# AP Inter 2nd Year Physics Study Material Chapter 4 Electric Charges and Fields

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

Question 1.
What is meant by the statement ‘charge is quantized’ ?
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 ?
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 ?
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 ?
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 ?
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 ?
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 ?
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 ?
It is conservative force.

Question 9.
State Gauss’s law in electrostatics.
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 ?
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.
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.
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.
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 1.
State and explain Coulomb’s inverse square law in electricity.
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.
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.
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.
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)
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.
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.

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.)
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.
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.
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, ER(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)
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,
ϕE = $$\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}$$ = $$\frac{1}{6} \times \frac{4 \pi r^2 \sigma}{\varepsilon_0}$$ = $$\frac{2 \pi \mathrm{r}^2 \sigma}{3 \varepsilon_0}$$

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 1st and 2nd wire = $$\frac{\mathrm{r}}{2}$$
Electric field intensity at P due 1st 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 2nd 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 me = m; q = e; d = x; ux = u; uy = 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 = uyt + $$\frac{1}{2} \mathrm{at}^2$$
⇒ y = 0 + $$\frac{1}{2}\left(\frac{e E}{m}\right)\left(\frac{x}{u}\right)^2$$
∴ y = $$\frac{\mathrm{eEx}^2}{2 \mathrm{mu}^2}$$

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 memp 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, Fe = $$\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}$$ = $$\frac{\mathrm{Ke}^2}{\mathrm{r}^2}$$ ……. (1)
Where q1 = q2 = e
In gravitation, Fg = $$\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}$$ = $$\frac{\mathrm{Gm}_{\mathrm{e}} \mathrm{m}_{\mathrm{p}}}{\mathrm{r}^2}$$ …. (2)
Where m1 = me ; m2 = mp 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 ?
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.
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 qA = 2 µC, qB = -5 µC, qC = 2 µC and qD = -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, FA = -FC and FD = -FB
∴ The net resultant force on 1 µC is
F = FA + FB + FC + FD
= -FC + FB + FC – FB
= 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 ?
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 qA = 3 μC and qB = -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 qA = 3 μC = 3 × 10-6 C; qB = -3 μC = -3 × 10-6C From fig. AO = OB = 10 cm = 0.1 m
Electric field at midpoint ‘O’ due to qA The direction of EA is A to O.
Electric field at midpoint ‘O’ due to qB at B is The direction of EB is O to B.
Now we see that EA and EB are in same direction. So, the resultant electric field at O is E. Hence,
E = EA + EB = 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 = Eq = 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 qA = 2.5 × 10-7 C, and qB = -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)
qA = 2.5 × 10-7C
qB = -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 = qA + qB = 2.5 × 10-7C – 2.5 × 10-7C = 0
The electric dipolemoment
P = Any charge (qA) × 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, me = 9.1 × 10-3 kg
Mass transferred from wool to polythene M = n × me
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, qA = 6.5 × 10-7C ; qB = 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 qA = 6.5 × 10-7C;
qB = 6.5 × 10-7 C; qC = 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 ? 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 ?
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 σA = 127.0 × 10-22 C/m2
σB = 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.
PZ = -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 pz, as the length of the dipole is dz.

∴ F = Pz. $$\frac{\mathrm{dE}}{\mathrm{dz}}$$ = -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, ax 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, bx 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 Vx (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 1st Year Textbook of Physics.
Solution:
Mass of particle = m
Charge of particle = -q
Speed of particle = Vx
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 Vx = 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, me = 9.1 × 10-31 kg.)
Solution:
Given Vx = 2 × 106 m/s; E = 9.1 × 102 N/C
q = e = 1.6 × 10-19 C; me = 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. 