NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 1
Chapter Name : Electric Charges And Fields

Q1.1 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?

Answer. Repulsive force of magnitude 6 × 10 N −3

Charge on the rst sphere, q = 2 × 10 C 1
−7

Charge on the second sphere, q = 3 × 10 C 2
−7

Distance between the spheres, r = 30cm = 0.3m
Electrostatic force between the spheres is given by the relation,
q1 q2
F = 2
4πϵ0 r

Where, ϵ 0 = Permittivity of free space
1 9 2 −2
= 9 × 10 Nm C
4πϵ0

Hence, force between the two small charged spheres is 6 × 10 −3
N . The charges are of same nature. Hence, force
between them will be repulsive.

Page : 46 , Block Name : Exercise

Q1.2 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 rst?

Answer. (a) Electrostatic force on the rst sphere, F = 0.2N
Charge on this sphere, q = 0.4μC = 0.4 × 10 C 1
−6

Charge on the second sphere, q = −0.8μ,C = −0.8 × 10 C 2
−6

Electrostatic force between the spheres is given by the relation,
q1 q2
F =
2
4πϵ0 r

Where, ϵ = Permittivity of free space
0

And, 1
= 9 × 10 Nm C
4πϵ0
9 2 −2

q1 q2
2
r =
4πϵ0 F
−6 −6 9
0.4×10 ×8×10 ×9×10
=
0.2
−4
= 144 × 10

−4
r = √144 × 10 = 0.12m

The distance between the two spheres is 0.12m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the rst is
0.2N.

Page : 46 , Block Name : Exercise

Q1.3 Check that the ratio ke /Gm m is dimensionless. Look up a Table of Physical Constants and determine the value
2
e ρ

of this ratio. What does the ratio signify?

Answer. The given ratio is .
2
ke

Gme mp

Where,
G = Gravitational constant

Its unit is Nm kg . 2 −2

m and m = Masses of electron and proton.
e p

Their unit is kg.
e = Electric charge.
Its unit is C.
k = A constant
1
=
4πϵ0

ϵ0 = Permittivity of free space
Its unit is Nm C . 2 −2

2 2 −2
[Nm C ][C ]

Therefore, unit of the given ratio
2
ke 0 0 0
= = M L T
Gme mp 2 −2
[Nm kg ][kg][kg]

Hence, the given ratio is dimensionless.
−19
e = 1.6 × 10 C
−11 2 −2
G = 6.67 × 10 Nm kg
−31
me = 9.1 × 10 kg
−27
mp = 1.66 × 10 kg

Hence, the numerical value of the given ratio is
∘ −19 2
2 9×10 ×(1.6×10 )
ke 39
= −11 −3 −22
≈ 2.3 × 10
Gme mp 6.67×10 ×9.1×10 ×1.67×10

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between
them constant.

Page : 46 , Block Name : Exercise

Q1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer. (a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be
transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only
in integral multiples of electric charge.
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge.
Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that
electric charge is continuous.

Page : 46 , Block Name : Exercise

Q1.5 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. Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are
created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed
bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a
silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of
conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Page : 46 , Block Name : Exercise

Q1.6 Four point charges q = 2μC, q = −5μC, q = 2μC and q = −5μC are located at the corners of a square
A B c D

ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?

Answer. The given gure shows a square of side 10 cm with four charges placed at its corners. O is the center of the
square.

Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10√2cm
AO = OC = DO = OB = 5√2cm
A charge of amount 1μC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction
relative to the force of repulsion between the charges placed at corner C and center O, Hence, they will cancel each other.
Similarly, force of attraction between charges placed at corner B and center O is equal in magnitude but opposite in
direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also
cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1μC charge at
centre O is zero.

Page : 46 , Block Name : Exercise

Q1.7 (a) An electrostatic eld line is a continuous curve. That is, a eld line cannot have sudden breaks. Why not?
(b) Explain why two eld lines never cross each other at any point?

