g – I₄ = 0 ……………. (2)
Applying Kirchhoff s second law to the closed path ADBA,
-I₁R₁ – I_gG + I₂R₂ = 0
or
⇒ I₁R₁ + I_gG = I₂R₂ …………….. (3)
applying kirchhoft’s second law to the closed path DCBD,
-I₃R₃ + I₄R₄ + I_gG = 0
⇒ I₃R₃ – I_gG = I₄R₄ ……………….. (4)

When the galvanometer shows zero deflection the points D and B are at the same potential. So I_g = 0.
Substituting this value in (1), (2), (3) and (4).
I₁ = I₃ ………………. (5)
I₂ = I₄ ……………… (6)
I₁R₁ = I₂R₂ ……………….. (7)
I₃R₃ = I₄R₄ …………….. (8)
Dividing (7) and (8)
\(\frac{I_1 R_1}{I_3 R_3}=\frac{I_2 R_2}{I_4 R_4} \Rightarrow \frac{R_1}{R_3}=\frac{R_2}{R_4}\) [∵ I₁ = I₃ & I₂ = I₄]
∴ Wheatstone’s Bridge principle : R₄ = R₃ × \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 2.
State the working principle of potentiometer explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer. [A.P. Mar. 16]
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant,
i.e. ε ∝ l ⇒ ε = Φl where Φ is potential gradient.

Comparing the emf of two cells ε₁ and ε₂ :

1. To compare the emf of two cells of emf E₁ and E₂ with potentiometer is shown in diagram.
   
2. The points marked 1, 2, 3 form a two way key.
3. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to ε₁.
4. The Jockey is moved along the wire till at a point N₁ at a distance l₁ from A, there is no deflection in the galvanometer. Then ε₁ ∝ l₁ ⇒ ε₁ = Φl₁ ………………. (1)
5. Similarly, if another emf ε₂ is balanced against
   l₂ (AN₂), then ε₂ ∝ l₂ ⇒ ε₂ = Φl₂ ……………. (2)
6. \(\frac{(1)}{(2)} \Rightarrow \frac{\varepsilon_1}{\varepsilon_2}=\frac{l_1}{l_2}\)

Question 3.
State the working principle of potentiometer. Explain with the help of circuit diagram how the potentiometer is used to determine the internal resistance of the given primary cell. [A.P. & T.S. Mar. 17, 15]
Answer:
Working principle of potentiometer : The potential difference across a length of the potentiometer wire is directly proportional to its length (or) when a steady current is passed through a uniform wire, potential drop per unit length or potential gradient is constant.
i.e. ε ∝ l ⇒ ε = Φl
where Φ is potential gradient.

Measurement of internal resistance (r) with potentiometer :

1. Potentiometer to measure internal resistance (r) of a cell (ε) is shown in diagram.
   
2. The cell (emf ε) whose internal resistance (r) is to be determined is connected across a resistance box (R.B) through a key K₂.
3. With key K₂ open, balance is obtained at length l₁ (AN₁). Then ε = Φl₁ …………… (1)
4. When key K₂ is closed, the cell sends a current (I) through the resitance box (R.B).
5. If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
   Then V = Φ l₂ ………………… (2)
6. \(\frac{(1)}{(2)} \Rightarrow \frac{\varepsilon}{V}=\frac{l_1}{l_2}\) ……………. (3)
7. But ε = I (r + R) and V = IR. This gives
   \(\frac{\varepsilon}{V}=\frac{(r+R)}{R}\)
   \(\frac{l_1}{l_2}=\left(\frac{r}{R}+1\right)\) [∵ from (3)]
   ∴ r = \(\mathrm{R}\left(\frac{l_1}{l_2}-1\right)\)

Question 4.
Under what condition is the heat produced in an electric circuit a)directly proportional b) inversely proportional to the resistance of the circuit ? Compute the ratio of the total quantity of heat produced in the two cases.
Answer:
Expression of heat produced by electric current:
Consider a conductor AB of resistance R.
Let V = P.D applied across the ends of AB.
I = current flowing through AB.
t = time for which the current is flowing.
∴ Total charge flowing from A to B in time t is q = It. By definition of P.D, work done is carrying unit charge from A to B = V
Total work done in carrying a charge q from A to B is
W = V × q = V It = I² Rt (∵V = IR)
This work done is called electric work done. If this electric work done appears as heat, then amount of heat produced (H) is given by
H = W = I² Rt Joule.
This is a statement of Joule’s law of heating.

a) If same current flows through an electric circuit, heat is developed.
i.e., H ∝ R.

b) If same P.D applied across the the electric circuit heat is developed.
i.e., H₂ ∝ \(\frac{1}{R}\)

c) The ratio of H₁ and H₂ is given by
\(\frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\mathrm{R}}{\frac{1}{\mathrm{R}}}\)
∴ \(\frac{\mathrm{H}_1}{\mathrm{H}_2}\) = R²

Problems

Question 1.
A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there is no change in its density on stretching, calculate the resistance of the stretched wire.
Solution:
Given R₁ = 10 Ω,
l₁ = l
l₂ = 3l, R₂ ?

R₁ = \(\frac{\rho}{\mathrm{V}} l_1^2\)
R₂ = \(\frac{\rho}{\mathrm{V}} l_2^2\)
\(\frac{\dot{\mathrm{R}}_2}{\mathrm{R}_1}=\left(\frac{l_2}{l_1}\right)^2 \Rightarrow \frac{\mathrm{R}_2}{10}=\left(\frac{3 l}{l}\right)^2\)
∴ R₂ = 10 × 9 = 90Ω.

Question 2.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter ? [T.S. Mar. 16; Mar. 14]
Solution:
Resistance of long wire = 4R
Hence the resistance of half wire = \(\frac{4 R}{2}\) = 2R

Now these two wire are connected in parallel. Hence the effective resistance between the ends of the diameter
R_P = \(\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}\) ⇒ R_P = \(\frac{2 R \times 2 R}{2 R+2 R}\)
∴ R_P = R.

Question 3.
Three resistors 3Ω, 6Ω and 9Ω are connected to a battery. In which of them will the power of dissipation be maximum if:
a) They all are connected in parallel
b) They all are connected in series ? Give reasons.
Solution:
Given R₁ = 3Ω, R₂ = 6Ω, R₃ = 9Ω
a) Effective resistance in parallel is given by

\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\)
\(\frac{1}{R_P}=\frac{6+3+2}{18}\)
∴ R_P = \(\frac{18}{11}\) Ω
∴ Dissipated power in parallel,
P_P ∝ \(\frac{1}{R_P}\) ⇒ P_P ∝ \(\frac{1}{\left(\frac{18}{11}\right)}\)
∴ P_P ∝ \(\frac{11}{18}\) …………….. (1)

b) Effective resistance in series is given by
R_s = R₁ + R₂ + R₃ = 3 + 6 + 9 = 18 Ω
∴ Dissipated power in series, P_S ∝ R_S ⇒ P_S ∝ 18
From equations (1) and (2) power dissipation is maximum in series and minimum in parallel.
Reasons:

1. In series connection, P ∝ R and V ∝ R. Hence dissipated power (P) and potential difference (V) is more because current is same across each resistor.
2. In parallel connection, P ∝ \(\frac{1}{R}\) and I ∝ \(\frac{1}{R}\) Hence dissipated power (P) and potential difference (V) is less because voltage is same across each resistor.

Question 4.
A silver wire has a resistance of 2.1Ω at 27.5°C and a resistance of 2.7Ω at 100°C. Determine the temperature coeff. of resistivity of silver.
Solution:
For silver wire, R₁ = 2.1Ω, t₁ = 27.5°C
R₂ = 2.7 Ω, t₂ = 100°C, α = ?
α = \(\frac{R_2-R_1}{R_1 \mathrm{t}_2-\mathrm{R}_2 \mathrm{t}_1}=\frac{2.7-2.1}{2.1 \times 100-2.7 \times 27.5}=\frac{0.6}{210-74.25}=\frac{0.6}{135.75}\)
∴Temperature coefficient of resistivity
α = 0.443 × 10^-2/°C

Question 5.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm. what is the emf of the second cell ? [A.P. Mar. 15
Solution:
Here ε₁ = 1.25 V, l₁ = 35.0 cm, ε₂ = ?. l₂ = 63.0 cm.
As \(\frac{\varepsilon_2}{\varepsilon_1}=\frac{l_2}{l_1}\) or
ε₂ = \(\frac{\varepsilon_1 \times l_2}{l_1}=\frac{1.25 \times 63}{35}\) = 2.25 V

Question 6.
A battery of emf 2.5 V and internal resistance r is connected in series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA Draw the circuit diagram and calculate the value of r. [T.S. Mar. 17]
Solution:
Circuit diagram for the given data is shown below.

Given, E = 2.5 V; R = 45Ω;
r_A = 1A; I = 50mA;
r = ?
E = I (R + r_A + r)
2.5 = 50 × 10^-3 (45 + 1 + r)
46 + r = \(\frac{2.5}{50 \times 10^{-3}}=\frac{2.5 \times 10^3}{50}\) = 50
∴ r = 50 – 46 = 4Ω.

Question 7.
The balancing point in meter bridge experiment is obtained at 30 cm from the left. If the right gap contains 3.5 £2, what is the resistance in the left gap? [BMP]
Solution:
\(\frac{x}{R}=\frac{I_1}{100-I_1} \Rightarrow \frac{x}{3.5}=\frac{30}{70} \Rightarrow x=\frac{30 \times 3.5}{70}=1.5\)
∴ x = 1.5 Ω
The resistance in the left gap in meter bridge is, x = 1.5 Ω

Question 8.
The storage battery of a car has an emf of 12 V If the internal resistance of the battery is 0.4Q, what is the maximum current that can be drawn from the battery ?
Solution:
Here E = 12 V, r = 0.4Ω
Maximum Current, I_max = \(\frac{E}{r}=\frac{12}{0.4}\) = 30 A

Question 9.
A battery of emf 10V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ? [T.S. Mar. 15]
Solution:
Here E = 10 V, r = 3Ω,1 = 0.5 A, R = ?, V = ?
I = \(\frac{E}{(R+r)}\) or (R + r) = \(\frac{\mathrm{E}}{\mathrm{I}}=\frac{10}{0.5}\)= 20 or
R = 20 – 3 = 17Ω
Terminal voltage V = IR = 0.5 × 17 = 8.5 Ω.

Question 10.
If the balancing point in a meter bridge from the left is 60 cm, compare the resistance in the left and right gaps of meter bridge.
Solution:
\(\frac{R_1}{R_2}=\frac{I_1}{100-I_1}=\frac{60}{100-60}=\frac{60}{40}=\frac{3}{2} \Rightarrow \frac{R_1}{R_2}=\frac{3}{2}\)

Question 11.
A potentiometer wire is 5m long and a potential difference of 6 V is maintained between its ends. Find the emf of a cell which balances against a length of 180cm of the potentiometer wire. [A.P. Mar. 17, 16]
Solution:
Length of potentiometer wire L = 5m
Potential difference V = 6 Volt
Potential gradient Φ = \(\frac{V}{L}=\frac{6}{5}\) = 1.2V/m
Balancing length l = 180cm
= 1.80m
Emf of the cell E = Φl
= 1.2 × 1.8 = 2.16V.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 5th Lesson Electrostatic Potential and Capacitance

Very Short Answer Questions

Question 1.
Can there be electric potential at a point with zero electric intensity ? Give an example.
Answer:
Yes, There can exist potential at a point where the electric intensity is zero.

Ex :

1. Between two similar charges intensity of electric field is zero. But potential is not zero.
2. Inside the charged spherical conductor electric field intensity is zero but potential is not zero.

