Ncert solutions physics mtg class 12

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NCERT
SOLUTIONS

PHYSICS
CLASS 12

MTG Learning Media (P) Ltd.
New Delhi | Gurgaon

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Contents


Chapters

Page No.

1.

Electric Charges and Fields

1

2.

Electrostatic Potential and Capacitance

15

3.

Current Electricity

33

4.

Moving Charges and Magnetism

46

5.

Magnetism and Matter

61

6.

Electromagnetic Induction

73

7.

Alternating Current

86

8.

Electromagnetic Waves

100

9.

Ray Optics and Optical Instruments

108

10.

Wave Optics

136

11.

Dual Nature of Radiation and Matter

145

12.

Atoms

164

13.

Nuclei

171

14.

Semiconductor Electronics: Materials, Devices

186



and Simple Circuits

15.

Communication Systems

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194

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Electric Charges and Fields

Chapter

1

1

Electric Charges
and Fields

1.

What is the force between two small charged spheres having charges
of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?
2 × 10−7 × 3 × 10−7
1 q1q2
9
=
9
×
10

×
or F = 6 × 10–3 N
4πε 0 r 2
( 30 × 10−2 )2
(repulsive in nature, as the two charges are like charges.)
Soln. F =

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?
Soln. (a) Given, F12 = 0.2 N, q1­= 0.4 × 10–6 C, q2 = –0.8 × 10–6 C
1 q1q2
1 q1q2
0.4 × 10 −6 × 0.8 × 10 −6
or r2 =
= 9 × 109 ×
2
4πε 0 r
4πε 0 F12
0.2
or r2 = 14.4 × 10–3 = 1.44 × 10–2
or r = 1.2 × 10–1 m = 12 cm
(b) By Newton’s third law, F21 = F12 = 0.2 N attractive

As, F12 =

Ke 2

is dimensionless. Look up a table of
Gme mp
Physical Constants and determine the value of this ratio. What does the
ratio signify?
Soln. The ratio of electrostatic force to the gravitational force between an
electron and a proton separated by a distance r from each other is
Felec
Ke . e / r 2
Ke 2
=
=

Fgrav Gm m / r 2
Gme mp
e p

3.

Check that the ratio

In terms of dimensions
 Ke 2   Felec  MLT −2
 =
=1
So, 
 = 
 Gme mp   Fgrav  MLT −2
Thus, the ratio

Ke 2

is dimensionless
Gme mp

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NCERT Solutions (Class XII)

9 × 109 × (1.6 × 10 −19 )2
Ke 2
=
−
11
Gme mp 6.67 × 10 × 9.1 × 10 −31 kg × 1.67 × 10 −27 kg

= 0.23 × 1040 = 2.3 × 1039
This ratio signifies that electrostatic forces are 1039 times stronger than
gravitational forces.

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?
Soln. (a) The net charge possessed by a body is an integral multiple of
charge of an electron i.e. q = ± ne, where n = 0, 1, 2, 3, ... is the number of
electrons lost or gained by the body and e = 1.6 × 10–19 C is charge of an
electron. This is called law of quantization of charge.

(b) At macroscopic level, charges are enormously large as compared to
the charge of an electron, e = 1.6 × 10–19C. Even a charge of 1 nC contains
nearly 1013 electronic charges. So, at this large scale, charge can have a
continuous value rather than discrete integral multiple of e, and hence,
the quantization of electric charge can be ignored.
5.

When a glass rod is rubbed with a silk cloth, charges appears 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.
Soln. When a glass rod is rubbed with a silk cloth, electrons
from the glass rod are transferred to the piece of silk cloth. Due to this,
the glass rod acquires positive (+) charge whereas the silk cloth acquires
negative (–) charge. Before rubbing, both the glass rod and silk cloth are
neutral and after rubbing the net charge on both of them is also equal
to zero. Such similar phenomenon is observed with many other pairs of
bodies. Thus, in an isolated system of bodies, charge is neither created
nor destroyed, it is simply transferred from one body to the other. So, it is
consistent with the law of conservation of charge.