Answer. (a) An electrostatic eld line is a continuous curve because a charge experiences a continuous force when traced
in an electrostatic eld. The eld line cannot have sudden breaks because the charge moves continuously and does not
jump from one point to the other.
(b) If two eld lines cross each other at a point, then electric eld intensity will show two directions at that point. This is
not possible. Hence, two eld lines never cross each other.

Page : 46 , Block Name : Exercise

Q1.8 Two point charges q = 3μC and q = −3μC are located 20 cm apart in vacuum. (a) What is the electric eld at
A B

the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10 is placed at this point, what is the force experienced by the test
−9

charge?

Answer. (a) The situation is represented in the given gure. O is the mid point of line AB.

Distance between the two charges, AB = 20 cm
∴ AO = OB = 10 cm

Net electric eld at point O = E
Electric eld at point O caused by +3μC charge,
−6 −6

along OB
3×10 3×10
E1 = 2
= N/C
2
4πϵ0 (AO) −2
4πϵ0 (10×10 )

Where,
E = Permittivity of free space
0

1 9 2 −2
= 9 × 10 Nm C
4πϵ0

Magnitude of electric eld at point O caused by −3μC charge,
−6 −6
∣ ∣
along OB
−3×10 3×10
E2 = = N/C
∣ 4πϵ0 (OB)2 ∣ −2
2
4πϵ0 (10×10 )

∴ E = E1 + E2

−6

= 2 × [(9 × 10 ) ×
9 3×10

−2 2
] [Since the values of E and E are same, the value is multiplied with 2]
1 2
(10×10 )

6
= 5.4 × 10 N/C along OB

Therefore, the electric eld at mid-point O is 5.4 × 10 NC along OB. 6 −1

(b) A test charge of amount 1.5 × 10 C is placed at mid-point O. −9

−9
q = 1.5 × 10 C

Force experienced by the test charge = F
∴ F = qE
−9 6
= 1.5 × 10 × 5.4 × 10
−3
= 8.1 × 10 N

The force is directed along line OA. This is because the negative charge is repelled by the charge placed at point B but
attracted towards point A. Therefore, the force experienced by the test charge is 8.1 × 10 N along OA. −3

Page : 46 , Block Name : Exercise

Q1.9 A system has two charges q = 2.5 × 10 C and q = −2.5 × 10 C located at points A: (0, 0, –15 cm) and B:
A
−7
B
−7

(0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Answer. Both the charges can be located in a coordinate frame of reference as shown in the given gure.

At A, amount of charge, q = 2.5 × 10 C
A
−7

At B, amount of charge, q = −2.5 × 10 C
B
−7

Total charge of the system,
q = qA + qB
7 −7
= 2.5 × 10 C − 2.5 × 10 C

= 0

Distance between two charges at point A and B,
d = 15 + 15 = 30cm = 0.3m

Electric dipole moment of the system is given by,
p = qA × d = qB × d
−7
= 2.5 × 10 × 0.3

= 7.5 × 10
−8
m long positive z-axis
C

Therefore, the electric dipole moment of the system is 7.5 × 10 −8
C m along positive z- axis.

Page : 46 , Block Name : Exercise

Q1.10 An electric dipole with dipole moment 4 × 10 C m is aligned at 30° with the direction of a uniform electric eld
−9

of magnitude 5 × 10 NC . Calculate the magnitude of the torque acting on the dipole.
4 −1

Answer. Electric dipole moment, p = 4 × 10 Cm −9

Angle made by p with a uniform electric eld, θ = 30 ∘

Electric eld, E = 5 × 10 NC 4 −1

Torque acting on the dipole is given by the relation,
T = pE sin θ
−9 4
= 4 × 10 × 5 × 10 × sin 30
−5 1
= 20 × 10 ×
2
−4
= 10 Nm

Therefore, the magnitude of the torque acting on the dipole is 10 .
−4
Nm

Page : 46 , Block Name : Exercise

Q1.11 A polythene piece rubbed with wool is found to have a negative charge of 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?

