NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Physics
Chapter : 2
Chapter Name : Electrostatic Potential And Capacitance

Q2.1 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 in nity to be zero.

Answer. There are two charges
q 1 = 5 × 10 − 8C
q 2 = − 3 × 10 − 8C
Distance between the two charges, d=16 cm = 0.16
Consider a point P on the line joining the two changes, as shown in the given gure

r= Distance of point from charge q 1
Let the electric potential (V) at point P be zero.
Potential at point P is the sum of potential caused by charges
q 1 and q 2 respectively.
q1 q2
∴ V = 4πϵ r + 4πϵ ( d − r ) ....(i)
0 0
where,
ϵ 0 = Permittivity of free space
For V=0, equation (i) reduces to
q1 q2
4πϵ r
= − 4πϵ ( d − r )
0 0
q1 − q2
r
= d−r

5 × 10 − 8 ( − 3 × 10 ) −8

r = − ( 0.16 − r )
0.16 3
r
−1= 5
0.16 8
r
= 5
∴ r = 0.1m = 10cm
Therefore the potential is zero at a distance of 10 cm from the positive charge between the charges,
Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential
is zero, as shown in the following gure.

For this arrangement, Potential is given by,
q1 q2
V = 4πϵ s + 4πϵ ( s − d ) ….(ii)
0 0
For V=0, equation (ii) reduces to
q1 q2
4πϵ s = − 4πϵ ( s − d )
0 0
q1 q2
4πϵ s
= − 4πϵ ( s − d )
0 0

5 × 10 − 8 ( − 3 × 10 ) −8

s
= − ( s − 0.16 )
0.16 3
1− s = 5
0.16 2
s
= 5

∴ s = 0.4m = 40cm
Therefore the potential is zero at a distance of 40 cm from the positive charge outside of the system of charges.

Page : 86 , Block Name : Exercise

Q2.2 A regular hexagon of side 10 cm has a charge 5μC at each of its vertices Calculate the potential of the
centre of the hexagon.

Answer. The given gure shows sequi nourt of wages, the vertices of a regular charge, to Site of die hexagon.

Where,
Charge, q=5 μC = 5 × 10 − 6C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
6×q
V = 4πϵ d
0
Where,
ϵ 0 = Permittivity of free space

1
4πϵ 0
= 9 × 10 9NC − 2m − 2
6 × 9 × 10 9 × 5 × 10 − 6
∴V= 0.1
6
= 2.7 × 10 V
Therefore, the potential of the centre of the hexagon is
= 2.7 × 10 6V.

Page : 86 , Block Name : Exercise

Q2.3 Two charges 2μC and − 2μC are placed at points A and B 6 m apart.
(a) Identity an equipotential surface of the system. b) What is the direction of the electric eld at every point
on this surface?

Answer. (a) The situation is represented in the given gure.

An equipotential surface is the plane on which total potential is zero everywhere. This place is normal to the
line AB. The plane s located at the midpoint of line AB because the magnitude of charges is the same.

(b) The direction of the electric eld at every point on this surface is normal to the plane in the direction of AB.

Page : 86 , Block Name : Exercise

Q2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10 − 7C distributed uniformly on its surface.
What is the electric eld.
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?

Answer.
(a) Radius of the spherical conductor, r = 12cm = 0, 12m
Charge is uniformly distributed over the conductor, q = 1.6 × 10 − 7C
Electric eld inside the conductor is zero. This is because if there is a eld inside the conductor. Then charges
will move to neutralize it.

(b) Electric eld E just outside the conductor is given by the relation,
q
E=
4πϵ 0r 2
Where,
ϵ 0 = Permittivity of free space
1
4πϵ 0
= 9 × 10 9Nm 2C − 2
1.6 × 10 − 7 × 9 × 10 − 9
∴E=
( 0.12 ) 2
= 10 5NC − 1

Therefore, the electric eld just outside the sphere is 10 5NC − 1.

(c) Electric Held at point 18 m from the centre of the sphere = E 1
Distance of the point from the centre, d = 18 cm = 0.18 m
q
E1 =
4πϵ 0d 2
9 × 10 9 × 1.6 × 10 − 7
=
( 18 × 10 ) −2 2

= 4.4 × 10 4N / C
Therefore, the electric eld at a point 18 cm from the centre of the sphere is 4.4 × 10 4N / C.

Page : 86 , Block Name : Exercise

Q2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF ( 1 pF = 10 − 12F ). What will
be the capacitance if the distance between the plates is reduced by half, and the space between them is lled
with a substance of dielectric constant 6?

Answer. Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was lled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
kϵ 0A
C= d
ϵ 0A
= d ….(i)
Where,
A = Area of each plate
ϵ 0 = Permittivity of free space
If distance between the plates is reduced to half, then new distance
d
d= 2
Dielectric constant of the substance lled in between the plates, k = 6
Hence,
Capacitance of the capacitor becomes
kϵ 0A 6ϵ 0A
C′ = d = d …(ii)
2
Taking ratio of equations (i) and (ii), we obtain
C ′ = 2 × 6C
= 12 C
= 12 x 8 = 96 pF
Therefore, the capacitance between the plates is 96 pF.

Page : 86 , Block Name : Exercise

Q2.6 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?