Question 2.
Can there be electric intensity at a point with zero electric potential ? Give an example.
Answer:
Yes, electic intensity need not be zero at a point where the potential is zero.

Ex :
1) At mid point between two equal opposite charges potential is zero. But intensity is not zero.

Question 3.
What are meant by equipotential surfaces ?
Answer:
Surface at every point of which the value of potential is the same is defined as equipotential surface
For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces.

Question 4.
Why is the electric field always at right angles to the equipotential surface ? Explain.
Answer:
No work is done in moving a charge from one point on equipotential surface to the other. Therefore, component of electric field intensity along the equipotential surface is zero. Hence, the surface is perpendicular to the field lines.

Question 5.
Three capacitors of capacitances 1μF, 2μF and 3μF, are connected in parallel
(a) What is the ratio of charges ?
(b) What is the ratio of potential difference ?
Answer:
When capacitors are connnected in parallel
(a) q₁ : q₂ : q₃ = V: C₂ V: C₃ V = 1μF : 2μF : 3μF

∴ q₁ : q₂ ; q₃ = 1 : 2 : 3
(b) V₁ : V₂ : V₃ = V : V : V = 1 : 1 : 1

Question 6.
Three capacitors of capacitances 1μE, 2μF and 3μF are connected in series
(a) What is the ratio of charges ?
(b) What is the ratio of potential differences ?
Answer:
When capacitors are connected in series

(a) q₁ : q₂ : q₃ = q : q : q = 1 : 1 ; 1
(b) V₁ : V₂ : V₃ = \(\frac{\mathrm{q}}{\mathrm{C}_1}: \frac{\mathrm{q}}{\mathrm{C}_2}: \frac{\mathrm{q}}{\mathrm{C}_3}\) = \(\frac{1}{1}: \frac{1}{2}: \frac{1}{3}\)
∴ V₁ : V₂ : V₃ = 6 : 3 : 2

Question 7.
What happens to the capacitance of a parallel plate capacitor if the area of its plates is doubled ?
Answer:
\(\frac{C_2}{C_1}\) = \(\frac{A_2}{A_1}\) [∵ C₂ = 2C₁]
Given A₂ = 2A₁ ; \(\frac{C_2}{C_1}\) = \(\frac{2 \mathrm{~A}_1}{\mathrm{~A}_1}\) [∴ C₂ = 2C₁]
Therefore capacity increases by twice.

Question 8.
The dielectric strength of air is 3 × 10⁶.Vm^-1 at certain pressure, A parallel plate capacitor with air in between the plates has a plate separation of 1 cm. Can you charge the capacitor
to 3 × 10⁶V?
Answer:
Dielectric strength of air E₀ = 3 × 10⁶ Vm^-1
Electric field intensity between the plates, E = \(\frac{E_0}{K}\) = 3 × 10⁶ Vm^-1 [∵ for air K = 1]
Distance between two plates, d = 1 cm = 102m
Electric potential difference between plates, V = Ed = 3 × 10⁶ × 10^-2
∴ V = 3 × 10⁴ Volt.
Hence we cant charge the capacitor upto 3 × 10⁶ Volt.

Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge. (T.S. Mar. ’16)
Answer:
Expression for the electric potential due to a point charge:

1. Electric potential at a point is defined as the amount of workdone in moving a unit +ve charge from infinity to that point.
   
2. Consider a point P at a distance r from the point charge having charge + q. The electric field at P = E = \(\frac{q}{4 \pi \varepsilon_0 x^2}\)
3. Workdone in taking a unit +ve charge from B to A = dV = -E.dx (-ve sign shows that the workdone is +ve in the direction B to A, Whereas the potential difference is +ve in, the direction A to B.
4. Therefore, potential at P = The amount of workdone in taking a unit +ve charge from infinity to P
   

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Expression for the electrostatic potential energy of a system of two point charges:

1. Let two point charges q₁ and q₂ are separated by distance ‘r’ in space.
2. An electric field will develop around the charge q₁
   
3. To bring a charge q2 from infinity to the point B some work must be done.
   workdone = q₂ V_B
   But V_B = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1}{\mathrm{r}}\)
   W = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
4. This amount workdone is stored as electrostatic potential energy (U) of a system of two charged particles. Its unit is joule.
   ∴ U = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}\)
5. If the two charges are similar then ‘U’ is positive. This is in accordance with the fact that two similar charges repel one another and positive work has to be done on the system to bring the charges nearer.
6. Conversely if the two charges are of opposite sign, they attract one another and potential energy is negative.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Expression for potential energy of an electric dipole placed in a uniform electric field:

1. Consider a electric dipole of length 2a having charges + q and -q.
2. The electric dipole is placed in uniform electric field E and its axis makes an angle θ with E.
   
3. Force on charges are equal but opposite sign. They constitute torque on the dipole.
   Torque \(\tau\) = one of its force (F) × ⊥^r distance (BC)
   F = qE and sinθ = \(\frac{\mathrm{BC}}{2 \mathrm{a}}\) ⇒ BC = 2a sinθ
   ∴ Torque \(\tau\) = qE × 2a sinθ = PE sin θ [∴ p = 2aq]
4. Suppose the dipole is rotated through an angle dθ, the workdone dw is given by
   dw = tdθ = PE sinθ dθ
5. For rotating the dipole from angle θ₁ to θ₂,
   workdone W = \(\int_{\theta_1}^{\theta_2} \mathrm{PE} \sin \theta d \theta\) = PE(cosθ₁ – cosθ₂)
6. This workdone (W) is then stored as potential energy(U) in the dipole.
   ∴ U = PE(cosθ₁ – cosθ₂)
7. If θ₁ = 90°and θ₂ = 0°, U = -PE cosθ.
   In vector form U = \(-\vec{P} \cdot \vec{E}\)

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor. (Mar.’16 (AP) Mar ’14)
Answer:
Expression for the capacitance of a parallel plate capacitor:

1. P and Q are two parallel plates of a capacitor separated by a distance of d.
2. The area of each plate is A. The plate P is charged and Q is earth connected.
3. The charge on P is + q and surface charge density of charge = σ
   ∴ q = Aσ
   
4. The electric intensity at poiñt x, E = \(\frac{|\sigma|}{\varepsilon_0}\)
5. Potential difference between the plates P and Q,
   
6. Capacitance of the capacitor Farads (In air)
   Note : Capacity of a capacitor with dielectric medium is C = \(\frac{\varepsilon_0 \mathrm{~A}}{\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathbf{k}}\right]}\) Farads.

Question 5.
Explain the behaviour of dielectrics in an external field. (A.P. Mar. ’19)
Answer:

1) When an external field is applied across dielectrics, the centre of positive charge distribution shifts in the direction of electric field and that of the negative charge distribution shifts

opposite to the electric field and induce a net electric field within the medium opposite to the external field. In such situation the molecules are said to be polarised.

2) Now consider a capacitor with a dielectric between the plates. The net field in the dielectric becomes less.

3) If E₀ the external field strength and E₁ is the electric field strength induced, then the net field \(\overrightarrow{\mathrm{E}}_{\text {net }}\) = \(\overrightarrow{\mathrm{E}}_0\) + \(\overrightarrow{\mathrm{E}}_1\)
(E_net) = E₀ – E_i = \(\frac{E}{K}\) where K is the dielectric constant of the medium.

Long Answer Questions

Question 1.
Define electric potential. Derive and expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) the axial line of electric dipole (b) on the equatorial line of electric dipole.
Answer:
Electric potential (V) : The workdone by a unit positive charge from infinite to a point in an electric field is called electric potential.
Expression for the potential at a point due to a dipole:

1. Consider A and B having -q and + q charges separated by a distance 2a.
2. The electric dipole moment P = q × 2a along AB
3. The electric potential at the point ‘P’ is to be calculated.
4. P is at a distance ‘r’ from the point ‘O’. θ is the angle between the line OP and AB.
5. BN and AM are perpendicular to OP.
6. Potential at P due to charge + q at B,
   
7. Potential at P due to charge -q at A, V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{AP}}\right]\)
   ∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\mathrm{q}}{\mathrm{MP}}\right]\) [∵ BP = NP]
8. Therefore, Resultant potential at P is V = V₁ + V₂
   V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{N P}-\frac{q}{M P}\right]\) …… (1)
9. In Δle ONB, ON = OB cosθ = a cosθ; ∴ NP = OP – ON = r – a cosθ ….. (2)
10. In Δle AMO, OM = AO cosθ = a cosθ; ∴ MP = MO + OP = r + a cos θ ….. (3)
11. Substituting (2) and (3) in (1), we get
    
12. As r > >a, a² cos²θ can be neglected with comparision of r².
    ∴ V = \(\frac{\mathrm{P} \cos \theta}{4 \pi \varepsilon_0 \mathbf{r}^2}\)
13. Electric potential on the axial line of dipole :

(i) When θ = 0°, point p lies on the side of + q
∴ V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 0° = 1]

ii) When θ = 180°, point p lies on the side of -q.
∴ V = \(\frac{-\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) [∵ cos 180° = -1]

b) Electric potential on the equitorial line of the diopole:
when θ = 90°, point P lies on the equitorial line.
∴ V = o [∵ cos 90° = 0]

Question 2.
Explain series and parallel combination of capacitors. Derive the formula for equivalent capacitance in each combination. (AP. & T.S. Mar. ‘15)
Answer:
Series combination : If a number of condensers are connected end to end between the fixed points then such combination is called series.

In this combination

1. Charge on each capacitor is equal.
2. PD’s across the capacitors is not equal.

Consider three capacitors of capacitanceš C₁, C₂ and C₃ are connected in series across a battery of P.D ‘V’ as shown in figure.

Let ‘Q’ be the charge on each capacitor.
Let V₁, V₂ and V₃ be the P.D’s of three
V = V₁ + V₂ + V₃ —— (1)
P.D. across I^st condenser V₁ = \(\frac{\mathrm{Q}}{\mathrm{C}_1}\)
P.D. across II^nd condenser V₂ = \(\frac{\mathrm{Q}}{\mathrm{C}_2}\)
RD across III^rd condenser V₃ = \(\frac{\mathrm{Q}}{\mathrm{C}_3}\)
∴ From the equation (1), V = V₁ + V₂ + V₃
= \(\frac{\mathrm{Q}}{\mathrm{C}_1}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_2}\) + \(\frac{\mathrm{Q}}{\mathrm{C}_3}\) = Q\(\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right]\)

For ‘n’ number of capacitors, the effective capacitance can be written as

Parallel Combination : The first plates of different capacitors are connected at one terminal and all the second plates of the capacitors are connected at another terminal then the two terminals are connected to the two terminals of battery is called parallel combination.

In this combination,

1. The PD’s between each capacitor is equal (or) same.
2. Charge on each capacitor is not equal. Consider three capacitors of capacitancé C₁, C₂ and C₃ are connected in parallel across a RD ‘V’ as shown in fig.