Answer. (a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene.
Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece, q = −3 × 10 C −7

Amount of charge on an electron, e = −1.6 × 10 C −19

Number of electrons transferred from wool to polythene = n
N can be calculated using the relation,
q = ne
q
n =
e
−7
−3×10
=
−19
−1.6×10

12
= 1.87 × 10

Therefore, the number of electrons transferred from wool to polythene is 1.87× 10 .
12

(b) Yes.
There is a transfer of mass taking place. This is because an electron has mass,
−3
me = 9.1 × 10 kg

Total mass transferred to polythene from wool,
m = me × n

−31 12
= 9.1 × 10 × 1.85 × 10

−18
= 1.706 × 10 kg

Hence, a negligible amount of mass is transferred from wool to polythene.

Page : 46 , Block Name : Exercise

Q1.12 (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 is 6.5 × 10 C ? The radii of A and B are negligible
−7

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?

Answer. (a) Charge on sphere A, q = Charge on sphere B, q = 6.5 × 10
A s
−7
C

Distance between the spheres, r = 50cm = 0.5m
Force of repulsion between the two spheres,
qλ qn
F =
4πϵ0 r2

Where,
E = Free space permittivity
0

1 9 2 −2
= 9 × 10 Nm C
4πϵ0

2
9 −7
9×10 ×(6.5×10 )
F =
2
(0.5)

−2
= 1.52 × 10 N

Therefore, the force between the two spheres is 1.52 × 10 −2
N .

(b) After doubling the charge, charge on sphere A, q A = Charge on sphere B, q B = 2 ×6.5 × 10
−7 −5
C = 1.3 × 10 C

The distance between the spheres is halved.
0.5
r=
∴ 2 = 0.25m

Force of repulsion between the two spheres,

qA qn
F =
2
4πϵ0 r

∘ −6 −6
9×10 ×1.3×10 ×1.3×10
= 2
(0.25)

−2
= 16 × 1.52 × 10

= 0.243N

Therefore, the force between the two spheres is 0.243 N.

Page : 46 , Block Name : Exercise

Q1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is
brought in contact with the rst, then brought in contact with the second, and nally removed from both. What is the
new force of repulsion between A and B?

Answer. Distance between the spheres, A and B, r = 0.5m
Initially, the charge on each sphere, q = 6.5 × 10 C −7

q
When sphere A is touched with an uncharged sphere C, amount of charge from A will transfer to sphere C. Hence,
2
q
charge on each of the spheres, A and C, is 2
.
q
When sphere C with charge 2
is brought in contact with sphere B with charge q, total charges on the system will divide
into equal halves given as,
q
+q 3q
2
=
2 4
3q
Each sphere will share each half. Hence, charge on each of the spheres, C and B, is 4
.
q
Force of repulsion between sphere A having charge 2
and sphere B having charge
q 3q

3q × 3q
2
2 4
= 2
= 2
4 4πϵ0 r 8×4πϵ0 r
2
−7
3×(6.5×10 )
9
= 9 × 10 ×
8×(0.5)2

−3
= 5.703 × 10 N

Therefore. The force of attraction between the two spheres is 5.703 × 10 −3
N .

Page : 47 , Block Name : Exercise

Q1.14 Figure shows tracks of three charged particles in a uniform electrostatic eld. Give the signs of the three charges.
Which particle has the highest charge to mass ratio?

Answer. Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2
both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two
particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and
repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of de ection for a given velocity.

Since the de ection of particle 3 is the maximum, it has the highest charge to mass ratio.

Page : 47 , Block Name : Exercise

Q1.15 Consider a uniform electric eld E = 3 × 10 ^iN/C. (a) What is the ux of this eld through a square of 10 cm on
3

a side whose plane is parallel to the yz plane? (b) What is the ux through the same square if the normal to its plane
makes a 60° angle with the x-axis?