Answer. (a) Capacitance of each of the three capacitors, C = 6 pF

( )
Equivalent capacitance C ′ of the combination of the capacitors is given by the relation,

1 1 1 1
= C + C + C
C′
1 1 1 3 1
= 9 + 9 + 9 = 9 = 3
∴ C ′ = 3μF
Therefore, total capacitance of the combination is 3μF.

(b) Supply voltage, V = 100 V
Potential difference ( V ′ ) across each capacitor is equal to one-third of the supply voltage.
V 120
∴ V ′ = 3 = 3 = 40V
Therefore, the potential difference across each capacitor is 40 V.

Page : 86 , Block Name : Exercise

Q2.7 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.

Answer. (a) Capacitance of the given capacitors are
C 1 = 2 pF
C 2 = 3 pF
C 3 = 4 pF
For the parallel combinations of the capacitors, equivalent capacitors 9C ′ is given by the algebraic sum,
C ′ = 2 + 3 + 4 = 9 pF
Therefore, total capacitance of the combination is 9pF.

(b) Supply voltage , V = 100 V
The voltage through all the three capacitors is Same = V = 100V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q =VC ….(i)
For C = 2 pF,
Charge = VC = 100 x 2 = 200 pC = 2 × 10 − 10C
For C = 3 pF,
Charge = VC = 100 x 3 = 200 pC = 3 × 10 − 10C
For C = 4 pF,
Charge = VC = 100 x 4 = 200 pC = 4 × 10 − 10C.

Page : 86 , Block Name : Exercise

Q2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10 − 3m 2 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?

Answer. Area of each plate of the parallel plate capacitor, A = 6 × 10 − 3m 2
Distance between the plates, d = 3 mm = 3 × 10 − 3m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by,
ϵ 0A
C= d
Where,

ϵ 0 = Permittivity of free space
= 8.854 × 10 − 12N − 1m − 2C − 2
8.854 × 10 − 12 × 6 × 10 − 3
∴C=
3 × 10 − 3
− 12
= 17.71 × 10 F
= 17.71 pF
Potential V is related with the charge q and capacitance C as
q
V= C
∴ q = VC
= 100 × 17.71 × 10 − 12
= 1.771 × 10 − 9C
Therefore, capacitance of the capacitor is 17.71 pF and the charge on each plate is = 1.771 × 10 − 9C.

Page : 86 , Block Name : Exercise

Q2.9 Explain what would happen if in the capacitor given in Exercise 2.8, 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.

Answer. (a) Dielectric constant of the mica sheet, k =6
Initial Capacitance, C = C = 1.771 × 10 − 11F
New capacitance, C ′ = kC = 6 × 1.771 × 10 − 11 = 106pF
Supply Voltage, V =100 V
New Charge, q ′ = C ′V = 6 × 1.771 × 10 − 9 = 1.06 × 10 − 8C
Potential across the plates remains 100 V.

(b) Dielectric constant, k =6
Initial Capacitance, C = C = 1.771 × 10 − 11F
New capacitance, C ′ = kC = 6 × 1.771 × 10 − 11 = 106pF

If supply Voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 1.771 × 10 − 9C
Potential across the plates is given by,
q
∴ V′ =
C′
1.771 × 10 − 9
=
106 × 10 − 12
= 16.7 V

Page : 87 , Block Name : Exercise

Q2.10 A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the
capacitor?

Answer. Capacitor of the capacitance, C = 12 pF = 12 × 10 − 12F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
1
E = 2 CV 2

= 1.5 × 10 − 8J
Therefore, the electrostatic energy stored in the capacitor is = 1.5 × 10 − 8J

Page : 87 , Block Name : Exercise

Q2.11 A 600 pF capacitor is charged by a 200 V 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?

Answer. Capacitance of the capacitor, C = 600 pF
Potential Difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
1
E = 2 CV 2

( )
1
= 2 × 600 × 10 − 12 × (200) 2
−5
= 1.2 × 10 J
If supply is disconnected from the capacitor and another capacitor of
capacitance C = 600 pF is connected to it, then equivalent capacitance
( C ′ ) of the combination is given by,
1 1 1
C ′ = C + C
1 1 2 1
= 600 + 600 = 600 = 300
∴ C ′ = 300pF
New electrostatic energy can be calculated as
1
E′ = 2 × C′ × V2
1
= 2 × 300 × (200) 2
= 0.6 × 10 − 5J
Loss in electrostatic energy lost = E − E ′
= 1.2 × 10 − 5 − 0.6 × 10 − 5
= 0.6 × 10 − 5
= 6 × 10 − 6J
Therefore, the electrostatic energy lost in the process is = 6 × 10 − 6J.

Page : 87 , Block Name : Exercise

Q2.12 A charge of 8 mC is located at the origin. Calculate the work done intaking a small charge of − 2 × 10 − 9
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).

Answer. Charge located at the origin, q = 8 mC = 8 × 10 − 3C
Magnitude of a small scharge, which is taken from a point P to point R to point Q, q 1 = − 2 × 10 − 9C
All the points are represented in the gure.