The charge on Ist capacitor Q₁ = C₁ V
The charge on IInd capacitor Q₂ = C₂ V
The charge on IIIrd capacitor Q₂ = C₃ V
∴ The total charge Q = Q₁ + Q₂ + Q₃
= C₁ V + C₂ V + C₃ V
Q = V(C₁ + C₂ + C₃) ⇒ \(\frac{Q}{V}\) = C₁ + C₂ + C₃
C = C₁ + C₂ + C₃ [ ∵ C = \(\frac{\mathrm{Q}}{\mathrm{V}}\)]
for ‘n’ number of capacitors connected in parallel, the equivalent capacitance can be written as C = C₁ + C₂ + C₃ + …. + C_n

Question 3.
Derive an expression for the energy stored in a capacitor. What is the energy stored when the space between the plates is filled with a dielectric
(a) with charging battery disconnected?
(b) with charging battery connected in the circuit?
Answer:
Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance ‘c’ and its initial will be zero. Now it is connected across a battery for charging then the final potential difference across the capacitor be V and final charge on the capacitor be ‘Q’
∴ Average potential difference V_A = \(\frac{\mathrm{O}+\mathrm{V}}{2}\) = \(\frac{\mathrm{V}}{2}\)
Hence workdone to move the charge Q = W = V_A × Q = \(\frac{\mathrm{VQ}}{2}\)
This is stored as electrostatic potential energy U’
∴ U = \(\frac{\mathrm{QV}}{2}\)
We know Q = CV then ‘U’ can be written as given below.
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)
∴ Energy stored in a capacitor
U = \(\frac{\mathrm{QV}}{2}\) = \(\frac{1}{2}\) CV² = \(\frac{1}{2} \frac{\mathrm{Q}^2}{\mathrm{C}}\)

Effect of Dielectric on energy stored :

Case (a) : When the charging battery is disconnected from the circuit:
Let the capacitor is charged by a battery and the disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant ‘K’ then potential decreases by \(\frac{1}{K}\) times and charge remains constant.
Capacity increases by ‘K times

∴ Energy stored decreases by \(\frac{1}{\mathrm{~K}}\) times.

Case (b): When the charging battery is connected in the circuit:
Let the charging battery is continue the supply of charge. When the dielectric is introduced then potential decreases by \(\frac{1}{\mathrm{~K}}\) times and charge on the plates increases until the potential difference attains the original value = V ,
New charge on the plates Q’ = KQ
Hence new capacity C’ = \(\frac{Q^{\prime}}{V}=\frac{K Q}{V}\) = KC
Energy stored in the capacitor U’ = \(\frac{1}{2}\)Q’V = \(\frac{1}{2}\)(KQ) V = KU
U’ = KU
∴ Energy stored in the capacitor increases by ‘K times.

Problems

Question 1.
An elementary particle of mass ‘m’ and charge +e initially at a very large distance is projected with velocity ‘v’ at a much more massive particle of charge + Ze at rest. The closest possible distance of approach of the incident particle is
Solution:
For an elementary particle, mass = m; charge = +e; velocity = v.
For much more massive particle, charge = + Ze
From law of conservation of energy, we have
K.E of elementary particles = Electrostatic potential energy of elementary particle at a closest distance (d)

∴ The closest possible distance of approach of the incident particle,
d = \(\frac{Z e^2}{2 \pi \varepsilon_0 m v^2}\)

Question 2.
In a hydrogen atom the electron and proton are at a distance of 0.5 A. The dipole moment of the system is
Answer:
In a hydrogen atom the charge of an electron = -1.6 × 10^-19C

In a hydrogen atom the charge of proton,
q_p = +1.6 × 10^-19C
The distance between the proton and an electron
2a = 0.5A = 0.5 × 10^-10 m
The dipole moment of the system,
P = 2a × q_p = 0.5 × 10^-10 × 1.6 × 10
∴ P = 8 × 10^-30cm

Question 3.
There is a uniform electric field in the XOY plane represented by (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1. If the electric potential at the origin is 200 V, the electric potential at the point with coordinates (2m, 1m) is
Answer:
Given, uniform Electric field intensity,
\(\overrightarrow{\mathrm{E}}\) = (40\(\hat{i}\) + 30\(\hat{j}\)) Vm^-1
Electric potential at the origin = 200V
Position vector d\(\begin{aligned}
&\rightarrow \\
&\mathrm{r}
\end{aligned}\) = (2\(\hat{i}\) + 1\(\hat{j}\)) m
We know that,

V_p – V_o = -(80 + 30) = -110Volt
V_p = V_o – 110 = (200 – 110) Volt = 90 Volt
∴ potential at point P, V_p = 90Volt.

Question 4.
An equilateral triangle has a side length L. A charge +q is kept at the centroid of the triangle. P is a point on the perimeter of the triangle. The ratio of the minimum and maximum possible electric potentials for the point P is
Answer:
Charge at the centroid of an equilateral triangle = +q
The charge + q divides the line segment in ratió 2 : 1.
That means r_max = 2 and r_min = 1
V_min = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_{\max }}\) and V_max = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_{\min }}\)

Question 5.
ABC is an equilateral triangle of side 2m. There is a uniform electric field of intensity 100V/m in the plane of the triangle and parallel to BC as shown. If the electric potential at A is 200 V, then the electric potentials at B and C.
Answer:
Given length of side of an equilateral triangle a = 2m
E = 100V/m; V_A = 200V
Let D be the mid point between B and C Potential at D = V_D = 200V

From fig V_B – V_D = Ed
⇒ V_B – 200 = 100 × 1
∴ Potential at B, V_B = 300 V And V_D – V_C = Ed
200 – V_C = 100 × 1
∴ Potential at C, V_C = 100V.

Question 6.
An electric dipole of moment p is placed in a uniform electric field E, with p parallel to E. It is then rotated by an angle q. The work done is
Solution:
Let AB be a electric dipole having charges -q and + q
Electric dipole moment of AB = p
Electric field = E The workdone by a dipole, when it is rotated through an angle q from E,

W = \(\int_0^q p E \sin \theta d \theta\)
⇒ W = pE \([\cos \theta]_0^q\) = pE (cos0° – cosq)
∴ W = pE(1 – cosq)

Question 7.
Three identical metal plates each of area ‘A’ are arranged parallel to each other, ’d’ is the distance between the plates as shown. A battery of V volts is connected; as shown. The charge stored in the system of plates is

Answer:
Area of each plate = A
distance between two plates = d capacity of each, parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Two capacitors are connected in parallel as shown in figure
Equivalent capacity of two capacitors connected in parallel, C_p = 2C = \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Charge stored in the system of plates,
q = C_pV = \(\frac{2 \varepsilon_0 A}{d} V\)
∴ q = \(\frac{2 A \varepsilon_0 \mathrm{~V}}{\mathrm{~d}}\)

Question 8.
Four identical metal plates each of area A are separated mutually by a distance d and are connected as shown. Find the capacity of the system between the terminals A and B.

Solution:
Area of each plate of a capacitor = A
Distance between two parallel plate capacitors = d
Capacity of each parallel plate capacitor,
C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
Given fig is

The equivalent circuit of the above fig is

Question 9.
In the circuit shown the battery of ‘V’ volts has no internal resistance. All three condensers are equal in capacity. Find the condenser that carries more charge ?

Answer:
The equivalent circuit to the given circuit is as shown

In series combination of capacitors, charge q flows through each capacitor. Then q₁ = q = C₁V₁; q₂ – q = C₂V₂; q₃ = q = C₃V₃
∴ q₁ = q₂ = q₃
Hence three capacitors C₁, C₂ and C₃ carry the same charge.

Question 10.
Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over, the battery is removed. Now a dielectric slab of K = 2 is inserted between the plates of A so as to fill the completely. The energy lost by the system during the sharing of charges is
Solution:
i) With battery of parallel combination:
C₁ = C;C₂ = 2C; V = V
C_p = C₁ + C₂ = 3C; q = 3Cv

ii) Without battery of parallel combination : .

Question 11.
A condenser of certain capacity is charged to a potential V and stores some energy. A second condenser of twice the capacity is to store half the energy of the first, find to what potential one must be charged?
Answer:
For first capacitor, C₁ = C; V₁ = V
And U₁ = \(\frac{1}{2}\) C₁V\(V_1^2\) = \(\frac{1}{2}\) CV² …….. (1)
For second capacitor, C₂ = 2C₁ = 2C;
U₂ = \(\frac{\mathrm{U}_1}{2}\) = \(\frac{1}{4} \mathrm{CV}^2\); Let potential difference across the capacitor = V₂

Textual Exercises

Question 1.
Two charges 5 × 10^-8 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.
Solution:
Here q1 = 5 × 10^-8C, q₂ = -3 × 10^-8C
Let the potential be zero at a distance x cm from the charge q₁ = 5 × 10^-8C.
∴ r₁ = x × 10^-2m
r₂ = (16 – x) × 10^-2m
Now V = \(\frac{\mathrm{q}_1}{4 \pi \varepsilon_0 \mathrm{r}_1}\) + \(\frac{\mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_2}\)
= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]\)
∴ \(\frac{\mathrm{q}_1}{\mathrm{r}_1}\) = \(\frac{-\mathrm{q}_2}{\mathrm{r}_2}\)
= \(\frac{5 \times 10^{-8}}{x \times 10^{-2}}\) = \(\frac{-\left(-3 \times 10^{-8}\right)}{(16-x) 10^{-2}}\) or \(\frac{5}{x}\) = \(\frac{3}{16-x}\)
3x = 80 – 5x
8x = 80, x = 10cm

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer:
In fig. O is centre of hexagon ABCDEFA of each side 10 cm, As it clear from the figure, OAB, OBC etc. are equilateral triangles.
Therefore
OA = OB = OC = OD = OE = OF = r = 10 cm = 10^-1m
As potential is scalar, there for C potential at O is

Question 3.
Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.
a) Identify an equipotential surface of the system.
b) What is the direction of the electric field at every point on this surface ?
Solution:
a) The plane normal to AB and passing through its middle point has zero potential everywhere.
b) Normal to the plane in the direction AB.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field.
a) inside the sphere
b) just outside the sphere
c) At a point 18 cm from the centre of the sphere ?
Solution:
a) Here r = 12 cm = 12 × 10^-2m,
q = 1.6 × 10^-7C. Inside the sphere, E = 0

b) Just coincide the sphere (say on the surface of the sphere)
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
= 9 × 10⁹ × \(\frac{1.6 \times 10^{-7}}{\left(12 \times 10^{-2}\right)^2}\) = 10⁵ N/c

c) At r = irm = 18 × 10^-2
E = \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}^2}\) = \(\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2}\)
= 4.4 × 10⁴ N/C

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
Solution:
C₁ = \(\frac{\varepsilon_0 A}{d}\) = 8pF
C₂ = k\(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}\) = \(\frac{6 \times 2 \varepsilon_0 A}{d}\) = 12 × 8 = 96pF.

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
a) What is the total capacitance of the combination ?
b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Solution:
a) \(\frac{1}{C_S}\) = \(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) = \(\frac{3}{9}\) = \(\frac{1}{3}\) ; C_s = 3pF
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

b) P.d across each capacitor =
\(\frac{\mathrm{V}}{3}\) = \(\frac{\mathrm{120}}{3}\) = 40V

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
a) What is the total capacitance of the combination ?
b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Solution:
a) C_p = 2 + 3 + 4 = 9pF
b) For each capacitor, V is same = 100 Volt
q₁ = C₁V₁ = 2 × 100 = 200pC
q₂ = C₂V = 3 × 100 = 300pC
q₃ = C₃V = 4 × 100 = 400pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Solution:
Here A = 6 × 10^-3m², d = 3mm = 3 × 10^-3m, C = ?
V = 100V, q = ?
C₀ = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{3 \times 10^{-3}}\)
= 1.77 × 10^-11F
q = C₀V= 1.77 × 10^-11 × 100
= 1.77 × 10^-9C

Question 9.
Explain what would happen if in the capacitor given in Exercise 8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.
a) While the voltage supply remained connected.
b) after the supply was disconnected.
Solution:
a) Capacity increases to C = KC₀
= 6 × 1.77 × 10^-11F
charge increases to
q¹ = C¹V = 6 × 1.77 × 10^-11 × 10²C

b) After the supply was disconnected new capacity C = KC₀ = 6 × 1.77 × 10^-11F
New voltage V¹ = \(\frac{q}{C^1}\) = \(\frac{1.77 \times 10^{-9}}{6 \times 1.77 \times 10^{-11}}\)
= 16.67V

Question 10.
A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor ?
Solution:
Here C = 12pF = 12 × 10^-12F, V = 50Volt, E = ?
E = \(\frac{1}{2} \mathrm{CV}^2\) = \(\frac{1}{2}\left(12 \times 10^{-12}\right)(50)^2\)
= 1.5 × 10^-8J

Question 11.
A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Solution:
Here C₁ = C₂ = 600pF = 600 × 10^-12
F = 6 × 10^-10F,
V₁ = 200 V, V₂ = o

Additional Exercises

Question 1.
A charge of 8mC is located at the origin. Calculate the workdone in taking a small charge of’ -2 × 10^-9 C from a point P(0, 0, 3 cm) to a point Q(0,4 cm, 0), Via a point R(0,6 cm, 9 cm).
Solution:
From fig. a charge q = 8mc = 8 × 10^-3C is located at the origin O. Charge to be carried is
q₀ = -2 × 10^-9C from P to Q
Where OP = r_p = 3 cm
= 3 × 10^-2m and OQ = r_Q = 4cm = 4 × 10^-2m
As electrostatic forces are conservative forces, workdone is independent of the path. Therefore there is no relevance of point R.