6.

Four point charges qA = + 2 µC, qB = –5µC, 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?
Soln. Forces of repulsion on 1 mC charge at O
due to 2 mC charge, at A and C are equal and
opposite. Therefore, they cancel. Similarly,
forces of attraction on 1 mC charge at O, due to
–5 mC charges at B and at D are also equal and

opposite. Therefore, these also cancel.
Hence, the net force on the charge of l mC at
O is zero.

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Electric Charges and Fields

3

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?
q = –5 C
Soln. (a) q = +2 C
B
A

A



O



FOD
C



FOC FOB

+1 C

q = +2 C
C

B



FOA
D
q = –5 C
D





As, F = − F  or  FOA + FOC = 0
OC
OA
So, the net force on charge of + 1 µC at O is







F = FOA + FOC + FOB + FOD or F = 0

...(i)

A field line cannot have sudden breaks because the moving test charge
never jumps from one position to the other.
(b) Two field lines never cross each other at any point, because if they
do so, we will obtain two tangents pointing in two different directions of
electric field at a point, which is not possible.

8.

Two point charges qA = 3 µC and qB = –3 µC are located 20 cm apart in
vacuum.
(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?
Soln. (a) Electric fields at O due to the charges at A and B are
EOA = EOB =

3 × 10 −6 C
1 q
9
=

9
×
10
×
( 20 / 2 × 10 −2 m )2
4πε 0 r 2

or EOA = EOB = 27 × 105 N C–1
+3 C

+1 EOA EOB
A 10 cm O 10 cm B

As, they are directed in same direction, so net electric field at midpoint
O is
E = EOA + EOB = 2EOB = 2 × 27 × 105
or E = 5.4 × 106 N C–1 directed along OB.
(b) When test charge q0 = –1.5 × 10–9 C is placed at point O, it experiences
a force
F = q0E = 1.5 ×10–9 × 5.4 × 106 or F = 8.1 × 10–3N
In direction opposite to that of E i.e. along OA.

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9.

NCERT Solutions (Class XII)
–7

A system has two charges qA = 2.5 × 10 C and qB = –2.5 × 10–7C located
at point A (0, 0, –15  cm) and B (0, 0, +15 cm), respectively. What are the total
charge and electric dipole moment of the system?
Soln. The total charge of the system is,
qnet = qA + qB or qnet = 2.5 × 10–7 – 2.5 × 10–7 or qnet = 0
and the total electric dipole moment of the system is
qB = – 2.5 × 10–7C

qA = 2.5 × 10–7C
A
(0,0, – 15 cm)

P

B
(0,0, + 15 cm)

p = q.2a = 2.5 × 10–7C × 2 × 15 × 10–2m or p = 75 × 10–9 C m
or p = 7.5 × 10–8 C m directed along BA

10. An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30°

with the direction of a uniform electric field of magnitude 5 × 104 N C–1.
Calculate the magnitude of the torque acting on the dipole.
Soln. Torque on dipole is, τ = pE sin 30°
1
or τ = 4 × 10–9 × 5 × 104 ×

or τ = 10 × 10–5 or τ = 1 × 10–4 N m.
2
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?
q
3 × 10 −7
= 1.875 × 1012
=
e 1.6 × 10 −19
so, 1.875 × 1012 electrons have transferred from wool to polythene, as
polythene acquires negative charge.
(b) Yes, mass of 1.875 × 1012 electrons
i.e., m = nme = 1.875 × 1012 × 9.1 × 10–31 kg

= 1.71 × 10–18 kg has transferred from wool to polythene.
Soln. (a) q = ne or n =

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 × 10–7C ? 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?
1 q1q2

Soln. (a)F =

50 cm
4πε 0 r 2
A
B
−7 2
9 (6.5 × 10 )
or F = 9 × 10
q1 = 6.5 × 10–7C
q 2 = 6.5 × 10–7C
(0.5)2
or F = 1.520 × 10–2 N

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Electric Charges and Fields

(b) q1′ = 2q1 , q2′ = 2q2 and r ′ =

5

r
2

1 2q1 × 2q2
1 q1′ × q2′
=
2
2

πε
4
4πε 0 (r ′ )
0
r
 2 
or F ′ = 16 × 1.520 × 10–2 = 24.32 × 10–2 N or F ′ = 0.24 N.