Point P at a distance, d 1 = 3cm, from the origin along z -axis.
Point P at a distance, d 2 = 4cm, from the origin along y -axis.
q
Potential at point P, V 1 = 4πϵ × d
0 1
q
Potential at point Q, V 2 = 4πϵ d
0 2
Work done (W) by the electrostatic force is independent of the path.
∴ W = q1 V2 − V1[ ]

[ q
= q 1 4πϵ d − 4πϵ d
0 2 0 1
q
]
qq 1
= 4πϵ
0 [ ]
1
d2

1
− d
1
1
….(i)

Where, 4πϵ = 9 × 10 9Nm 2C − 2
0

( )[ ]
1 1
∴ W = 9 × 10 9 × 8 × 10 − 3 × − 2 × 10 − 9 0.04
− 0.03

= − 144 × 10 − 3 × ( ) − 25
3

= 1.27 J
Therefore, work done during the process is 1.27 J.

Page : 87 , Block Name : Additional Exercise

Q2.13 A cube of side b has a charge q at each of its vertices. Determine the potential and electric eld due to
this charge array at the centre of the cube.

Answer. Length of the side of a cube = b
Charge at each of its vertices = q
A cube of side b is shown in the following gure.

d = Diagonal of one of the six faces of the cube
d2 = √b 2 + b 2 = √2b 2
d = b√2
l = Length of the diagonal of the cube
√d 2 + b 2
l2 =

= √(√2b) 2 + b 2 = √2b 2 + b 2 = √3b 2
l = b√3
l b√ 3
r = 2 = 2 is the distance between the centre of the cube and one of the eight vertices.
8q
V = 4πϵ
0
8q
=
4πϵ 0 b 2
( ) √3

4q
=
√3πϵ 0b
4q
Therefore, the potential at the centre of the cube is = .
√3πϵ 0b
The electric eld at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges
are distributed symmetrically with respect to the centre of the cube. Hence, The electric eld is zero at the
centre.

Page : 87 , Block Name : Additional Exercise

Q2.14 Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and
electric eld:
(a) at the midpoint 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 midpoint.

Answer. Two charges placed at points A and B are represented in the given qure. O is the midpoint of the line
joining the two charges.

Magnitude of charge located at A, q 1 = 1.5μC
Magnitude of charge located at B, q 2 = 2.5μC
Distance between the two charges, d = 30 cm = 0.3 m

(a) Let V 1 and E 1 are the electric potential and electric eld respectively at O.
V 1 = Potential due to charge at A + Potential due to charge B
q1 q2 1
V1 = + = (q 1 + q 2 )
4πϵ 0 ()
d
2 4πϵ 0 () d
2 4πϵ 0 () d
2

Where,
ϵ 0 = Permittivity of free space
1
4πϵ 0
= 9 × 10 9NC 2m − 2
9 × 10 9 × 10 − 6
∴ V1 = (2.5 + 1.5) = 2.4 × 10 5V
( ) 0.30
2

E 1 = Electric field due to q 2 − Electric field due to q 1
q2 q1
= −
4πϵ 0 () d
2
2
4πϵ 0 () d
2
2

= 4 × 10 5Vm − 1
Therefore, the potential at midpoint is 2.4 × 10 5V and the electric eld at mid point is 4 × 10 5Vm − 1. The eld is
directed from the larger charge to the sameller charge.

(b) Consider a point Z such that the normal distance OZ = 10 cm = 0.1 m, as shown in the following gure.

V 2 and E 2 are the electric potential and electric eld at Z.
It can be observed from the gure that distance,
BZ = AZ = √(0.1) 2 + (0.15) 2 = 0.18m
V 2 = Electric potential due to A + Electric Potential due to B
q1 q1
= 4πϵ ( AZ ) + 4πϵ ( BZ )
0 0
9 × 10 9 × 10 − 6
= 0.18
(1.5 + 2.5)
= 2 × 10 5V
Electric eld due to q at Z,
q1
EA =
4πϵ 0 ( AZ ) 2
9 × 10 9 × 1.5 × 10 − 6
=
( 0.18 ) 2
= 0.69 × 10 Vm − 1
6

The resultant eld intensity at Z,
2 2
E=
√E + E + 2E E cos2θ
A B A B

Where, 2θ is the angle, ∠AZB

From the gure we obtain
0.10 5
cosθ = 0.18 = 9 = 0.5556
θ = cos − 1 ∘ 0.5556 = 56.25
∴ 2θ = 112.5 ∘
cos2θ = − 0.38

E= √ (0.416 × 106 )2 × (0.69 × 106 )2 + 2 × 0.416 × 0.69 × 1012 × ( − 0.38)
= 6.6 × 10 3Vm − 1
Therefore , the potential at a point 10 cm (perpendicular to the midpoint) is 2.0 × 10 5V and electric eld is
= 6.6 × 10 3Vm − 1.

Page : 87 , Block Name : Additional Exercise

Q2.15 A spherical conducting shell of inner radius r 1 and outer radius r 2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer
surfaces of the shell?

(b) Is the electric eld inside a cavity (with no charge) zero, even if the shell is not spherical, but has any
irregular shape? Explain.