Question 2.
A cube of side b has a charge q at each of its vertices, Determine the potential and electric field due to this charge array at the centre of the cube.
Answer:
We know that the length of diagonal of thè cube of each side b is \(\sqrt{3 b^2}\) = \(\mathrm{b} \sqrt{3}\)
Distance between centre of the cube and each vertex r = \(\frac{b \sqrt{3}}{2}\)
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}\)
and 8 charges each of valué q are present at the eight vertices, of the cube therefore

∴ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{8 q}{b \sqrt{3} / 2}\) or V = \(\frac{4 \mathrm{q}}{\sqrt{3} \pi \varepsilon_0 \mathrm{~b}}\)
Further electric field intensity at the centre due to all the eight charges is zero because the fields due to individual charges cancel in pairs.

Question 3.
Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field
a) at the mid-point of the line joining the two charges and
b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Solution:
Here q₁ = 1.5,C = 1.5 × 10^-6 C,
q₁ = 2.5μC = 2.5 × 10^-6C
Distance between the two spheres = 30cm
from figure

b) Let P be the point in a plane normal to the line passing through the mid point,
where OP 10cm = 0. 1m
From figure,
Now PA = PB

Resultant field intensity at P is
E = \(\sqrt{\mathrm{E}_1^2+\mathrm{E}_2^2+2 \mathrm{E}_1 \mathrm{E}_2 \cos \theta}\)
E = 6.58 × 10⁵ Vm^-1
Let θ be the angle which resultant intensity \(\overrightarrow{\mathrm{E}}\) makes with \(\overrightarrow{\mathrm{E}_1}\).

Question 4.
A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
a) A charge q is placed at the centre of the shell. What ¡s the surface charge density on the inner 4nd outer surfaces of the shell?
Answer:
a) The charge of + Q resides on the Outer surface of the shell. The charge q placed at the centre of the shell induces charge -q on the inner surface and charge + q on the outer surface of the shell, from figure.
∴ Total charge on inner surface of the shell is -q and total charge on the outer surface of the shell is (Q + q)
σ₁ = \(\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}\) and σ₁ = \(\frac{\mathrm{Q}_1+\mathrm{q}}{4 \pi \mathrm{r}_2^2}\)

b) Is the electric field inside a cavity (with no charge) zero, even If the shell is not spherical, but has any irregular shape? Explain.
Solution:
Electric field intensity inside a cavity with no charge is zero, eveñ when the shell has

any irregular shape. If we were to take a closed loop part of which is inside the cavity along a field line and the rest outside it then network done by the field in carrying a rest charge over the closed loop will not be zero. This is impossible for an electrostatic field. Hence electric field intensity inside a cavity with no charge is always zero.

Question 5.
a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to
another given by (E₂ – E₁). \(\hat{\mathbf{n}}\) = \(\frac{\sigma}{\varepsilon_0}\)
Where \(\hat{\mathbf{n}}\) is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of \(\hat{\mathbf{n}}\) is from side 1 to side 2.)
Hence show that just out side a conductor, the electric field is σ\(\hat{\mathrm{n}} / \varepsilon_0\)
b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
(Hint: for (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.)
Answer:
a) Normal component of electric field intensity due to a thin infinite plane sheet of charge on left side.
\(\overrightarrow{\mathrm{E}}_1\) = –\(\frac{\sigma}{2 \varepsilon_0} \hat{\mathbf{n}}\)
and on right side 2 = \(\overrightarrow{\mathrm{E}_2}\) = \(\frac{\sigma}{2 \varepsilon_0} \hat{\mathrm{n}}\)
Discontinuity in the normal component from one side to the other is

b) To show that the tangential component of electostatic field is continous from one side of a charged surface to another, we use the fact that workdone by electrostatic field on a closed loop is zero.

Question 6.
A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Solution:
From figure A is a long charged cylinder of linear charge density λ, lengh l and radius a. A hollow co-axial conducting cylinder B of length L and radius b surrounds A.

The charge q = λL spreads uniformly on the outer surface of λ. It induces – q charge on the cylinder B. which spreads on the inner surface of B. An electric field \(\overrightarrow{\mathrm{E}}\) is produced in the space between the two cylinders which is directed radically outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The Electric flux through the end faces of the cylindrical Gaussian surface is zero as \(\overrightarrow{\mathrm{E}}\) is parallel to them. According to Gauss’s theorem

Question 7.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53Å:
a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation ?
Solution:
a) Here q₁ = -1.6 × 10^-19C; q₂ = + 1.6 × 10^-19C.
r = 0.53λ = 0.53 × 10^-10m
Potential energy = P.E at ∞ -P. E at r
= 0 – \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r}\) = \(\frac{-9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{0.53 \times 10^{-10}}\)
= -43.47 × 10^-19 Joule
= \(\frac{-43.47 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\) = -27.16 eV

b) K.E in the Orbit = \(\frac{1}{2}\) (27.16) eV
Total energy = K.E + P.E
= 13.58 – 27.16 = – 13.58 eV
Work required to free the electron = 13.58eV

c) Potential energy at a seperation of r₁ (= 1.06A) is
= \(\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}_1}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10}}\)
= 21.73 × 10^-19J = 13.58eV
Potential energy of the system, when zero of P.E is taken at r₁ = 1.06A is
= RE at r₁ -P.E at r = 13.58 – 27.16 = -13.58eV.
By shifting the zero of potential energy work required to free the electron is not affected. It continues to be the same, being equal to + 13.58 eV

Question 8.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion \(\mathbf{H}_2^{+}\). In the ground state of an \(\mathbf{H}_2^{+}\), the two protons are separated by roughly 1.5A. and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
Here q₁ = charge on electron (= -1.6 × 10^-19C)
q₂, q₃ = charge on two protons, each = 1.6 × 10^-19C
r₁₂ = distance between
q₁ and q₂ = 1A = 10^-10m
r₂₃ = distance between
q₂ and q₃ = 1.5A = 1.5 × 10^-10m
r₃₁ = distance between
q₃ and q₁ = 1A = 10^-10m.
Taking zero of potential energy at infinity, we have

Question 9.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends’ of a conductor is higher than on its flatter portions.
Solution:
The charge flows from the sphere at higher potential to the other at lower potential till their potentials become equal. After sharing the charges on two spheres would be.
\(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{C_1 V}{C_2 V}\) where C₁. C₂ are the capacities of two spheres.
But \(\frac{\mathrm{C}_1}{\mathrm{C}_2}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\) ∴ \(\frac{\mathrm{Q}_1}{\mathrm{Q}_2}\) = \(\frac{a}{b}\)
Ratio of surface density of charge on the two spheres

Hence ratio of electric fields at the surface of two spheres
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{\sigma_1}{\sigma_2}\) = \(\frac{\mathrm{b}}{\mathrm{a}}\)
A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius. Therefore, charge density on sharp and pointed ends of conductor is much higher than on its flatter portions.

Question 10.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> I.
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Here -q is at (0, 0, -a) and +q is at (0, 0, a)
a) Potential at (0, 0, z) would be

Potential at (x, y, 0) i.e at a point 1 to z-axis where charges are located is zero

b) we have proved that
V = \(\frac{P \cos \theta}{4 \pi \varepsilon_0\left(r^2+a^2 \cos ^2 \theta\right)}\)
If \(\frac{r}{a}\) >> 1 then a << r ∴ V = \(\frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}\)
∴ V = \(\frac{1}{\mathrm{r}^2}\)
i.e potential is inversely proportional to square of the distance

c) Potential at (5, 0, 0) is

As work done = charge (V₂ – V₁)
W = zero

As work done by electrostatic field is independent of the path connecting the two points therefore work done will
continue to be zero along every path.

Question 11.
Figure shows a charge array known as an electric quadrupole, For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1. and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer:
As is clear from figure an electric quadrupole may be regarded as a system of three charges +q, -2q and + q at A, B and C respectively.
Let AC = 2a we have to calculate electric potential at any point P where BP = r, using superposition principle. Potential at p is given by

∴ \(\frac{\mathrm{a}^2}{\mathrm{r}^2}\) is negligibly small V = \(\frac{\mathrm{q} \cdot 2 \mathrm{a}^2}{4 \pi \varepsilon_0 \mathrm{r}^3}\)
clearly V ∝ \(\frac{1}{\mathrm{r}}\)
In case of an electric dipole V ∝ \(\frac{1}{\mathrm{r}^2}\) and in case of an electric monopole
(i.e a single charge), V ∝ \(\frac{1}{\mathrm{r}}\)

Question 12.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Here total capacitance, C = 2μF
Potential difference v = 1KV = 1000 volt
Capacity of each capacitor C₁ = 1μF
Maximum potential difference across each V = 400 volt

Let n capacitors of 1μF each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row
= 1000 Volt
∴ Potential difference across each capacitor = \(\frac{1000}{\mathrm{n}}\) = 400
∴ n = \(\frac{1000}{400}\) = 2.5
As n has to be a whole number (not less than 2.5) therefore n = 3
capacitance of each row of 3 condensors of 1μF
Each is series = 1/3
Total capacitance of m such rows in parallel = \(\frac{\mathrm{m}}{3}\)
∴ \(\frac{\mathrm{m}}{3}\) = 2(μf) or m = 6
∴ Total number of capacitors =
n × m = 3 × 6 = 18.
Hence 1μF capacitors should be connected in six parallel rows, each row containing three capacitors in series.