→ = 3 × 10 ^iN/C
Answer. (a) Electric eld intensity, E
3

→
Magnitude of electric eld intensity, |E| 3
=3×10 N/C

Side of the square, s = 10cm = 0.1m
Area of the square, A = s = 0.01m 2 2

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and
electric eld, θ = 0 ∘

Flux (ϕ) through to the plane is given by the relation,
→
Φ = |E|A cos θ
3 ∘
= 3 × 10 × 0.01 × cos 0
2
= 30Nm /C

(b) Plane makes an angle of 60 with the x-axis. Hence, θ = 60
∘ ∘

→ cos θ
Flux, Φ = |E|A
3 ∘
= 3 × 10 × 0.01 × cos 60
1 2
= 30 × = 15Nm /C
2

Page : 47 , Block Name : Exercise

Q1.16 What is the net ux of the uniform electric eld of Exercise 1.15 through a cube of side 20 cm oriented so that its
faces are parallel to the coordinate planes?

Answer. All the faces of a cube are parallel to the coordinate axes. Therefore, the number of eld lines entering the cubes
is equal to the number of eld lines piercing out of the cube. As a result, net ux through the cube is zero.

Page : 47 , Block Name : Exercise

Q1.17 Careful measurement of the electric eld at the surface of a black box indicates that the net outward ux through
the surface of the box is 8.0 × 10 Nm /C. (a) What is the net charge inside the box? (b) If the net outward ux through
3 2

the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Answer. (a) Net outward ux through the surface of the box, Φ = 8.0 × 10 Nm /C 3 2

For a body containing bet charge q, ux is given by the relation,
q
ϕ =
ϵ0

E0 = Permittivity of free space
−12 −1 2 −2
= 8.854 × 10 N C m

q = ϵ0 Φ
−12 3
= 8.854 × 10 × 8.0 × 10
−8
= 7.08 × 10

= 0.07μC

Therefore. The net charge inside the box is 0.07 μC.

(b) Net ux piercing out through a body depends on the net charge contained in the body. If net ux is zero, then it can
be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Page : 47 , Block Name : Exercise

Q1.18 A point charge +10 µC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Figure.
What is the magnitude of the electric ux through the square? (Hint: Think of the square as one face of a cube with edge
10 cm.)

Answer. The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed.
According to Gauss’s theorem for a cube, total electric ux is through all its six faces.
q
ϕ rout =
ϵ0

ϕ thet q
Hence, electric ux through one face of the cube i.e., through the square, ϕ = 6
=
1

6 ϵ0

Where,
ϵ = Permittivity of free space
0

−12 −1 2 −2
= 8.854 × 10 N C m

−6
q = 10μC = 10 × 10 C
−6
1 10×10
ϕ = × −12
6 8.854×10

5 2 −1
= 1.88 × 10 Nm C

Therefore, electric ux through the square is 1.88 × 10 Nm C 5 2 −1
.

Page : 47 , Block Name : Exercise

Q1.19 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 ux
through the surface?

Answer. Net electric ux (Φ Net ) through the cubic surface is given by,
q
ϕN a =
ϵ0

Where,
E = Permittivity of free space
0

−12 −1 2 −2
= 8.854 × 10 N C m

q = Net charge contained inside the cube
−6
2.0μC = 2 × 10 C
−6
2×10
ϕNa = −12
8.854×10

5 2 −1
= 2.26 × 10 Nm C

The net electric ux through the surface is 2.26 × 10 Nm C 5 2 −1
.

Page : 48 , Block Name : Exercise

Q1.20 A point charge causes an electric ux of −1.0 × 10 Nm /C to pass through a spherical Gaussian surface of 10.0
3 2

cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much ux would pass
through the surface? (b) What is the value of the point charge?