So, F ′ =

13. Suppose the spheres A and B in previous question have identical sizes.
A third sphere of the same size but uncharged is brought in contact with the
first, then brought in contact with the second, and finally removed from
both. What is the new force of repulsion between A and B?
Soln.

B
A C
q1 = 6.5 × 10–7 C, q2 = 6.5 × 10–7 C
A
q1 = 3.25 ×

10–7

C B
–7
q
C, net = (3.25 + 6.5) × 10 C
= 9.75 × 10–7 C

A

B

q1 = 3.25 × 10–7 C, q2 = 4.875 × 10–7 C

The new force of repulsion between A and B is
1 q1′′× q2′′
4πε 0 (r ′′ )2



F ″ =

or

F ″ = 9 × 109 ×

3.25 × 10 −7 × 4.875 × 10 −7
(0.5)

2

or F ″ = 5.7 × 10–3 N

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?

Soln. Particles 1 and 2 are negatively charged
F = +q E
as they experience forces in direction opposite

to that of electric field E , whereas particle 3 is
positively charged as it experience force in the E

direction of electric field E .
Particle-3 has the highest charge to mass ratio, as it shows maximum
deflection in the electric field.

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NCERT Solutions (Class XII)


15. Consider a uniform electric field E = 3 × 103 i N C−1.
(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?
Soln. (a) Area of square, A = a2 = 102 = 100 cm2 = 100 × 10–4 m2 = 10–2 m2

As the plane is along yz plane, so area vector A is directed along x-axis

^

i.e.,  A = (10 −2 i ) m2
∴ Electric Flux through the square is
 
^
^
φ = E . A = (3 × 103 i ). (10–2 i ) = 3 × 101 or φ = 30 V-m


(b) When normal to plane i.e., A makes an angle of 60° with E , then
φ′ = EA cos 60° = 3 × 103 × 10–2 × 1/2 = 1.5 × 101 or φ′ = 15 V-m.

16. What is the net flux of the uniform electric field in previous question

through a cube of side 20 cm oriented so that its faces are parallel to the
coordinate planes?
Soln. φnet = 0, As the net electric flux with closed surface like cube in
uniform electric field is equal to zero, because the number of lines entering
the cube is the same as the number of lines leaving the cube.

17. Careful

measurement of the electric field at the surface of a black
box indicates that the net outward flux though the surface of the box is
8.0 × 103 N m2 C–1.
(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?
q
= 8 × 103 N m2 C–1
Soln. (a) φ = 8 × 103 N m2 C–1 or

ε0
or q = 8 × 103ε0 = 8 × 103 × 8.85 × 10–12 or q = 7.08 × 10–8 C
(b) As φ = 0 or

qnet
= 0 or qnet = 0
ε0

So, the net charge enclosed by that closed surface is zero, although it may
have some charges inside it.

18. A point charge +10 µC is at a distance of 5 cm directly

above the centre of a square of side 10 cm, as shown in
figure. What is the magnitude of the electric flux though
the square?
Soln. Let us assume that the given square be one face of the cube of edge
10 cm.

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Electric Charges and Fields

7

As charge of +10 µC is at a distance of 5 cm above
the centre of a square, so it is enclosed by the cube.
Hence by Gauss’s theorem, electric flux linked with

the cube is
q
10 × 10 −6
f=
=
= 1.13 × 106 N m2 C–1
ε 0 8.85 × 10 −12
So, the magnitude of the electric flux through the
square is
φ
1.13
fsq =
=
× 106 or fsq= 1.9 × 105 N m2 C–1
6
6

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?
Soln. f =

q
2 × 10 −6
=
= 0.23 × 106
ε0
8.85 × 10 −12

or f = 2.3 × 105 N m2 C–1.