Answer. (a) charge placed at the centre of a shell is +q. Hence, a charge of magnitude -q will be induced to the
inner surface of the shell Therefore, total charge on the inner surface of the shell is -q.
Surface charge density at the inner surface of the shell is given by the relation,
Total charge −q
σ 1 = Inner surface area = ….(i)
4πr 21

A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer
surface of the shell. Therefore, total charge on the outer surface of the shells Q + q. Surface charge density at
the outer surface of the shell,
Total charge Q+q
σ 2 = Outer surface area = 2 ….(ii)
4πr 2

(b) Yes
The electric eld intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.
Take a closed loop such that a part of it is inside the cavity along a eld line while the rest is inside the
conductor. Net work done by the eld in carrying a test charge over a closed loop is zero because the eld
inside the conductor is zero. Hence, electric eld zero, whatever is the shape.

Page : 87 , Block Name : Additional Exercise

Q2.16 (a) Show that the normal component of electrostatic eld has a discontinuity from one side of a charged
σ
surface to another given by E 2 − E 1 ⋅ n̂ = ε (
0
)
where n̂ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The
direction of n̂ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric eld is σ n̂ /ε0.
(b) Show that the tangential component of electrostatic eld 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 eld on a closed
loop is zero.]

Answer. (a) Electric Field on one side of a charged body is E 1 and electric eld on the other side of the same
body is E 2. If in nite plane charged body has a uniform thickness, then electric eld due to one surface of the
charged body is given by
→ σ
E 1 = − 2ϵ n̂ …(i)
0
Where,
n̂ - Unit vector normal to the surface at a point
σ - Surface charge density at that point
Electric eld due to the surface of the charged body,
→ σ
E 2 = − 2ϵ n̂ …(ii)
0
Electric feld at way point due to the two surfaces,
→ → σ σ σ
E 2 − E 1 = 2ϵ n̂ + 2ϵ n̂ = ϵ n̂
0 0 0
σ
(E 2 − E 1 ) ⋅ n̂ = ϵ ..(iii)
→ →

0

→
Since inside a closed conductor, E 1 = 0

→ σ
→
∴ E = E 2 = − 2ϵ n̂
0
σ
Therefore, the eld just outside the conductor is ϵ n̂
0

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the
electrostatic eld is zero. Hence, the tangential component of electrostatic eld is continuous from one side of
charged surface to the other.

Page : 87 , Block Name : Additional Exercise

Q2.17 A long charged cylinder of linear charged density λ is surrounded by a hollow coaxial conducting
cylinder. What is the electric eld in the space between the two cylinders?

Answer. Charge density of the long charged cylinder of length L and radius r is λ.
Another cylinder of same length surrounds the previous cylinder. The radius of this cylinder is R.
Let E be the electric eld produced in the space between the two cylinders.
Electric ux through the Gaussian surface is given by the Gauss’s theorem as,
ϕ = E(2πd)L
Where, d = Distance of a point from the common axis of the cylinders
Let q be the total charge on the cylinder.
It can be written as
q
∴ ϕ = E(2πdL) = ϵ
0
Where,
Q = charge on the inner surface of the outer cylinder
ϵ 0 = Permittivity of free space
λL
E(2πdL) = ϵ
0
λ
E = 2πϵ d
0
λ
Therefore, te electric led in the space between the cylinder is 2πϵ d .
0

Page : 88 , Block Name : Additional Exercise

Q2.18 In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at in nite
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 Å separation?

Answer. The distance between electron-proton of a hydrogen atom, d = 0.53 Å
Charge on an electron, q 1 = − 1.6 × 10 − 19C
Charge on a proton, q 2 = + 1.6 × 10 − 19C

(a) Potential at in nity is zero.
Potential energy of the system, p-e = Potential energy at in nity - Potential energy at distance, d.
q 1q 2
= 0 − 4πϵ d
0

Where,
ϵ 0 = Permittivity of free space

1
4πϵ 0
= 9 × 10 9Nm 2C − 2

(
9 × 10 9 × 1.6 × 10 − 19 ) 2

∴ Potential energy = 0 − = − 43.7 × 10 − 19J
0.53 × 10 10

since 1.6 × 10 − 19J = 1eV
− 43.7 × 10 − 19
∴ Potential energy = − 43.7 × 10 − 19 = = − 27.2eV
1.6 × 10 − 19
Therefore, the potential energy of the system is -27.2 eV.

(b) Kinetic energy is half of the magnitude of the potential energy.
1
Kinetic Energy = = 2 × ( − 27.2) = 13.6eV
Total energy = 13.6 - 27.2 = 13.6 eV.
Therefore minimum work required to free the electron is 13.6 eV.

(c) When zero of potential energy is taken, d 1 = 1.06 Å
∴ Potential energy of the system = Potential energy at d 1 - Potential energy at d
q 1q 2
= 4πϵ d − 27.2eV
0 1

(
9 × 10 9 × 1.6 × 10 − 19 ) 2

= − 27.2eV
1.06 × 10 − 10

= 21.73 × 10 − 19J − 27.2eV
= 13.58eV − 27.2eV
= − 13.6eV

Page : 88 , Block Name : Additional Exercise

+
Q2.19 If one of the two electrons of a H 2 molecule is removed, we get a hydrogen molecular ion H 2 . In the
ground state of an H 2+ , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from
each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer. The system of two protons and on electron is represented in the given gure.