Question 13.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Answer:
Here A = ? C = 2F,
d = 0.5 cm = 5 × 10^-3m
As C = \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\)
∴ A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
= 1.13 × 10⁹m²
Which is too large.
That is why ordinary capacitors are in the range of μF or less. However in electrolytic capacitors d is too small. Therefore their capacitance is much larger (=0.1F)

Question 14.
Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer:
Here C₂ and C₃ are in series
A = \(\frac{c d}{\varepsilon_0}=\frac{2 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}\)
∴ \(\frac{1}{C_1}\) = \(\frac{1}{200}\) + \(\frac{1}{200}\) = \(\frac{2}{200}\) = \(\frac{1}{100}\)
C_S = 100pF
Now C_S and C₁ are in parallel
∴ C_p = C_s + C₁ = 100 + 100 = 200pF
Again C_p and C₄ are in series
∴ \(\frac{1}{\mathrm{C}_{\mathrm{s}}}\) = \(\frac{1}{\mathrm{C}_{\mathrm{p}}}\) + \(\frac{1}{\mathrm{C}_4}\) = \(\frac{1}{200}\) + \(\frac{1}{100}\) = \(\frac{3}{200}\)
∴ C = \(\frac{200}{3}\)pF = 66.7 × 10^-12F
As C_p and C₄ are in series
∴ V_p + V₄ = 300
Charge on C₄ is q₄
= CV = \(\frac{200}{3}\) × 10^-12 × 300 = 2 × 10^-8C
Potential difference across
C₄ is V₄ = \(\frac{\mathrm{q}_4}{\mathrm{C}_4}\) = \(\frac{2 \times 10^{-8}}{100 \times 10^{-12}}\) = 200V
from (i) V_p = 300 – V₄ = 300 – 200 = 100
Potential difference across
C₁ is V₁ = V_p = 100V
Charge on C₁ = q₁ = C₁V₁
= 100 × 10^-12 × 100 = 10^-8C.
Potential diff across
C₂ and C₃ in series = 100V
charge on C₂;
q₂ = C₂ V₂ = 200 × 10^-12 × 50 = 10^-8C
Charge on C₃;
q₃ = C₃V₃ = 200 × 10^-12 × 50 = 10^-8C

Question 15.
The plates of parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u, Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:
a) Here A = 90 cm² = 90 × 1o^-4m²
= 9 × 10^-3m²
d = 2.5mm = 2.5 × 10^-3m

Question 16.
A 4µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Here C₁ = 4µF = 4 × 10^-6F, V₁ = 200volt. Initial elctrostatic energy stored in C₁ is E₁
= \(\frac{1}{2} C_1 V_1^2\) = \(\frac{1}{2}\) × 4 × 10^-6 × 200 × 200
E₁ = 8 × 10^-2 Joule
When 4µF capacitor is connected to uncharged capacitor of 2µF charge flows and both acquire a common potential.

∴ final electrostatic energy of both capacitors
E₂ = \(\frac{1}{2}\)(C₁ + C₂)V²
= \(\frac{1}{2}\) × 6 × 10^-6 × \(\frac{800}{6}\) × \(\frac{800}{6}\)
E₂ = 5.33 × 10^-2Joule.
Energy dissipated in the form of heat and electro magnetic radiation.
E₁ – E₂ = 8 × 10^-2 – 5.33 × 10^-2
= 2.67 × 10^-2 Joule.

Question 17.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Answer:
If F is the force on each plates of parallel plate capacitor, then work done in increasing the seperation between the plates by Δx = fΔx

This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is = A Δx
If U = energy density = energy stored/ volume then the increase in potential energy = U.AΔx
∴ fΔx = U. AΔx

The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is. E. Therefore the average value of the field (i.e E/2) contributes to the force.

Question 18.
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.) Show that the capacitance of a spherical capacitor is given by C = \(\frac{4 \pi \varepsilon_0 r_1 r_2}{r_1-r_2}\)

where r₁ and r₂ are the radii of outer and inner spheres, respectively.
Solution:
As is clear from the figure +Q charge spreads uniformly on inner surface of outer sphere of radius r₁. The induced charge – Q spreads uniformly on the outer surface of inner sphere of radius r₂. The outer surface of outer sphere is earthed. Due to electrostatic shielding E = 0 for r < r₂ and E = 0 for r < r₂ and E = 0 for r > r₁

In the space between the two spheres electric intensity E exists as shown. Potential difference between the two spheres.

Question 19.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere ?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Solution:
Here r_a = 12cm = 12 × 10^-2m
r_b = 13cm = 13 × 10^-2m
q = 2.5μC = 2.5 × 10^-6C E_r = 32
(a) C = ?

(b)
V = ?
V = \(\frac{q}{c}\) = \(\frac{2.5 \times 10^{-6}}{5.5 \times 10^{-9}}\) = 4.5 × 10²Volt

(c) Capacity of an isolated sphere of radius R
R = 12 × 10^-2m is
C¹ = \(4 \pi \varepsilon_0 R\) = \(\frac{1}{9 \times 10^9}\) × 12 × 10^-12
= 1.33 × 10^-11 Farad.

The capacity of an isolated sphere is much smaller because in a capacitor outer sphere is earthed potential difference decreases and capacitance increases.

Question 20.
Answer carefully:
(a) Two large conducting sphers carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁ Q₂/\(4 \pi \varepsilon_0 \mathbf{r}^2\). where r is the distance between their centres ?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point ?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacitance of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer:
a) When the charged spheres are brought close together the charge distributions on them become non-uniform. Therefore, coloumb’s law is not valid hence the magnitude of force is not given exactly by this formula.
b) No Gauss’s law will not be true if coloumb’s law involved 1/r³ dependence instead of 1/r² dependence.
c) The line of force gives the direction of accelaration of charge. If the electric line of force is linear the test charge will move along the line if the line of force is not linear the charge will not go along the line.
d) As force due to the field is discreted towards the nucleus and the electron does not move in the direction of this force, therefore work done is zero when the orbit is circular. This is true even when orbit is elliptical as electric forces are conservative forces.
e) No electric potential is continuous.
f) The capacity of a single conductor implies that the second conductor is infinity.
g) This is because a molecule of water in its normal state has an unsymmetrical shape and therefore it has a permanent dipole moment.

Question 21.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effets (i.e., bending of field lines at the ends).
Answer:
Here L = 15cms = 15 × 10^-2m
r_a = 1.4cm = 1.4 × 10^-2m,
r_b = 1.5cm = 1.5 × 10^-2m
q = 3.5 μC = 3.5 × 10^-6 coloumb, C = ? V = ?

Question 22.
A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e, without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?
Answer:
Here V = 1kV= 1000Volt; K = ε_r = 3
Dielectric strength = 10⁷V/m
As electric field at the most should be 10% of dielectric strength due to reasons of safety.
E = 10% of 10⁷V/m = 10⁶V/m A = ?
C = 50pF = 50 × 10^-12F

Question 23.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z – direction.
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
By definition an equipotential surfaces is that every point of which potential is the same. In the four cases given above:
a) Equipotential surfaces are planes parallel to x – y plane. These are equidistant.
b) Equipotential surfaces are planes parallel to x – y plane. As the field increases uniformly distance between the planes decreases.
c) Equipotential surfaces concentric spheres with origin at the centre.
d) Equipotential surfaces have the shape which changes periodically at far off distances from the grid.

Question 24.
In a Van de Graff type generator a spherical metal shell is to be a 15 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm^-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build and electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer:
Here V = 15 × 10⁶ volt .
Dielectric strength = 5 × 10^-7 Vm^-1
minimum rodius, r = ?
max. Electric field E = 10% (dielectric strength)
E = \(\frac{10}{100}\) × 5 × 10⁷ = 5 × 10⁶VM^-1
As E = \(\frac{\mathrm{V}}{\mathrm{r}}\) ∴ r = \(\frac{\mathrm{V}}{\mathrm{E}}\) = \(\frac{15 \times 10^6}{5 \times 10^6}\) = 3m
obviously we cannot build an electrostatic generator, using a very small shell.

Question 25.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radium r₂ and charge q₂. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.
Answer:
As the charge resides always on the outer surface of the shell therefore, when the sphere and shell are connected by a wire, charge will flow essentially from the sphere to the shell, whatever be the magnitude and sign of charge q₂.

Question 26.
Answer the following :
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^-1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house on evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge it self completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lighting?
(Hint : The earth has and electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9Cm^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Answer:
a) since our body and the surface of earth both are conducting therefore our body and the ground form an equipotential surface. As we step out into the open from our house the original equipotential surfaces of open air change, keeping out body and the ground at the same potential that is why we do not get an electric shock.

b) Yes, the man will get a shock This is because the steady discharging current of the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the condenser formed by the aluminium sheet and the ground and the insulating slab.

c) The atmosphere is being discharged continuously by understorms and lightning all over the globe. It is also discharging due to the small conductivity of air. The two opposing processes, on an overage, are in equilibrium. Therefore the atmosphere. Keeps charged.

d) During lightning the electric energy of the atmosphere is dissipated in the form of light, heat and sound.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(c)

I.

Question 1.
Three screws are drawn at random from a lot of 50 screws, 5 of which are defective. Find the probability of the event that all 3 screws are non-defective assuming that the drawing is (a) with replacement (b) without replacement.
Solution:
Let S be the sample space
∴ The total number of screws = 50
The number of defective screws is 5 and the remaining 45 screws are non-defective.
Let A be the event of getting a drawing of the 3 screws is non-defective.
(a) With replacement

(b) Without replacement

Question 2.
If A, B, C are three independent events of an experiment such that P(A ∩ B^C ∩ C^C) = \(\frac{1}{4}\), P(A^C ∩ B ∩ C^C) = \(\frac{1}{8}\), P(A^C ∩ B^C ∩ C^C) = \(\frac{1}{4}\), then find P(A), P(B) and P(C).
Solution:
Since A, B, C are independent events.

Question 3.
There are 3 black and 4 white balls in one bag. 4 black and 3 white balls in the second bag. A die is rolled and the first bag is selected if it is 1 or 3 and the second bag for the rest. Find the probability of drawing a black ball from the bag thus selected.
Solution:
Probability of selecting first bag = \(\frac{2}{6}=\frac{1}{3}\)
Probability of selecting second bag = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Probability of getting a black ball from first bag = \(\frac{3}{7}\)
Probability of getting a black ball from the second bag = \(\frac{4}{7}\)
Probability of drawing a black ball = \(\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}\) = \(\frac{11}{21}\)

Question 4.
A, B, C are aiming to shoot a balloon, A will succeed 4 times out of 5 attempts. The chance of B shooting the balloon is 3 out of 4 and that of C is 2 out of 3. If three aim at the balloon simultaneously, then find the probability that atleast two of them hit the balloon.
Solution:

Question 5.
If A, B are two events, then show that \(P\left(\frac{A}{B}\right) P(B)+P\left(\frac{A}{B^C}\right) P\left(B^C\right)=P(A)\)
Solution:

Question 6.
A pair of dice are rolled. What is the probability that they sum to 7, given that neither die shows a 2?
Solution:
Let A be the event that the sum of the two dice is 7, then
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Let B be the event that neither die shows a 2
B = {(1, 1), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 25
A ∩ B = {(1, 6), (3, 4), (4, 3), (6, 1)}
n(A ∩ B) = 4
Required probability
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
= \(\frac{n(A \cap B)}{n(B)}\)
= \(\frac{4}{25}\)

Question 7.
A pair of dice are rolled. What is the probability that neither die shows a 2, given that they sum to 7?
Solution:
Let A be the event that the sum on two dice is 7
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ n(A) = 6
Let B be the event that neither die shows a 2
B = {(1, 1), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1),(4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 3), (6, 4), (6, 5), (6, 6)}
A ∩ B = {(1, 6), (3, 4), (4, 3), (6, 1)}
n(A ∩ B) = 4
Required probability
\(P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}\)
= \(\frac{n(A \cap B)}{n(A)}\)
= \(\frac{4}{6}\)
= \(\frac{2}{3}\)

Question 8.
If A, B are any two events, in an experiment, and P(B) ≠ 1 Show that \(P\left(\frac{A}{B^C}\right)=\frac{P(A)-P(A \cap B)}{1-P(B)}\)
Hint: P(A ∩ B^C) = P(A) – P(A ∩ B)
Solution:
By definition of condition probability
\(P\left(\frac{A}{B^C}\right)=\frac{P\left(A \cap B^C\right)}{P\left(B^C\right)}\)
= \(\frac{P(A)-P(A \cap B)}{1-P(B)}\)
∵ P(B^C) = \(P(\bar{B})\) = 1 – P(B)

Question 9.
An urn contains 12 red balls and 12 green balls. Suppose two balls are drawn one after another without replacement. Find the probability that the second ball drawn is green given that the first ball drawn is red.
Solution:
Total number of balls in an urn n(S) = 24
Let E₁ be the event of drawing a red ball in the first draw
P(E₁) = \(\frac{{ }^{12} C_1}{24}=\frac{1}{2}\)
Now the number of balls remaining is 23
Let \(\frac{E_2}{E_1}\) be the events of drawing a green ball in the second drawn
P(E₂/E₁) = \(\frac{12}{23}\)
∴ Required probability
P(E₁ ∩ E₂) = P(E₁) . P(E₂/E₁)
= \(\frac{1}{2} \times \frac{12}{23}\)
= \(\frac{6}{23}\)
∴ The probability of the second ball drawn is green given that the first ball drawn is red = \(\frac{6}{23}\)

Question 10.
A single die is rolled twice in succession. What is the probability that the number showing on the second toss is greater than that on the first rolling?
Solution:
A single die is rolled twice.
Let S be the sample space n(S) = 6² = 36
Let A be the event of getting the required event.
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}
n(A) = 15
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

Question 11.
If one card is drawn at random from a pack of cards then show that events of getting an ace and getting a heart are independent events.
Solution:
Suppose A is the event of getting an ace and B is the event of getting a heart.
∴ P(A) = \(\frac{4}{52}=\frac{1}{13}\)
P(B) = \(\frac{13}{52}=\frac{1}{14}\)
A ∩ B is the event of getting a Heart’s ace
P(A ∩ B) = \(\frac{1}{52}=\frac{1}{13} \cdot \frac{1}{4}\) = P(A) . P(B)
∴ A and B are independent events.