20. A point charge causes an electric flux of –1.0 × 103 Nm2 C–1 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?
Soln. (a) On increasing the radius of the Gaussian surface, charge enclosed
by it remains the same and hence the electric flux linked with Gaussian
surface also remains the same.
q
(b) φ =
or q = φε0
ε0
or q = –1 × 103 × 8.85 × 10–12 or q = –8.85 × 10–9 C = –8.85 nC.

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–1 and points
radially inward. What is the net charge on the sphere?
Soln. r = 20 cm = 0.20 m, E = 1.5 × 103 N C–1, R = 10 cm = 0.1 m
q
1 q
or 1.5 × 103 = 9 × 109 ×
2
4πε 0 r
(0.20)2
1.5 × 10 3 × 0.04
or q =
= 6.67 × 10–9
9 × 109

Since electric field points radially inwards, so charge is negative i.e.

q = –6.67 nC.

E=

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NCERT Solutions (Class XII)

22. A uniformly charged conducting sphere of 2.4 m diameter has a surface

charge density of 80.0 µC m–2.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?
2.4 m
= 1.2 m, σ = 80.0 µ C m–2 = 80 × 10–6 C m–2
Soln. R =
2
q
(a) As σ =
4π R2
2
So, q = 4πR × σ = 4 × 3.14 × (1.22) × 80 × 10–6 or q = 1.45 × 10–3 C
(b) φ =

q
1.45 × 10 −3
=
= 1.64 × 108 N m2 C–1
ε0
8.85 × 10 −12

23. An infinite line charge produces a field of 9 × 104 N C–1 at a distance of
2 cm. Calculate the linear charge density.
Soln. E = 9 × 104 N C–1, r = 2 cm = 2 × 10–2 m
As E =

λ
1
, So, λ
= 2πε 0r . E =
× 9 × 104 × 2 × 10–2
2 πε0r
2 × 9 × 109

or λ = 1 × 10–7 C m–1 = 0.1 m C m–1.

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
magnitudes 17.0 × 10–22 C m–2. What is Electric field in
(a) the outer region of the first plate.
(b) the outer region of the second plate, and
(c) between the plates?
+

+
Soln. As both the plates have same surface charge
+
+
density σ, so
+
EA +
EA
EA
+
σ
I +
II
III
+
EA = EB =
+
+
2ε 0
+
+
σ
σ
+
(a) In region-I, EI = EB – EA =
−
or EI = 0
+
EB +
EB

EB
2ε 0 2ε 0
+
+
σ
σ
A
B
(b) In region-III, EIII = EA – EB =
−
or EIII = 0
2ε 0 2ε 0
(c) In region-II, EII = EA + EB =


or EII =

17.0 × 10 −22
8.85 × 10 −12

σ
σ
σ
+
=
2ε 0 2ε 0 ε 0

= 1.92 × 10–10 N C–1.

25. An oil drop of 12 excess electrons is held stationary under a constant

electric field of 2.55 × 104 N C–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 m s–2; e = 1.6 × 10–19C)

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Electric Charges and Fields

9

Soln. In equilibrium, force due to electric field on the drop balances the
weight mg of drop
i.e. qE = mg or qE = Vrg
3qE 3 × 12 e × E
4
=
or qE = π r 3 ρ g or r3 =
4
πρ g
4πρ g
3

or r3 =

36 × 1.6 × 10 −19 × 2.55 × 10 4

= 0.947 × 10–18
4 × 3.14 × 1.26 × 10 3 × 9.8

or r = 0.981 × 10–6 m = 9.81 × 10–7 m = 9.81 × 10–4 mm.

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











Soln. (a) It is wrong, because electric field lines must be normal to the

surface of conductor outside it.
(b) It is wrong because field lines cannot start or originate from negative
charge, and also cannot end or submerge into positive charge.
(c) It is correct
(d) It is wrong because electric field lines never intersect each other.
(e) It is wrong because electric field lines cannot form closed loops.