Charge on proton 1, q 1 = 1.6 × 10 − 19C
Charge on proton 2, q 2 = 1.6 × 10 − 19C
Charge on electron, q 3 = − 1.6 × 10 − 19C
Distance between protons 1 and 2, d 1 = 1.5 × 10 − 10m
Distance between proton 1 and electron, d 2 = 1 × 10 − 10m
Distance between proton 2 and electron, d 3 = 1 × 10 − 10m
The potential energy at in nity is zero.
Potential energy of the system,
q 1q 2 q zq 3 q 1q 1
V = 4πϵ d + 4πϵ d + 4πϵ d
0 1 0 3 0 2
1
Substituting 4πϵ = 9 × 10 ∘ Nm 2C − 2, we obtain
0

V=
9 × 10 9 × 10 − 19 × 10 − 19
10 − 19 [ − (16) 2 +
( 1.6 ) 2
1.5
+ − (1.6) 2
]
= − 30.7 × 10 − 10J
= − 19.2eV

Therefore the potential energy of the system is - 19.2 eV.

Page : 88 , Block Name : Additional Exercise

Q2.20 Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio
of electric elds 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 atter portions.

Answer. Let a be the radius of a sphere A, Q A be the charge on the sphere, and C A be the capacitance of the
sphere. Let b be the radius of a sphere B, Q B be the charge on the sphere, and C B be the capacitance of the
sphere. Since the two spheres are connected with a wire, there potential (V) will become equal.
Let E A be the electric eld of sphere A and E B be the electric eld of sphere B. Therefore the ratio,

EA QA b 2 × 4πϵ 0
EB
= 4πϵ × a × QB
0 2
EA QA b2
EB
= Q × …(1)
B a2

QA C AV
However, Q = C V
B B

CA a
And, C = b
B
QA a
∴ Q = b …(2)
s
Putting the values of (1) and (2), we obtain
EA a b2 b
∴ E = b 2 = a
B a
b
Therefore, the ratio of the electric elds at the surface is a .

Page : 88 , Block Name : Additional Exercise

Q2.21 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 >> 1.
(c) How much work is done in moving a small test charge from thepoint (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?

Answer. (a) Zero at both the points
Charge -q is located at (0, 0, -a) and charge +q is located (0, 0, a). Hence, they form a dipole. Point( 0, 0, z ) is
on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole Hence, electrostatic potential at
point (x, y, 0) is zero.
Electrostatic potential at point (0, 0, z) is given by,

V=
1
( )
q
4πϵ 0 z − a
+
1
4πϵ 0 ( −
q
z+a )
q(z + a − z + a)
=
(
4πϵ 0 z 2 − a 2 )
2qa p
= =
(
4πϵ 0 z 2 − a 2 ) 4πϵ (z − a )
0
2 2

Where,
ϵ 0 = permittivity of free space
p = Dipole moment of the system of two charges = 2 qa

(b) Distance r is much greater than half of the distance between the two charges. Hence the potential (V) at a
distance r is inversely proportional to square of the distance
1
i.e., V ∝
r2

(c) Zero
The answer does not change if the oath of the test is not along the x-axis.

A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x axis.

( )
Electrostatic potential V 1 at point (5, 0, 0) is given by,
−q 1 q 1
V1 = +
4πϵ 0 (5 − 0) + ( − a)
2 2 4πϵ 0 (5 − 0) 2 + a 2
√
−q q
25 2 + a 2 +
4πϵ 0 √
=
4πϵ 0 √25 + a 2
=0

Electrostatic potential, V 2 , at point (-7, 0, 0) is given by,
−q 1 q 1
V2 = +
4πϵ 0 ( − 7) 2 + ( − a) 2 4πϵ 0 ( − 7) 2 + (a) 2
√ √
−q q 1
= +
4πϵ 0
√
4πϵ 0 49 + a 2 √49 + a 2
=0

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.
The answer does not change because work done by the electrostatic eld in moving a test charge between the
two points is independent of the path connecting the two points.

Page : 88 , Block Name : Additional Exercise

Q2.22 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. Four charges of same magnitude are placed at points X, Y, Z and P respectively, as shown in the
following gure.

A point is located at P, which is r distance away from point Y.
The system of charges forms an electric quadrupole.
It can be considered that the system of the electric quadrupole has three charges.
Charge +q placed at point X
Charge - 2q placed at point Y
Charge +q placed at point Z

XY = YZ = a
YP = r
PX = r + a
PZ = r + a
Electrostatic potential caused by the system of three charges at point P is given by,

V=
1
4πϵ 0 XP[ q
−
2q
YP
+
q
ZP ]
=
1
[
4πϵ 0 r + a
q
−
2q
r
+
r−a
q
]
=
q
4πϵ 0 [ r(r − a) − 2(r + a)(r − a) + r(r + a)
r(r + a)(r − a) ]
=
q
4πϵ 0 [ r 2 − ra − 2r 2 + 2a 2 + r 2 + ra

(
r r2 − a2 ) ] [ ( )]
=
q
4πϵ 0
2a 2

r r2 − a2

2qa 2
=

4πϵ 0r 3 1 −
( ) a2
r2

r
Since, a >> 1
a
∴ r << 1
a2
is taken as negligible.
r2

2qa 2
∴V=
4πϵ 0r 3
1
It can be inferred that potential, V ∝
r3
1
However, it is known that for a dipole, V ∝
r2
1
And, for a monopole, V ∝ r