Question 12.
The probability that boy A will get a scholarship is 0.9 and that another boy B will get one is 0.8. What is the probability that atleast one of them will get the scholarship?
Solution:
Suppose E₁ is the event of a boy ‘A’ getting a scholarship and E₂ is the event of another boy B getting the scholarship.
Given P(E₁) = 0.9, P(E₂) = 0.8
E₁ and E₂ are independent events.
P(E₁ ∩ E₂) = P(E₁) . P(E₂)
= (0.9) (0.8)
= 0.72
The probability that atleast one of them will get a scholarship = P(E₁ ∪ E₂)
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= 0.9 + 0.8 – 0.72
= 1.7 – 0.72
= 0.98

Question 13.
If A, B are two events with P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then find the value of P(A^C) + P(B^C).
Solution:
By addition theorem on probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A) + P(B) = P(A ∪ B) + P(A ∩ B)
= 0.65 + 0.15
= 0.8
P(A) + P(B) = 0.8 ……..(1)
P(A^C) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 0.8 [∵ by (1)]
= 1.2

Question 14.
If A, B, C are independent events, show that A ∪ B, and C are also independent events.
Solution:
∵ A, B, C are independent events.
⇒ A, B; B, C; C , A are also independent events
P(A ∩ B ∩ C) = P(A) P(B) P(C)
P(A ∩ C) = P(A) . P(C)
P(B ∩ C) = P(B) . P(C)
P(A ∩ B) = P(A) . P(B)
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)]
= P(A ∩ C) + P(B ∩ C) – P[(A ∩ C) ∩ (B ∩ C)]
= P(A) . P(C) + P(B) . P(C) – P(A ∩ B ∩ C)
= P (A) . P(C) + P(B) . P(C) – P(A) . P(B) . P(C)
= [P(A) + P(B) – P(A) . P(B)] P(C)
= P(A ∪ B) . P(C)
∴ A ∪ B and C are independent events.

Question 15.
A and B are two independent events such that the probability of both the events occurring is \(\frac{1}{6}\) and the probability of both the events do not occur is \(\frac{1}{3}\). Find the probability of A.
Solution:
A and B are independent events.

Question 16.
A fair die is rolled. Consider the events. A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}. Find
(i) P(A ∩ B), P(A ∪ B)
(ii) P(\(\frac{A}{B}\)), P(\(\frac{B}{A}\))
(iii) P(\(\frac{A}{C}\)), P(\(\frac{C}{A}\))
(iv) P(\(\frac{B}{C}\)), P(\(\frac{C}{B}\))
Solution:
A fair die is rolled
P(A) = \(\frac{3}{6}=\frac{1}{2}\)
P(B) = \(\frac{2}{6}=\frac{1}{3}\)
P(C) = \(\frac{4}{6}=\frac{2}{3}\)
n(S) = 6¹ = 6
Given A = {1, 3, 5}, B = {2, 3}, C = {2, 3, 4, 5}
(i) A ∩ B = {3}
P(A ∩ B) = P{3} = \(\frac{1}{6}\)
∴ P(A ∩ B) = \(\frac{1}{6}\)
(A ∪ B) = {1, 2, 3, 5}
n(A ∪ B) = 4
n(S) = 6
P(A ∪ B) = \(\frac{4}{6}=\frac{2}{3}\)

Question 17.
If A, B, C are three events in a random experiment, prove the following:
(i) P(\(\frac{A}{A}\)) = 1
Solution:
\(P\left(\frac{A}{A}\right)=\frac{P(A \cap A)}{P(A)}=\frac{P(A)}{P(A)}=1\)

(ii) \(\mathbf{p}\left(\frac{\phi}{A}\right)=0\)
Solution:
\(P\left(\frac{\phi}{A}\right)=\frac{P(\phi \cap A)}{P(A)}=\frac{0}{P(A)}=0\)

(iii) A ⊂ B ⇒ P(\(\frac{A}{C}\)) ≤ P(\(\frac{B}{C}\))
Solution:

(iv) P(A – B) = P(A) – P(A ∩ B)
Solution:
A – B = {x / x ∈ A a x ∉ B}
A – B = A – (A ∩ B)

P(A – B) = P[A – (A ∩ B)] = P(A) – P(A ∩ B)

(v) If A, B are mutually exclusive and P(B) > 0, then P(\(\frac{A}{B}\)) = 0.
Solution:
We know \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
= \(\frac{0}{\mathrm{P}(\mathrm{B})}\)
= 0 [∵ A, B are mutually exclusive events]
Hint: A, B are mutually exclusive then A ∩ B = φ ⇒ P(A ∩ B) = 0.

(vi) If A, B are mutually exclusive then P(A/B^C) = \(\frac{\mathrm{P}(\mathrm{A})}{1-\mathrm{P}(\mathrm{B})}\); when P(B) ≠ 1.
Solution:
Given P(A ∩ B) = 0 (∵ A and B are mutually exchanging)

(vii) If A, B are mutually exclusive and P(A ∪ B) ≠ 0, then \(P\left(\frac{A}{A \cup B}\right)=\frac{P(A)}{P(A)+P(B)}\)
Solution:
Hint: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Question 18.
Suppose that a coin is tossed three times. Let A be “getting three heads” and B be the event of “getting a head on the first toss”. Show that A and B are dependent events.
Solution:
Let event A be “getting their heads”, and B be the event of “getting a head on the first toss” when a coin is tossed three times.
∴ A = {HHH}
n(A) = 1
P(A) = \(\frac{1}{8}\)
B = {HTT, HTH, HHT, HHH}
n(B) = 4
P(B) = \(\frac{4}{8}\)
A ∩ B = {HHH}
n(A ∩ B) = 1
P(A ∩ B) = \(\frac{1}{8}\)
P(A) . P(B) = \(\frac{1}{8} \cdot \frac{4}{8}=\frac{1}{16}\) ≠ P(A ∩ B)
∴ P(A ∩ B) ≠ P(A) . P(B)
Hence A, B are dependent events.

Question 19.
Suppose that an unbiased pair of dice is rolled. Let A denote the event that the same number shows on each die. Let B denote the event that the sum is greater than 7. Find (i) P(\(\frac{A}{B}\)) (ii) P(\(\frac{B}{A}\))
Solution:
Given a denote the event that the same number on pair of dice is rolled.
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) (6, 6)}
n(A) = 6
P(A) = \(\frac{6}{36}=\frac{1}{6}\)
Given B denote the event that the sum is greater than 7 when pair of dice is rolled.
∴ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ n(B) = 15
P(B) = \(\frac{15}{36}\)
A ∩ B = {(4, 4), (5, 5), (6, 6)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)

Question 20.
Prove that A and B are independent events if and only if \(P\left(\frac{A}{B}\right)=P\left(\frac{A}{B^C}\right)\)
Solution:
Let A and B be independent

⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(A)-P(A \cap B)}{1-P(B)}\)
⇒ P(A ∩ B) – P(B) P(A ∩ B) = P(A) P(B) – P(B) . P(A ∩ B)
⇒ P(A ∩ B) = P(A) . P(B)
∴ A, B are independent.
Hence, A, B are independent iff
\(P\left(\frac{A}{B}\right)=P\left(\frac{A}{B^C}\right)\)

II.

Question 1.
Suppose A and B are independent events with P(A) = 0.6, P(B) = 0.7 then compute
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(\(\frac{B}{A}\))
(iv) P(A^C ∩ B^C)
Solution:
Given A, B are independent events and
P(A) = 0.6, P(B) = 0.7
(i) P(A ∩ B) = P(A) . P(B)
= 0.6 × 0.7
= 0.42
(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.7 – 0.42
= 1.3 – 0.42
= 0.88
(iii) P(\(\frac{B}{A}\)) = P(B) = 0.7
(iv) P(A^C ∩ B^C) = P(A^C) . P(B^C) (∵ A^C & B^C are also independent events)
= [1 – P(A)] [1 – P(B)]
= (1 – 0.6) (1 – 0.7)
= 0.4 × 0.3
= 0.12

Question 2.
The probability that Australia wins a match against India in a cricket game is given to be \(\frac{1}{3}\). If India and Australia play 3 matches, what is the probability that,
(i) Australia will lose all three matches?
(ii) Australia will win atleast one match?
Solution:
Suppose A is the event of Australia winning the match.
Given P(A) = \(\frac{1}{3}\)
∴ P(\(\overline{\mathrm{A}}\)) = 1 – P(A)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
(i) Probability that Australia will loose all the three matches = P(\(\overline{\mathrm{A}}\))³
= \(\left(\frac{2}{3}\right)^3\)
= \(\frac{8}{27}\)
(ii) Probability that Australia will win atleast one match = 1 – P(\(\overline{\mathrm{A}}\))³
= 1 – \(\frac{8}{27}\)
= \(\frac{19}{27}\)

Question 3.
Three boxes numbered I, II, III contain balls as follows:

One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that it is from box II.
Solution:
Let B₁, B₂, B₃ be the events of selecting the Ist, IInd and IIIrd boxes respectively.
Then P(B₁) = P(B₂) = P(B₃) = \(\frac{1}{3}\)
Probability of selecting a red ball from the first box = \(\frac{3}{6}\) = P(R/B₁)
Probability of selecting a red ball from the second box = \(\frac{1}{4}\) = P(R/B₂)
Probability of selecting a red ball from the third box = \(\frac{3}{12}\) = P(R/B₃)
Assuming that the ball is red, the probability it is from box II,

Question 4.
A person secures a job in a construction company in which the probability that the workers go on strike is 0.65 and the probability that the construction job will be completed on time if there is no strike is 0.80. If the probability that the construction job will be completed on time even if there is a strike is 0.32, determine the probability that the constructed job will be completed on time.
Solution:
Let P(S) = Probability of the workers going on strike = 0.65
P(\(\bar{S}\)) = Probability on the workers go on strike
= 1 – P(S)
= 1 – 0.65
= 0.35
\(P\left(\frac{E}{S}\right)\) = Probability that the job completed if there is no strike = 0.32
\(\mathrm{P}\left(\frac{\mathrm{E}}{\overline{\mathrm{S}}}\right)\) = Probability that the job completed if there is a strike = 0.80
P(E) = Probability that the construction job will be completed on time
= P(S) \(P\left(\frac{E}{S}\right)\) + P(\(\bar{S}\)) \(\mathrm{P}\left(\frac{\mathrm{E}}{\overline{\mathrm{S}}}\right)\)
= (0.65) (0.32) + (0.35) (0.08)
= 0.2080 + 0.2800
= 0.4880

Question 5.
For any two events A, B show that
P(A ∩ B) – P(A) P(B) = P(A^C) P(B) – P(A^C ∩ B) = P(A) P(B^C) – P(A ∩ B^C)
Solution:
R.H.S. I = P(A^C) P(B) – P(A^C ∩ B)
= [(1 – P(A)] P(B) – [P(B) – P(A ∩ B)]
= P(B) – P(A) P(B) – P(B) + P(A ∩ B)
= P(A ∩ B) – P(A) P(B)
= L.H.S
R.H.S. II = P(A) P(B^C) – P(A ∩ B^C)
= P(A) [1 – P[P(B)] – [P(A) – P(A ∩ B)]
= P(A) – P(A) P(B) – P(A) + P(A ∩ B)
= P(A ∩ B) – P(A) P(B)
= L.H.S
∴ L.H.S = R.H.S I = R.H.S II
Hence P(A ∩ B) – P(A) . P(B) = P(A^C) P(B) – P(A^C ∩ B) = P(A) P(B^C) – P(A ∩ B^C)

III.