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NCERT Solutions (Class XII)

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 N C–1
per metre. What are the force and torque experienced by a system having a
total dipole moment equal to 10–7 C m in the negative z-direction?
Soln. As electric field increases in positive z-direction from A to B, So

dE
⋅ 2a
dr
So, the net force on the electric dipole in electric field is
EB = EA +


dE

Fnet = FB – FA = q(EB – EA) or Fnet = q  EA + .2 a − EA 
dr


dE
dE
–7
5
=p.
or Fnet = 10 × 10 or Fnet = 10–2 N

Fnet = q . 2a
dr
dr

directed in direction of FB i.e., from B to A or along –z direction.
As the two forces on charges of electric dipole are collinear and opposite,
so net torque on it is equal to zero.

28. (a)

A conductor A with a cavity is 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
fields in its environment. Suggest a possible way.
Q
Soln. (a) Let us consider a closed surface inside the
conductor enclosing the cavity. As electric field
inside the conductor is zero, so
Surfaces

 

φ = E . ds = 0
...(i)

∫ s

By Gauss’s theorem,
q

...(ii)
φ = enclosed
ε0
By equations (i) and (ii), we found that
qenclosed

= 0 or qenclosed = 0
ε0

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Electric Charges and Fields

11

This shows that there cannot be any charge on the inner surface of
conductor at the cavity, so any charge Q given to the conductor will
appear only on its outer surface.
(b) Let us consider that the charge –q1 is induced on the inner surface
of conductor A at the cavity, when conductor B of charge q is kept in
its cavity. Due to this charge of Q + q1 is induced on the outer surface of
conductor.

Q + q1
Let us construct a closed Gaussian surface S inside
the conductor A enclosing its cavity. As, electric
q1
field inside conductor is zero, so
B q A

 
Surfaces
.
φ= 
...(iii)
∫ s E ds = 0

But, by Gauss’s theorem
q − q1
φ=
...(iv)

ε0
By equations (iii) and (iv), we get
q − q1
= 0 or q – q1 = 0 or q1 = q
ε0


i.e., equal and opposite charge, –q is induced on inner surface of conductor
A at the cavity, when conductor B of charge, + q is kept is its cavity. Hence,
the charge on outer surface of conductor is Q
+ q1 = Q + q

(c) As we found that electric field inside the cavity of conductor is
zero, even on charging the conductor, so the sensitive instrument can
be shielded from the strong electrostatic fields in its environment, by
covering it with a metallic cover.

29. A hollow charged conductor has a tiny hole cut into its surface. Show
 σ 
n , where n is the unit vector in
that the electric field in the hole is 
 2ε 0 
the outward normal direction, and σ is the surface charge density near the
hole.
Soln. Inside a charged conductor, the electric field is zero.


σ
But a uniformly charged flat surface provide an electric field
normal
2ε 0
to its plane.

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NCERT Solutions (Class XII)

20

20

If we consider a small flat part on the surface of charged conductor, it
certainly provides an electric field σ inside the conductor, which is
2ε 0

nullified by an equal field due to rest of charged conductor.

20

0
0

Now if a hole is made in charged conductor, the field due to small flat
part is absent but the field due to rest of charged conductor is present i.e.,
σ ^
n.
2ε 0

equal to

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.
Soln. Consider a point P, a unit away from the long charged wire. Electric
field due to element dy,
Q
or Q = λdy

dEp =
4 πε0r 2
r2 = y2 + a2
⇒ dEp =

λdy

4 πε0 ( y + a ) ⇒ Ep =
2

2

∞

∫ dEp

−∞

Vertical components cancel out and horizontal components are added
due to symmetry
∞

2 λdy

⇒ Ep = ∫

2
2
0 4 πε0 ( y + a )