Answer. (a) Electric ux, Φ = −1.0 × 10 Nm /C 3 2

Radius of the Gaussian surface,
r = 10.0cm

Electric ux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the
size of the body. If the radius of the Gaussian surface is doubled, then the ux passing through the surface remains the
same i.e., −10 Nm /C.
3 2

(b) Electric ux is given by the relation,
q
ϕ =
ϵ0

Where,
q = Net charge enclosed by the spherical surface

ϵ = Permittivity of free space = 8.854 × 10
−12 −1 2 −2
0 N C m

∴ q = ϕϵ0

3 −12
= −1.0 × 10 × 8.854 × 10

−9
= −8.854 × 10 C

= −8.854nC

Therefore, the value of the point charge is −8.854nC.

Page : 48 , Block Name : Exercise

Q1.21 A conducting sphere of radius 10 cm has an unknown charge. If the electric eld 20 cm from the centre of the
sphere is 1.5 × 10 N/C and points radially inward, what is the net charge on the sphere?
3

Answer. Electric eld intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the
relation,
q
E =
2
4πϵ0 d

Where,
3
q = Net charge = 1.5 × 10 NC

d = Distance from the centre = 20cm = 0.2m

ϵ0 = Permittivity of free space
And, 4πϵ0
1
= 9 × 10 Nm C
9 2 −2

2
q=E(4πϵ0 )d
∴
3 2
1.5×10 ×(0.2)
= 9
9×10

9
= 6.67 × 10 C

= 6.67nC

Therefore, the net charge on the sphere is 6.67 nC.

Page : 48 , Block Name : Exercise

Q1.22 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0μC/m . (a) Find the
2

charge on the sphere. (b) What is the total electric ux leaving the surface of the sphere?

Answer. (a) Diameter of the sphere, d = 2.4m
Radius of the sphere, r = 1.2m
Surface charge density, σ = 80.0μC/m = 80 × 10 2 −6
C/m
2

Total charge on the surface of the sphere,
Q = Charge density x Surface area
2
= σ × 4πr

−6 2
= 80 × 10 × 4 × 3.14 × (1.2)

−3
= 1.447 × 10 C

Therefore, the charge on the sphere is 1.447 × 10 C. −3

(b) Total electric ux (ϕ ) leaving out the surface of a sphere containing net charge Q is given by the relation,
rout

Q
ϕrow =
ϵ0

Where,
ϵ0 = Permittivity of free space

−12 −1 2 −2
= 8.854 × 10 N C m

−3
Q = 1.447 × 10 C
−3
1.44×10
ϕ Total = −12
8.854×10

8 −1 2
= 1.63 × 10 NC m

Therefore, the total electric ux leaving the surface of the sphere is 1.63 × 10 NC .
8 −1 2
m

Page : 48 , Block Name : Exercise

Q1.23 An in nite line charge produces a eld of 9 × 10 N/C at a distance of 2 cm. Calculate the linear charge density.
4

Answer. Electric eld produced by the in nite line charges at a distance d having linear charge density λ is given by the
relation,
λ
E =
2πϵ0 d

λ = 2πϵ0 dE

Where,
d = 2cm = 0.02m

4
E = 9 × 10 N/C

ϵ0 = Permittivity of free space

1 9 2 −2
= 9 × 10 Nm C
4πϵ0
4
0.02×9×10
λ =
9
2×9×10

= 10μC/m

Therefore, the linear charge density is 10μC/m.

Page : 48 , Block Name : Exercise

Q1.24 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 C/m . What is E: (a) in the outer region of the rst
−22 2

plate, (b) in the outer region of the second plate, and (c) between the plates?

Answer. The situation is represented in the following gure.

A and B are two parallel plates close to each other. Outer region of a plate A is labelled as I, outer region of plate B is
labelled as III , and the region between the plates, A and B, is labelled as II.
Charge density of plate A, σ = 17.0 × 10 C/m −22 2

Charge density of plate B, σ = −17.0 × 10 C/m −22 2

In the regios, I and III, electric eld E is zero. This is because charge is not enclosed by the respective plates.
Electric eld E in region II is given by the relation,
σ
E =
ϵ0

Where,
ϵ = Permittivity of free space = 8.854 × 10
−12 −1 2 −2
0 N C m
−22
17.0×10
E = −12
8.854×10
−10
= 1.92 × 10 N/C

Therefore, electric eld between the plates is 1.92 × 10 −10
N/C .