Question 1.
Three Urns have the following composition of balls.
Urn I: 1 White, 2 black
Urn II: 2 White, 1 black
Urn III: 2 White, 2 balck
One of the Urn is selected at random and a ball is drawn. It turns out to be white. Find the probability that it comes from Urn III.
Solution:
Let E_i be the event of Choosing the Urn i = 1, 2, 3, and P(E_i) be the probability of choosing the Urn i = 1, 2, 3
Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
Having choosen the Urn i, the probability of drawing a white ball, P(W/E_i), is given by
P(W/E₁) = \(\frac{1}{3}\)
P(W/E₂) = \(\frac{2}{3}\)
P(W/E₃) = \(\frac{2}{4}\)
We have to find the probability P(E₃/W) by Baye’s theorem.

Question 2.
In a shooting test the probability of A, B, C hitting the targets are \(\frac{1}{2}\), \(\frac{2}{3}\) and \(\frac{3}{4}\) respectively. If all of their fire is at the same target. Find the probability that
(i) Only one of them hits the target
(ii) At atleast one of them hits the target
Solution:
The probabilities that A, B, C hit the targets are denoted by

Question 3.
In a certain college, 25% of the boys and 10% of the girls are studying mathematics. The girls constitute 60% of the student’s strength. If a student selected at random is found studying mathematics. Find the probability that the student is a girl.
Solution:
The probability that a student selected to be a girl
P(G) = \(\frac{60}{100}=\frac{6}{10}\)
The probability that a student selected to be a boy
P(B) = \(\frac{40}{100}=\frac{4}{10}\)
The probability that a boy studying mathematics
P(M/B) = \(\frac{25}{100}=\frac{1}{4}\)
Similarly probability that a girl studying mathematics
P(M/G) = \(\frac{10}{100}=\frac{1}{10}\)
We have to find P(G/M) By Baye’s theorem

Question 4.
A person is known to speak the truth 2 out of 3 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.
Solution:
P(T) = Probability that a person speaks truth 2 out of 3 times = \(\frac{2}{3}\)
P(F) = 1 – P(T)
= 1 – \(\frac{2}{3}\)
= \(\frac{1}{3}\)
After he reports that it is 1, it is true if it actually shows 1 otherwise false if does not show 1.
P(1) = \(\frac{1}{6}\) and P(T) = \(\frac{5}{6}\)
P(T/1) = P(reports true if it is 1) = \(\frac{2}{3}\)
P(F/T) = P(report False if it is T) = \(\frac{1}{3}\)
By Baye’s theorem

∴ The probability that reports that it is 1 and actually it is 1 is \(\frac{2}{7}\).

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(b)

I.

Question 1.
If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear.
Solution:
4 coins are tossed simultaneously.
Total number of ways = 2⁴ = 16
n(S) = 16
From 4 heads we must get 2 heads.
Number of ways of getting 2 heads = ⁴C₂
= \(\frac{4.3}{1.2}\)
= 6
∴ n(E) = 6
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
∴ Probability of getting 2 heads and 2 tails = \(\frac{3}{8}\)

Question 2.
Find the probability that a non-leap year contains
(i) 53 Sundays
(ii) 52 Sundays only
Solution:
A non-leap year contains 365 days 52 weeks and 1 day more.
(i) We get 53 Sundays when the remaining day is Sunday.
Number of days in the week = 7
∴ n(S) = 7
The number of ways getting 53 Sundays.
n(E) = 1
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{1}{7}\)
∴ Probability of getting 53 Sundays = \(\frac{1}{7}\)
(ii) Probability of getting 52 Sundays
P(\((\overline{\mathrm{E}})\)) = 1 – P(E)
= 1 – \(\frac{1}{7}\)
= \(\frac{6}{7}\)

Question 3.
Two dice are rolled, what is the probability that none of the dice shows the number 2?
Solution:
The random experiment is rolling 2 dice.
n(S) = 6² = 36
Let E be the event of not getting 2
n(E) = 5 × 5 = 25
∴ P(E) = \(\frac{\mathrm{n}(E)}{\mathrm{n}(s)}=\frac{25}{36}\)

Question 4.
In an experiment of drawing a card at random from a pack, the event of getting a spade is denoted by A, and getting a pictured card (King, Queen, or Jack) is denoted by B. Find the probabilities of A, B, A ∩ B, and A ∪ B.
Solution:
A is the event of getting a spade from the pack
∴ P(A) = \(\frac{13}{52}=\frac{1}{4}\)
B is the event of getting a picture card
P(B) = \(\frac{4 \times 3}{52}=\frac{3}{13}\)
A ∩ B is the event of getting a picture card in spades.
n(A ∩ B) = 3, n(s) = 52
P(A ∩ B) = \(\frac{3}{52}\)
A ∪ B is the event of getting a spade or a picture card.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{4}+\frac{3}{13}-\frac{3}{52}\)
= \(\frac{13+12-3}{52}\)
= \(\frac{11}{26}\)

Question 5.
In a class of 60 boys and 20 girls, half of the boys and half of the girls know cricket. Find the probability of the event that a person selected from the class is either a boy or a girl who knows cricket.
Solution:
Let A be the event that the selected person is a boy and B be the event that the selected person knows a cricket when a person is selected from the class and S be the sample space.
Now, n(S) = ⁸⁰C₁ = 80
n(A) = ⁶⁰C₁ = 60
n(B) = ⁴⁰C₁ = 40
n(A ∩ B) = ³⁰C₁ = 30
∴ P(A) = \(\frac{60}{80}\), P(B) = \(\frac{40}{80}\), P(A ∩ B) = \(\frac{30}{80}\)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{60}{80}+\frac{40}{80}-\frac{30}{80}\)
= \(\frac{70}{80}\)
= \(\frac{7}{8}\)

Question 6.
For any two events A and B, show that P(A^C ∩ B^C) = 1 + P(A ∩ B) – P(A) – P(B).
Solution:
A^C ∩ B^C = \(\overline{A \cup B}\)
P(A^C ∩ B^C) = P(\(\overline{A \cup B}\))
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 + P(A ∩ B) – P(A) – P(B)

Question 7.
Two persons A and B are rolling die on the condition that the person who gets 3 will win the game. If A starts the game, then find the probabilities of A and B respectively to win the game.
Solution:
p = Probability of getting 3 = \(\frac{1}{6}\)
q = 1 – p
= 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Probability success (p) = \(\frac{1}{6}\)
Probability of failure (q) = \(\frac{5}{6}\)
A may win the game either in I trial or in trial or in V trial etc.
Probability of A win = p + q . q . p + q . q . q . q . p + ……..

Question 8.
A, B, C are 3 newspapers from a city. 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C, and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = \(\frac{20}{100}\) = 0.2
P(B) = \(\frac{16}{100}\) = 0.16
P(C) = \(\frac{14}{100}\) = 0.14
P(A ∩ B) = \(\frac{8}{100}\) = 0.08
P(B ∩ C) = \(\frac{4}{100}\) = 0.04
P(A ∩ C) = \(\frac{5}{100}\) = 0.05
P(A ∩ B ∩ C) = \(\frac{2}{100}\) = 0.02

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= 0.2 + 0.16 + 0.14 – 0.08 – 0.04 – 0.05 + 0.02
= 0.52 – 0.17
= 0.35
Percentage of population who read atleast one newspaper = 0.35 × 100 = 35%

Question 9.
If one ticket is randomly selected from tickets numbered 1 to 30. Then find the probability that the number on the ticket is
(i) a multiple of 5 or 7
(ii) a multiple of 3 or 5
Solution:
(i) Number of ways drawing one ticket = n(S) = ³⁰C₁ = 30
Suppose A is the event of getting a multiple of 5 and B is the event of getting a multiple of 7.
A = {5, 10, 15, 20, 25, 30}
B = {7, 14, 21, 28}
A ∩ B = φ ⇒ A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
= \(\frac{6}{30}+\frac{4}{30}\)
= \(\frac{1}{3}\)
Probability of getting a multiple of 5 or 7 = \(\frac{1}{3}\)
(ii) Suppose A is the event of getting a multiple of 3 and B is the event of getting a multiple of 5.
A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
B = {5, 10, 15, 20, 25, 30}
A ∩ B = {15, 30}
P(A) = \(\frac{10}{30}\)
P(B) = \(\frac{6}{30}\)
P(A ∩ B) = \(\frac{2}{30}\)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{10}{30}+\frac{6}{30}-\frac{2}{30}\)
= \(\frac{14}{30}\)
= \(\frac{7}{15}\)
Probability of getting a multiple of 3 or 5 = \(\frac{7}{15}\)

Question 10.
If two numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is (i) an even number (ii) an odd number.
Solution:
(i) Let A be the event that the sum of the numbers is even when two numbers are selected out of 20 consecutive natural numbers.
In 20 consecutive natural numbers, we have 10 odd and 10 even natural numbers.
∵ The sum of two odd natural numbers is an even number and the sum of two even natural numbers is also an even number

(ii) Probability that the sum of two numbers is an odd number
P(\(\overline{\mathrm{A}}\)) = 1 – P(A)
= 1 – \(\frac{9}{19}\)
= \(\frac{10}{19}\)

Question 11.
A game consists of tossing a coin 3 times and noting its outcome. A boy wins if all tosses give the same outcome and lose otherwise. Find the probability that the boy loses the game.
Solution:
Let A be the event that the boy wins the game getting the same outcome when a coin is tossed 3 times and S be the sample space.
∴ n(S) = 2³ = 8
A = {HHH, TTT}
n(A) = 2
P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{8}=\frac{1}{4}\)
∴ The probability that the boy loses the game = P(\(\overline{\mathrm{A}}\))
= 1 – P(A)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Question 12.
If E₁, E₂ are two events with E₁ ∩ E₂ = φ then show that \(P\left(E_1^c \cap E_2^c\right)=P\left(E_1^c\right)-P\left(E_2\right)\)
Solution:
Given E₁ ∩ E₂ = φ
∴ P(E₁ ∩ E₂) = 0
\(P\left(E_1^c \cap E_2^c\right)=P\left(\overline{E_1 \cup E_2}\right)\)
= 1 – P(E₁ ∪ E₂)
= 1 – [P(E₁) + P(E₂) – P(E₁ ∩ E₂)]
= 1 – P(E₁) – P(E₂) + P(E₁ ∩ E₂)
= \(P\left(E_1^c\right)\) – P(E₂) + 0
∴ \(P\left(E_1^c \cap E_2^c\right)=P\left(E_1^c\right)-P\left(E_2\right)\)

II.