Page : 88 , Block Name : Additional Exercise

Q2.23 An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A
large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not
more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer. Total required capacitance, C = 2 μF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitors, C 1 = 1μF
Each capacitor can withstand a potential difference, V 1 = 400V
Suppose a number of capacitors are connected in series and these circuits are connected in parallel( row) to
each other. The potential difference across each row must be 1000 V and potential difference across each
capacitor must be 400 V. Hence, number of capacitors in each row is given as

1000
400
= 2.5
Hence, there are three capacitors in each row.
1 1
Capacitance of each row = 1 + 1 + 1 = 3 μF
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent
capacitance of the circuit is given as
1 1 1
3
+ 3 + 3 + ……………n terms
n
= 3
However, capacitance of the circuit is given as 2 μF.
n
∴ 3 =2

n=6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x 3 i.e., 18 capacitors are required
for the given arrangement.

Page : 89 , Block Name : Additional Exercise

Q2.24 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 orless.
However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation
between the conductors.]

Answer. Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5cm = 0.5 × 10 − 2m
Capacitance of a parallel plate capacitor is given by the relation,
ϵ 0A
C=
d
Cd
A=
ϵ0
Where,
ϵ 0 = Permittivity of free space = 8.85 × 10 − 12C 2N − 1m − 2
2 × 0.5 × 10 − 2
∴A= = 1130km 2
8.85 × 10 − 12
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of μF.

Page : 89 , Block Name : Additional Exercise

Q2.25 Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and
voltage across each capacitor.

Answer. Capacitance of capacitor C 1 is 100 pF.
Capacitance of capacitor C 2 is 200 pF.
Capacitance of capacitor C 3 is 200 pF.
Capacitance of capacitor C 4 is 100 pF.
Supply potential, V = 300 V
Capacitors C 2 and C 3 are connected in series.
Let the equivalent capacitance be C ′.

1 1 1 2
∴ = 200 + 200 = 200
C′

C ′ = 100pF

Capacitors C 1 and C ′ are in parallel. Let the equivalent capacitance be C ′′.
∴ C ′′ = C ′ + C 1
= 100 + 100 = 200pF
C ′′ and C 4 are connected in series. Let their equivalent capacitance be C.
1 1 1
∴ = ′′ +
C C C4
1 1 2+1
= + =
200 100 200
200
C= pF
3
Hence,
200
the equivalent capacitance of the circuit is 3 pF.
Potential difference across C ′′ = V ′′
Potential difference across C 4 = V 4
∴ V ′′ + V 4 = V = 300V
Charge on C 4 is given by,
Q 4 = CV

200
= × 10 − 12 × 300
3
= 2 × 10 − 8C
Q4
∴ V4 =
C4
2 × 10 − 8
= = 200V
100 × 10 − 12
Voltage across C 1 is given by,
V1 = V − V4
= 300 − 200 = 100V

Hence, Potential difference, V 1, acrossC 1 is 100V
Charge on C 1 is given by,
Q 1 = C 1V 1
= 100 × 10 − 12 × 100
= 10 − 8C

C 2 and C 3 having same capacitances have a potential difference of 100 V together. Since C 2 and C 3 are in
series, the potential difference across C 2 and C 3 is given by,
V 2 = V 3 = 50 V
Therefore, charge on C 2 is given by,
Q 2 = C 2V 2
= 200 × 10 − 12 × 50
= 10 − 3C
And, charge on C 3 is given by,
Q 3 = C 3V 3
= 200 × 10 − 12 × 50
= 10 − 3C

200
Hence, the equivalent capacitance of the given circuit is 3 pF with,
Q 1 = 10 − 8C, V 1 = 100V

Q 2 = 10 − 8C, V 2 = 50V

Q 3 = 10 − 8C, V 3 = 50V

Q 4 = 2 × 10 − 8C, V 4 = 200V

Page : 89 , Block Name : Additional Exercise

Q2.26 The plates of a parallel plate capacitor have an area of 90cm 2 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 eld between the plates, and obtain the energy per unit
volume u. Hence arrive at a relation between u and the magnitude of electric eld E between the plates.

Answer. Area of the plates of a parallel capacitor, A = 90cm 2 = 90 × 10 − 4m 2

Distance between the plates, d = 2.5 mm = 2.5 × 10 − 3m
Potential difference across the plates, V = 400 V

(a) Capacitance of the capacitor is given by the relation,
ϵ 0A
C= d
1
Electrostatic energy stored in the capacitor is given by the relation, E 1 = 2 CV 2
1 ϵ 0A
= 2 d V2
Where,
ϵ 0 = Permittivity of free space = 8.85 × 10 − 12C 2N − 1m − 2
1 × 8.85 × 10 − 12 × 90 × 10 − 4 × ( 400 ) 2
∴ E1 = = 2.55 × 10 − 6J
2 × 2.5 × 10 − 3
Hence, the electrostatic energy stored by the capacitor is 2.55 × 10 − 6J.