Page : 48 , Block Name : Exercise

Q1.25 An oil drop of 12 excess electrons is held stationary under a constant electric eld of 2.55 × 10 NC 4 −1
(Millikan’s
oil drop experiment). The density of the oil is 1.26 g cm . Estimate the radius of the drop. (
−3

C).
−2 −19
g = 9.81ms ; e = 1.60 × 10

Answer. Excess electrons on an oil drop, n = 12
Electric eld intensity, E = 2.55 × 10 NC 4 −1

Density of oil, ρ = 1.26gm/cm = 1.26 × 10 kg/m3 3 3

Acceleration due to gravity, g = 9.81ms −2

Charge on an electron, e = 1.6 × 10 C −19

Radius of the oil drop = r
Force ( F) due to electric eld E is equal to the weight of the oil drop (W)
F=W
Eq = mg
Ene = = πr × ρ × g4

3
3

Where,
q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

4 3
= πr × ρ
3

3
3Ene
∴ r = √
4πρg

4 −19
3 3×2.55×10 ×12×1.6×10
= √
3
4×3.14×1.26×10 ×9.81

3
√ −21
= 946.09 × 10

−7
= 9.82 × 10 m
−4
= 9.82 × 10 mm

Therefore, the radius of the oil drop is 9.82 × 10
−4
mm

Page : 48 , Block Name : Additional Exercise

Q1.26 Which among the curves shown in Figure cannot possibly represent electrostatic eld lines?

Answer. (a) The eld lines showed in (a) do not represent electrostatic eld lines because eld lines must be normal to
the surface of the conductor.
(b) The eld lines showed in (b) do not represent electrostatic eld lines because the eld lines cannot emerge from a
negative charge and cannot terminate at a positive charge.
(c) The eld lines showed in (c) represent electrostatic eld lines. This is because the eld lines emerge from the positive
charges and repel each other.
(d) The eld lines showed in (d) do not represent electrostatic eld lines because the eld lines should not intersect each
other.
(e) The eld lines showed in (e) do not represent electrostatic eld lines because closed loops are not formed in the area
between the eld lines.

Page : 48 , Block Name : Additional Exercise

Q1.27 In a certain region of space, electric eld is along the z-direction throughout. The magnitude of electric eld is,
however, not constant but increases uniformly along the positive z-direction, at the rate of 10 NC per metre. What
5 −1

are the force and torque experienced by a system having a total dipole moment equal to 10 Cm in the negative z-
−7

direction ?

Answer. Dipole moment of the system, p = q × dl = −10 −7
C m
Rate of increase of electric eld per unit length,
dE +5 −1
= 10 NC
dl

Force ( F) experienced by the system is given by the relation,
F = qE
dE
F = q × dl
dl
dE
= p ×
dl
−7 −5
= −10 × 10

−2
= −10 N

The force is −10 N in the negative direction i.e., opposite direction of electric eld. Hence, the angle between electric
−2

eld and dipole moment is 180 . ∘

Torque (T ) is given by the relation,
∘
T = pE sin 180

=0
Therefore, the torque experienced by the system is zero.

Page : 49 , Block Name : Additional Exercise

Q1.28 (a) A conductor A with a cavity as shown in Figure (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 [Figure (b)]. (c) A sensitive instrument is
to be shielded from the strong electrostatic elds in its environment. Suggest a possible way.

Answer. (a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The
electric eld intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and E is the permittivity of free space. According to Gauss’s law,
0

→ →
ϕ = E ⋅ ds =
q

ϵ0

Flux,
Here, E = 0
q
= 0
ϵ0

∵ ϵ0 ≠ 0

∴ q = 0

Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
(b) The outer surface of the conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside
conductor A and it is insulated from A. Hence, a charge of amount -q will be induced in the inner surface of conductor A
and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q
+q.
(c) A sensitive instrument can be shielded from the strong electrostatic eld in its environment by enclosing it fully
inside a metallic surface. A closed metallic body acts as an electrostatic shield.