Question 1.
A pair of dice rolled 24 times. A person wins by not getting a pair of 6’s on any of the 24 rolls. What is the probability of his winning?
Solution:
Random experiment is tossing two dice 24 times = 36 × 36 × ……. 36 = (36)²⁴
∴ n(S) = (36)²⁴
Let A be the event of not getting a pair of 6’s on any of the 24 rolls.
∴ number of ways favourable to an event A = 35 × 35 × ……. × 35 = (35)²⁴
n(A) = (35)²⁴
∴ P(A) = \(\frac{(35)^{24}}{(36)^{24}}=\left(\frac{35}{36}\right)^{24}\)

Question 2.
If P is a probability function, then show that for any two events A and B.
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)
Solution:
For any sets A, B we have

A ∩ B ≤ A ≤ A ∪ B
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B)
By Addition theorem of probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ≤ P(A) + P(B) ……….(2)
From (1), (2) we get
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)

Question 3.
In a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, find the probability of the event, that
(i) None of them is defective
(ii) Only one of them is defective
(iii) Atleast one of them is defective
Solution:
Out of 15 bulbs, 5 are defective
probability of selecting a defective bulb = P
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
We are selecting 5 bulbs
n(S) = ¹⁵C₅
(i) None of them is defective.
All 5 bulbs must be selected from 10 good bulbs. This can be done in ¹⁰C₅ ways.
P(A) = \(\frac{{ }^{10} C_5}{{ }^{15} C_5}=\frac{10.9 .8 .7 .6}{15.14 .13 .12 .11}=\frac{12}{143}\)
(ii) Only one of them is defective in 4 good and 1 defective ball.
This can be done in (¹⁰C₄) (⁵C₁) = \(\frac{10.9 .8 .7}{1.2 \cdot 3.4} \cdot 5\)
= 210 × 5
= 1050
Probability of selecting one defective = \(\frac{1050}{{ }^{15} C_5}\)
= \((1050) \frac{1.2 .3 .4 .5}{15.14 .13 .12 .11}\)
= \(\frac{50}{143}\)
(iii) Probability atleast one of them is defective = P(\(\overline{A}\))
= 1 – P(A)
= 1 – \(\frac{12}{143}\)
= \(\frac{131}{143}\)

Question 4.
A and B are seeking admission into I.I.T. If the probability for A to be selected is 0.5 and that both to be selected is 0.3. Is it possible that the probability of B being selected is 0.9?
Solution:
Given P(A) = 0.5; P(A ∩ B) = 0.3
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.5 + P(B) – 0.3
= 0.2 + P(B)
P(A ∪ B) ≤ 1
0.2 + P(B) ≤ 1
P(B) ≤ 0.8
∴ It is not possible to have P(B) = 0.9

Question 5.
The probability for a contractor to get a road contract is \(\frac{2}{3}\) and to got a building contract is \(\frac{5}{9}\). The probability to get atleast on contract is \(\frac{4}{5}\). Find the probability to get both contracts.
Solution:
Suppose A is the event of getting a road contract
B is the event of getting a building contract
Given P(A) = \(\frac{2}{3}\); P(B) = \(\frac{5}{9}\); P(A ∪ B) = \(\frac{4}{5}\)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{2}{3}+\frac{5}{9}-\frac{4}{5}\)
= \(\frac{30+25-36}{45}\)
= \(\frac{19}{45}\)
∴ Probability to get both contracts = \(\frac{19}{45}\)

Question 6.
In a committee of 25 members, each member is proficient either in Mathematics or Statistics or in both. If 19 of these are proficient in Mathematics, and 16 in Statistics, find the probability that a person selected from the committee is proficient in both.
Solution:
When a person is chosen at random from the academy consisting of 25 members
Let A be the event that the person is proficient in Mathematics
B be the event that the person is proficient in Statistics and
S is the sample space.
Since 19 members are proficient in Mathematics and 16 members are proficient in Statistics.
P(A) = \(\frac{19}{25}\), P(B) = \(\frac{16}{25}\)
Since everyone is either proficient in Mathematics or Statistics or in both
A ∪ B = S
⇒ P(A ∪ B) = P(S)
⇒ P(A) + P(B) – P(A ∩ B) = P(S)
⇒ \(\frac{19}{25}+\frac{16}{25}\) – P(A ∩ B) = 1
⇒ P(A ∩ B) = \(\frac{19}{25}+\frac{16}{25}-1\) = \(\frac{10}{25}\)
∴ P(A ∩ B) = \(\frac{2}{5}\)

Question 7.
A, B, C are three horses in a race. The probability of A winning the race is twice that of B and the probability of B is twice that of C. What are the probabilities of A, B, and C winning the race?
Solution:
Let A, B, C be the events that horses A, B, and C win the race respectively.
Given P(A) = 2P(B), P(B) = 2P(C)
∴ P(A) = 2P(B) = 2[2P(C)] = 4P(C)
Since the horses A, B and C run the race,
A ∪ B ∪ C = S and A, B, C are mutually disjoint.
P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
⇒ P(S) = 4P(C) + 2P(C) + P(C)
⇒ 1 = 7P(C)
⇒ P(C) = \(\frac{1}{7}\)
P(A) = 4P(C) = 4 × \(\frac{1}{7}\) = \(\frac{4}{7}\)
P(B) = 2P(C) = 2 × \(\frac{1}{7}\) = \(\frac{2}{7}\)
∴ P(A) = \(\frac{4}{7}\), P(B) = \(\frac{2}{7}\), P(C) = \(\frac{1}{7}\)

Question 8.
A bag contains 12 two rupee coins, 7 one rupee coins, and 4 half rupee coins. If 3 coins are selected at random find the probability that
(i) The sum of the 3 coins is maximum
(ii) The sum of the 3 coins is minimum
(iii) Each coin is of a different value
Solution:
In the bag, there are 12 two rupees, 7 one rupees, and 4 half rupee coins.
Total number of coins = 12 + 7 + 4 = 23
Number of ways drawing 3 coins ²³C₃
n(S) = ²³C₃
(i) We get the maximum amount if the coins are 2 rupee coins.
Number of drawing 3 two rupee coins = ¹²C₃
n(E₁) = ¹²C₃
P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{{ }^{12} C_3}{{ }^{23} C_3}\)
(ii) We get a minimum amount if 3 coins are taken from 4 half rupee coins.
Number of ways of drawing 3 half rupee coins = ⁴C₃
n(E₂) = ⁴C₃
P(E₂) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{{ }^4 C_3}{{ }^{23} C_3}\)
(iii) Each coin is of different value we must draw one coin each.
This can be done in ¹²C₁, ⁷C₁, ⁴C₁ ways
n(E₃) = ¹²C₁ × ⁷C₁ × ⁴C₁ = 12 × 7 × 4
P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{12 \times 7 \times 4}{{ }^{23} C_3}\)

Question 9.
The probabilities of three events A, B, C are such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09 and P(A ∪ B ∪ C) ≥ 0.75. Show that P(B ∩ C) lies in the interval [0.23, 0.48]
Solution:
P(A ∪ B ∪ C) ≥ 0.75
0.75 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0.75 ≤ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) ≤ 1
⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 – 0.28 – P(B ∩ C) + 0.09 ≤ 1
⇒ 0.75 ≤ 1.23 – P(B ∩ C) ≤ 1
⇒ -0.75 ≥ P(B ∩ C) – 1.23 ≥ -1
⇒ 0.48 ≥ P(B ∩ C) ≥ 0.23
⇒ 0.23 ≤ P(B ∩ C) ≤ 0.48
∴ P(B ∩ C) lies in the interval [0.23, 0.48]

Question 10.
The probabilities of three mutually exclusive events are respectively given as \(\frac{1+3 P}{3}, \frac{1-P}{4}, \frac{1-2 P}{2}\). Prove that \(\frac{-1}{3} \leq P \leq \frac{1}{2}\)
Solution:
Three mutually exclusive events probabilities are given.
∴ 0 ≤ \(\frac{1+3 P}{3}\) ≤ 1
⇒ 0 ≤ 1 + 3P ≤ 3
⇒ -1 ≤ P ≤ 2
⇒ \(\frac{-1}{3} \leq P \leq \frac{2}{3}\) …….(1)
Also 0 ≤ \(\frac{1-\mathrm{P}}{4}\) ≤ 1
⇒ 0 ≤ 1 – P ≤ 4
⇒ -1 ≤ -P ≤ 3
⇒ 1 ≥ P ≥ -3
⇒ -3 ≤ P ≤ 1 …….(2)
Also 0 ≤ \(\frac{1-2 P}{2}\) ≤ 1
⇒ 0 ≤ 1 – 2P ≤ 2
⇒ -1 ≤ -2P ≤ 1
⇒ 1 ≥ 2P ≥ -1
⇒ -1 ≤ 2P ≤ 1
⇒ \(\frac{-1}{2} \leq P \leq \frac{1}{2}\) ………(3)
From (1), (2), (3) we have \(\frac{-1}{3} \leq P \leq \frac{1}{2}\)

Question 11.
On a Festival day, a man plans to visit 4 holy temples A, B, C, D in random order. Find the probability that he visits (i) A before B (ii) A before B and B before C.
Solution:
(i) It is nothing but arranging A, B, C, D in 4 chains so that A seats before B.
For this, first, we arrange C, D is the 4 chains it can be done in ⁴P₂ = 12 ways and the remaining 2 seats A & B can be sit in only one way (A before B).
Also n(S) = 24
∴ The probability that he visit 4 temples, A before B = \(\frac{12 \times 1}{24}\) = \(\frac{1}{2}\)
(ii) Similarly to the above problem
First, we arrange D in any one of 4 chains it can be done in 4 ways. Then the remaining 3 seats A, B, C can sit in only one way (A before 3 and B before C)
Also n(A) = 4! = 24
The probability that he visit 4 temples, A before B and B before C = \(\frac{4 \times 1}{24}\) = \(\frac{1}{6}\)

Question 12.
From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. The particulars of 5 persons are as follows.

A person is selected at random from this group to act as a spokesperson. Find the probability that the spokesperson will be either male or above 35 years.
Solution:
Let A be the event that the selected person is male and B be the event that the selected person is above 35 years when a person is selected at random from the group of 5 persons to act as a spokesperson and S be the sample space.
∴ n(S) = ⁵C₁ = 5
n(A) = ³C₁ = 3
P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{5}\)
n(B) = ²C₁ = 2
P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{2}{5}\)
n(A ∩ B) = ¹C₁ = 1
P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}\) = \(\frac{1}{5}\)
By Addition theorem on probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\)
= \(\frac{4}{5}\)

Question 13.
Out of 100 students, two sections of 40 are 60 and formed. If you and your friend are among the 100 students, find the probability that
(i) You both enter the same section.
(ii) You both enter the different sections.
Solution:
Let S be the sample space
n(S) = no.of ways of doing 100 students into 2 sections of 40 and 60 = \(\frac{100 !}{40 ! 60 !}\)
(i) You both enter the same section:
First, you and your friend both enter the section the remaining 98 students and I can be divided into two sections (First section 38 and second section 60) is \(\frac{98 !}{38 ! 60 !}\)
You and your friend both enter section II and the remaining 98 students can be divided into two sections (First section 40 and second section 58) is \(\frac{98 !}{40 ! 58 !}\)

= \(\frac{26}{165}+\frac{59}{165}\)
= \(\frac{17}{33}\)
(ii) You both enter the different sections:
The probability that the both in different sections = 1 – P(E)
= 1 – \(\frac{17}{33}\)
= \(\frac{16}{33}\)