(b) Volume of the given capacitor,
V′ = A × d
= 90 × 10 − 4 × 25 × 10 − 3
= 2.25 × 10 − 4m 3
Energy stored in the capacitor per unit volume is given by,
E1
u= ′
V
2.55 × 10 − 6
= = 0.113Jm − 3
2.25 × 10 − 6
E1
u=
V′
Again, 1 ϵ 0A

()
2
CV 2 2d
V2 1 V 2
= Ad = Ad = 2 ϵ0 d

Where,
V
d
= Electric intensity = E
1 .
∴ u = 2 ϵ 0E 2

Page : 89 , Block Name : Additional Exercise

Q2.27 A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected
to another uncharged 2 μFcapacitor. How much electrostatic energy of the rst capacitor is lost in the form of
heat and electromagnetic radiation?

Answer. Capacitance of a charged capacitor, C 1 = 4μF = 4 × 0 − 6F
Supply voltage, V 1 = 200V
Electrostatic energy stored in C 1 is given by,
1 2
E 1 = C 1V 1
2
1
= × 4 × 10 − 6 × (200) 2
2
= 8 × 10 − 2J

Capacitance of an uncharged capacitor, C 2 = 2μF = 2 × 0 − 6F
When C 2 is connected to the circuit, the potential acquired by it is V 2.
According to the conversation of charge, initial charge on capacitor C 1 is equal to the nal charge on
capacitors, C 1 and C 2.

( )
∴ V 2 C 1 + C 2 = C 1V 1

V 2 × (4 + 2) × 10 − 6 = 4 × 10 − 6 × 200
400
V2 = 3 V
Electrostatic energy for the combination of two capacitors is given by,

1
E2 =
2 (C1 + C2 )V22
=
1
2
(2 + 4) × 10 − 6 × ( )
400 2
3
= 5.33 × 10 − 2J
Hence, amount of electrostatic energy lost by capacitor C 1
= E1 − E2
= 0.08 − 0.0533 = 0.0267 .
= 2.67 × 10 − 2J

Page : 89 , Block Name : Additional Exercise

Q2.28 Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q
is the charge on the capacitor, and E is the magnitude of electric eld between the plates. Explain the origin of
the factor ½.

Answer. Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx .
hence, work done by the force to do so = F Δx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates

The work done will be equal to the increase in the potential enery i.e.,
FΔx = uAΔx

F = uA =
( )
1
ϵ E
2 0
2 A

Electric intensity is given by,
V
E= d

1
() V
∴ F = 2 ϵ 0 d EA = 2 ϵ 0A d E
1
( ) V

ϵ 0A
However, capacitance, C = d
1
∴ F = 2 (CV)E
Charge on the capacitor is given by,
Q = CV
1
∴ F = 2 QE

The physical origin of the factor, ½ , in the force formula lies in the fact that just outside the conductor, eld is
E and inside it is zero. Hence, it is the average value, E/2 , of the eld that contributes to the force.

Page : 89 , Block Name : Additional Exercise

Q2.29 A spherical capacitor consists of two concentric spherical conductors,held in position by suitable
insulating supports (Fig). Show

4πε 0r 1r 2
that the capacitance of a spherical capacitor is given by C = r − r where r 1 and r 2 are the radii of outer and
1 2
inner spheres, respectively.

Answer.
Radius of the outer shell = r 1
Radius of the inner shell = r 2
The inner surface of the outer shell has charge + Q
The outer surface of the outer shell has induced charge - Q.
Potential difference between the two shells is given by,
Q Q
V = 4πϵ r − 4πϵ r
0 2 0 1
Where,

ϵ 0 = Permittivity of free space

V=
Q
[ ]
4πϵ 0 r 2
1
−
1
r1

(
Q r1 − r2 )
=
4πϵ 0r 2r 2
Capacitance if the given system is given by,
Charge (Q)
C=
Potential difference (V)
4πϵ 0r 1r 2
=
r1 − r2
Hence, Proved.

Page : 90 , Block Name : Additional Exercise

Q2.30 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
lled 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 anisolated sphere of radius 12 cm. Explain why the
latter is much smaller.

Answer. Radius of the inner sphere, r 2 = 12cm = 0.12m
Radius of the outer sphere r 1 = 13cm = 0.13m
Charge on the inner sphere, q = 2.5μC = 2.5 × 0 − 6C
Dielectric constant of a liquid, ϵ r = 32

(a) Capacitance of the capacitor is given by the relation.
4πϵ 0 ∈ rr 1r 2
C= r1 − r2
Where,

ϵ 0 = Permittivity of free space = 8.85 × 10 − 12C 2N − 1m − 2
1
4πϵ 0
= 9 × 10 9Nm 2C − 2
32 × 0.12 × 0.13
∴C=
9 × 10 9 × ( 0.13 − 0.12 )

≈ 5.5 × 10 − 9F
Hence,
the capacitance of the capacitor is approximately 5.5 × 10 − 9F

(b) Potential of the inner sphere is given by,
q
V=
C
2.5 × 10 − 6 2
= − 9 = 4.5 × 10 V
5.5 × 10
Hence,
the potential of the inner sphere is 4.5 × 10 2V

(c) Radius of an isolated sphere r = 12 × 10 − 2m
Capacitance of the sphere is given by the relation,
C ′ = 4πϵ 0r
= 4π × 8.85 × 10 − 12 × 12 × 10 − 12
= 1.33 × 10 − 11F
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the
outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is
more than the isolated sphere.