Page : 49 , Block Name : Additional Exercise

Q1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric eld in the hole is ( σ

2ϵ0
) ,
^
n

where n
^ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Answer. Let us consider a conductor with a cavity or a hole. Electric eld inside the cavity is zero.
Let E is the electric eld just outside the conductor, q is the electric charge, σ is the charge density, and E is the
0

permittivity of free space.
Charge |q| = σ→×α → s→
According to Gauss’s law,
→ ⋅ ds
Flux, ϕ = E → = |q|

ϵ0

→
σ×ds →
Eds =
ϵ0

σ
∴ E = ^
n
ϵ0

Therefore, the electric eld just outside the conductor is σ

ϵ0
^
n . This eld is a superposition of eld due to the cavity (E)
and the eld due to the rest of the charged conductor (E). These elds are equal and opposite inside the conductor, and
equal in magnitude and direction outside the conductor.
′ ′
∴ E + E = E

E
′
E =
2
σ
= ^
n
2ϵ0

Therefore, the eld due to the rest of the conductor is σ

ϵ0
^
n .
Hence, proved.

Page : 49 , Block Name : Additional Exercise

Q1.30 Obtain the formula for the electric eld due to a long thin wire of uniform linear charge density E without using
Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Answer. Take a long thin wire XY (as shown in gure) of uniform linear charge density λ.

Consider a point A at a perpendicular distance / from the mid-point O of the wire, as shown in the following gure.

Let E be the electric eld at point A due to the wire, XY.
Consider a small length element dx on the wire section with OZ = x
Let q be the charge on this piece.
∴ q = λdx

Electric eld due to the piece,
1 λdx
dE = 2
4πϵ0 (AZ)

However, AZ = √(l 2 2
+ x )
λdx
∴ dE = 2 2
4πϵ0 (l +x )

The electric eld is resolved into two rectangular components. dE cos θ is the perpendicular component and dE sin θ is
the parallel component.
When the whole wire is considered, the component dE sin θ is cancelled.
On;y the perpendicular component dE cos θ affects point A.
Hence, effective electric eld at point A due to the element dx is dE . 1

∴ dE1 =
λdx cos θ

2 2
… (1)
4πϵ0 (x +l )

ln Δ AZO.
x
tan θ =
l

x = l tan θ (2) …

On differentiating equation (2), we obtain
dx 2
= l sin θ
dθ

dx = l sin
2
θdθ … (3)
From equation (2),
2 2 2 2 2
x + l = l + l tan θ
2 2 2 2
∴ l (1 + tan θ) = l sec θ

x
2
+ l
2
= l (4) 2
sin
2
θ …

Putting equations (3) and (4) in equation (1), we obtain
2
λl sec dθ
∴ dE1 = × cos θ
2 2
4πϵ0 l sec θ

(5)
λ cos θdθ
∴ dE1 = …
4πϵ0 l

The wire is so long that θ tends from − π

2
+
π

2
.
to

By integrating equation (5), we obtain the value of eld E as, 1
π π

2 2 λ
∫ π dE1 = ∫ π cos θdθ
− 4πϵ0 l
2 2
π
λ
2
E1 = [sin θ] π
−
4πϵ0 / 2

λ
= × 2
4πϵ0 l
λ
E1 =
2πϵ0 l

Therefore, the electric eld due to long wire is λ

2πϵ0 l
.

Page : 49 , Block Name : Additional Exercise

Q1.31 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.

Answer. A proton has three quarks. Let there be n up quarks in a proton, each having a charge of + 2

3
e .
Charge due to n up quarks = ( 2

3
e) n

Number of down quarks in a proton = 3 - n
Each down quark has a charge of − e 1

3

1
= (− e) (3 − n)
3

Charge due to (3 - n) down quarks
Total charge on a proton = + e
2 1
∴ e = ( e) n + (− e) (3 − n)
3 3

2ne ne
e = ( ) − e +
3 3

2e = ne

n = 2

Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3 - n = 3 - 2 = 1
Therefore, a proton can be represented as ‘udd’.
A neutron also has three quarks. Let there be n up quarks in a neutron, each having a charge of + 3

2
e .