Page : 90 , Block Name : Additional Exercise

Q2.31 Answer carefully:
(a) Two large conducting spheres carrying charges Q 1 and Q 2 are brought close to each other. Is the magnitude
of electrostatic force between them exactly given by Q 1Q 2 / 4πε 0r 2, where r is the distance between their
centres?
(b) If Coulomb’s law involved 1 / r 3 dependence (instead of 1 / r 2 ), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic eld con guration. Will it travel along
the eld line passing through that point?
(d) What is the work done by the eld of a nucleus in a complete circular orbit of the electron? What if the
orbit is elliptical?
(e) We know that electric eld 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) The force between two conducting spheres is not exactly given by the expression,
ϵ
Q 1Q 2 / 4π 0 r 2 because there is a non-uniform charge distribution on the spheres.

(b) Gauss's law will not be true, if Coulomb's law involved
1 / r 3 dependence, instead of 1 / r 2, on r.

(c) Yes, If a small test charge is released at rest at a point in an electrostatic eld con guration, then it will
travel along the eld lines passing through the point, only if the eld lines are straight. This is because the
eld lines give the direction of acceleration and not of velocity.

(d) Whenever the electron completes an orbit, either circular or elliptical,the work done by the eld of a
nucleus is zero.

(e) No Electric eld is discontinuous across the surface of a charged conductor. However, electric potential is
continuous.

(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at
in nity.

(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a
greater dielectric constant than mica.

Page : 90 , Block Name : Additional Exercise

Q2.32 A cylindrical capacitor has two coaxial 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 3.5 μC. Determine the capacitance of the system
and the potential of the inner cylinder. Neglect end effects (i.e., bending of eld lines at the ends).

Answer. Length of a coaxial cylinder, I = 15 cm = 0.15 m
Radius of outer cylinder, r 1 = 1.5cm = 0.015m
Radius of inner cylinder, r 2 = 1.4cm = 0.014m
Charge on the inner cylinder, q = 3.5μC = 3.5 × 10 − 6C
Capacitance of a co-axial cylinder of radii r 1 and r 1 is given by the relation.
2πϵ 0l
C= r1
log e r
2

Where,
ϵ 0 = Permittivity of free space = 8.85 × 10 − 12N − 1m − 2C 2
2π × 8.85 × 10 − 12 × 0.15
∴C=

( )
0.15
2.3026log 10 0.14

2π × 8.85 × 10 − 12 × 0.15
= = 1.2 × 10 − 10F
2.3026 × 0.0299
Potential difference of the inner cylinder is given by,
q
V=
C
3.5 × 10 − 6 4
= − 6 = 2.92 × 10 V
1.2 × 10

Page : 91 , Block Name : Additional Exercise

Q2.33 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric
constant 3 and dielectric strength about 10 7Vm − 1. (Dielectric strength is the maximum electric eld a material
can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For

safety, we should like the eld 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. Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, ϵ r = 3
Dielectric strength = 10 7v / m
For safety, the eld intensity never exceeds 10% of the dielectric strength.
Hence,
electric eld intensity, E = 10% of 10 7 = 10 6V / m
Capacitance of the parallel plate capacitor, C = 50pF = 50 × 10 − 12F
Distance between the plates is given by,
V
d=
E
1000
= 6
= 10 − 3m
10
Capacitance is given by the relation,
ϵ 0ϵ eA
C=
d
Where, A = Area of each plate
ϵ 0 = Permittivity of free space = 8.85 × 10 − 12N − 1C 2m − 2
Cd
∴A=
ϵ0 ∈
50 × 10 − 12 × 10 − 3
= ≈ 19cm 2
8.85 × 10 − 12 × 3
Hence, the area of each plate is about 19cm 2.

Page : 91 , Block Name : Additional Exercise

Q2.34 Describe schematically the equipotential surfaces corresponding to
(a) a constant electric eld in the z-direction,
(b) a eld 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. (a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.
(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get
closer, the eld increases.
(c) Concentric spheres centered at the origin are equipotential surfaces.
(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches
the shape of planes parallel to the grid at a larger distance.

Page : 91 , Block Name : Additional Exercise

Q2.35 A small sphere of radius r 1 and charge q 1 is enclosed by a spherical shell of radius r 2 and charge q 2. Show
that if q 1 is positive, charge will necessarily ow from the sphere to the shell (when the two are connected by a
wire) no matter what the charge q 2 on the shell is.

Answer. According to Gauss's law, the electric eld between a sphere and a shell is determined by the charge 41

on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of
charge 92. For positive charge 91, potential difference V is always positive.

Page : 91 , Block Name : Additional Exercise

Q2.36 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 eld that decreases with altitude. Near the surface of the earth, the eld is about 100Vm − 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 eld inside!)
(b) A man xes outside his house one evening a two metre high insulating slab carrying on its top a large
aluminium sheet of area 1m 2. 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 itself 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
lightning?(Hint: The earth has an electric eld of about 100Vm − 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 + 1800 C 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) We do not get an electric shock as we step out of our house because the original equipotential
surfaces of open air changes, keeping our body and the ground at the same potential.

(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging
current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in
the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the
presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing
currents are in equilibrium and the atmosphere remains electrically neutral.

(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the
atmosphere.

Page : 91 , Block Name : Additional Exercise