Charge on a neutron due to n up quarks = (+ 3

2
e) n

Number of down quarks is 3 - n,each having a charge of (− 1

3
)e

Charge on a neutron due to (3 − n) down quarks = (− 1

3
e) (3 − n)

Total charge on a neutron = 0
2 1
0 = ( e) n + (− e) (3 − n)
3 3

2 ne
0 = en − e +
3 3

e = ne

n = 1

Number of up quarks in a neutron, n = 1
Number of down quarks in a neutron = 3 - n = 2
Therefore, a neutron can be represented as ‘udd’.

Page : 49 , Block Name : Additional Exercise

Q1.32 (a) Consider an arbitrary electrostatic eld con guration. A small test charge is placed at a null point (i.e., where E
= 0) of the con guration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the
simple con guration of two charges of the same magnitude and sign placed a certain distance apart.

Answer. (a) Let the equilibrium of the test charge be stable. If a test is in equilibrium and displaced from its position in
any direction, then it experiences a restoring force towards a null point, where the electric eld is zero. All the eld lines
near the null point are directed inwards towards the null point. There is a net inward ux of electric eld through a
closed surface around the null point. According to Gauss’s law, the ux of electric eld through a surface, which is not
enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of
the charges is the nul point. When a test charged is displaced along the long line, it experiences a restoring force. If it is
displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable
because stability of equilibrium requires restoring force in all directions.

Page : 50 , Block Name : Additional Exercise

Q1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-
axis with speed v (like particle 1 in Figure). The length of plate is L and an uniform electric eld E is maintained
x

between the plates. Show that the vertical de ection of the particle at the far edge of the plate is qEL / (2m .
2
2 vr
)

Answer. Charge on a particle of mass m = - q
Velocity of the particle = = v x

Length of the plates = L
Magnitude of the uniform electric eld between the plates = E
Mechanical force, F = Mass ( m) x Acceleration (a)
F
a =
m

However, electric force, F = qE
Therefore, acceleration, a = (1)
qE
…
m

Time taken by the particle to cross the eld of length L is given by,
t= (2)
Length of the plate L
= …
Velocity of the particle vx

In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical de ection s of the particle can be obtained as,
1
2
s = ut + at
2
2
1 qE L
s = 0 + ( )( )
2 m vx

2

...(3)
qEL
s =
2
2mVx

Hence, vertical de ection of the particle at the far edge of the plate is
qEL / (2mv ). This is similar to the motion of horizontal projectiles under gravity.
2 2
x

Page : 50 , Block Name : Additional Exercise

Q1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity v v = 2.0 × 10 ms . If E
6 −1
x

between the plates separated by 0.5 cm is 9.1 × 10 N/C, where will the electron strike the upper plate? (
2

|e| = 1.6 × 10
−19
C, me = 9.1× 10
−31
kg .)

Answer. Velocity of the particle, V = 2.0 × 10 m/s x
6

Seperation of the two plates, d = 0.5cm = 0.005m
Electric eld between the two plates, E = 9.1 × 10 N/C 2

Charge on an electron, q = 1.6 × 10 C −19

Mass of an electron, m = 9.1 × 10 kg e
−31

Let the electron strike the upper plate at the end of the plate L, when de ection is s.
Therefore,
2
qEL
s =
2
2mvx

2
2dmvr
L = √
qE

2
−31 6
2×0.005×9.1×10 ×(2.0×10 )
√
= −10 2
1.6×10 ×9.1×10

√ −2 √ −4
= 0.025 × 10 = 2.5 × 10

−2
= 1.6 × 10 m

= 1.6cm

Therefore, the electron will strike the upper plate after travelling 1.6 cm.

Page : 50 , Block Name : Additional Exercise