=

2
4 πε0

∞

λdy

∫ y 2 + a2 ×
0

× (cos θ)
y
y 2 + a2

=

∞

λydy
2
∫
2
4 πε0 0 ( y + a2 )3/ 2

Take y2 + a2 = x ⇒ 2ydy = dx
Taking proper limits

∞
∞
λ
λ
dx
 −2 λ 1 
Ep =
⇒
 = 2 πε a
 4 πε
4 πε0 ∫2 x 3 / 2
x
 a2

0
0
a

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Electric Charges and Fields

13

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.
2
1
Soln. As u = + e and d = − e
3
3
2
2
1
Charge of proton = + e = e + e − e
3
3
3
So, configuration of proton is ‘uud’
2
1
1
Charge of neutron = + e − e − e
3
3
3
So, configuration of neutron is ‘udd’.

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 at certain distance apart.
Soln. (a) Let us assume that a radial electric field is present with origin as

centre. The E is 0 at origin and a small test charge ‘q0’ is placed at origin.
Force on test charge at origin is zero.


F = qE = 0
But as the test charge is displaced a little, the
electrostatic force will drift it further away so
charge is in unstable equilibrium.
(b) A test charge placed at mid point of equal charges is in stable
equilibrium, for lateral displacement.
Initially at mid point position of test charge, two
repulsion forces on test charge are equal and
opposite.
FA = FB so q0 is in equilibrium
But on displacing q0 close to one of the charge,
one of the forces become stronger and pushes the
charge towards another.
FA > FB so q0 moves towards B.

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. The length of

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14

NCERT Solutions (Class XII)


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/(2mvx2).
Soln. Let the point at which the charged
particle enters the electric field, be origin
O (0, 0), then after travelling a horizontal
displacement L, it gets deflected by
displacement y in vertical direction as it
comes out of electric field.
So, co-ordinates of its initial position are x1 = 0 and y1­= 0
and final position on coming out of electric field are
x2 = L and y2 = y
qE
F
Components of its acceleration are ax = 0 and ay =
=
m
m
and of initial velocity are ux = vx and uy = 0
nd
so, by 2 equation of motion in horizontal direction,
1

y2 – y1 = uxt + axt2 or L – 0 = uxt + 0
2
L
ux

...(i)

t=
and by 2nd equation of motion in vertical direction,
1
y2 – y1= uyt +
a yt 2

2
2
1 . qE .  L 
or y – 0 = 0 +
...(i)
2 m  v  
qEL2

or y =

x

2 mvx2
This gives the vertical deflection of the particle at the far edge of the plate.

34. Suppose

that the particle in question 33 is an electron projected
with velocity vx = 2.0 × 106 m s–1. If E between the plates separated by
0.5 cm is 9.1 × 102 N C–1, where will the electron strike the upper plate?
(|e| = 1.6 × 10–19 c, me = 9.1 × 10–31 kg)
Soln. If the electron is released just near the negatively charged plate, then
y = 0.5 cm and hence
y=

qEL2
2 mvx2  

or  L2 =

or L2 =

2 mvx2 y
qE

2 × 9.1 × 10 −31 kg × ( 2 × 106 m s −1 )2 × 0.5 × 10 −2 m

1.6 × 10 −19 C × 9.1 × 10 2 NC −1
or L2 = 2.5 × 10–4 or L = 1.6 × 10–2 m = 1.6 cm

a 
b
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Electrostatic Potential and Capacitance

Chapter

2

15

Electrostatic Potential
and Capacitance

1.

Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what
point(s) on the line joining the two charges is the electric potential zero?
Take the potential at infinity to be zero.
Soln. (i) Let C be the point on the line joining the two charges, where
electric potential is zero, then
–8
–8
–3 × 10 C
5 × 10 C
VC = 0
A
C (16 – x) B
x
or VCA + VCB = 0 or VCA = – VCB
or 

1 qA

5 × 10 −8 C
( −3 × 10 −8 C)
1 qB
= −
or
=
−
4 πε0 rCA
4 πε0 rCB
x × 10 −2 m
[(16 − x) × 10 −2 m]

5
3
=
or 80 – 5x = 3x or 80 = 8x
x 16 − x
80
 or x = 10 cm
or x =
8
So, electric potential is zero at distance of 10 cm from charge of 5 × 10–8 C
on line joining the two charges between them.
If point C is not between the two charges, then
VCA + VCB = 0 or VCA = –VCB

or 

or 

1 qA
−1 qB
=
4 πε0 rCA
4 πε0 rCB

A 16 cm B

x

C

5 × 10 −8 C
−( −3 × 10 −8 C)
or  
=
(16 + x) × 10 −2 m 
[ x × 10 −2 m]


5
3
=
16 + x x
or 5x = 48 + 3x or 2x = 48 or x = 24 cm
So, electric potential is also equal to zero at a distance of 24 cm from
charge of –3 × 10–8 C and at a distance of (24 + 16) = 40 cm from charge of
5 × 10–8 C, on the side of charge of –3 × 10–8 C.

2.

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

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Soln. V = 6 ×

1 q
4 πε0 r

or V = 6 × 9 × 109 ×

NCERT Solutions (Class XII)
+q
r r
r
+q
+q
r r r
+q +q
+q

5 × 10 −6

10 × 10 2

orV = 2.7 ì 106 volts.

3.

Two charges 2 àC and –2 µC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this
surface?
Soln. (a) Since it is an electric dipole, so a plane normal to AB and passing
through its mid-point has zero potential everywhere.
(b) Normal to the plane in the direction AB.

4.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C
distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?
Soln. R = 12 cm, q = 1.6 × 10–7C
(a) Ein = 0
(b) Eout =
(c) Eon =

1.6 × 10 −7
1 q
9
=
9
×

10
×
or Eout = 1.0 × 105 N C–1
4 πε0 r 2
(12 × 10 −2 )2

1.6 × 10 −7
1 q
= 9 × 109 ×
or Eon = 4.44 × 104 N C–1
2
4 πε0 R
(18 × 10 −2 )2

5.

A parallel plate capacitor with air between the plates has a capacitance
of 8 pF. What will be the capacitance if the distance between the plates is
reduced by half, and the space between them is filled with a substance of
dielectric constant 6?
ε A
= 8 pF
Soln. C0 = 0
d
K ε0 A
2 Kε 0 A
C=
=
= 2 KC0 = 2 × 6 × 8 pF or C = 96 pF
d/2

d

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 capacitors if the
combination is connected to a 120 V supply?
C
9
Soln. (a)C = 1 = pF or C = 3 pF
3 3

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Electrostatic Potential and Capacitance

17

(b) Net charge stored in combination of capacitors is
Q = CV = 3 × 10–12 × 120 = 360 pC
So, potential difference across each capacitor is
Q 360 pC
V1 = C = 9 pF or V1 = 40 volts.
1

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.
Soln. (a) C = C1 + C2 + C3 = 2 + 3 + 4 or C = 9 pF
(b) Since the capacitors are in parallel, so potential difference across each
of them is same i.e.
V1 = V2 = V3 = 100 V
So, charges stored on capacitors are
Q1 = C1V1 = 2 × 100 = 200 pC
Q2 = C2V2 = 3 × 100 = 300 pC
Q3 = C3V3 = 4 × 100 = 400 pC

8.

In a parallel plate capacitor with air between the plates, each plate has
an area of 6 × 10–3m2 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?
Soln. A = 6 × 10–3 m2, d = 3 mm = 3 × 10–3 m, V = 100 V
C0 =

ε0 A
8.85 × 10 −12 × 6 × 10 −3
= 17.7 × 10–12 F or C0 = 17.7 pF = 18 pF
=
d
3 × 10 −3

On connecting the capacitor across 100 V supply, charge on each plate of
the capacitor is
Q0 = C0V = 18 × 10–12 × 100 or Q0 = 18 × 10–9 C.

9.

Explain what would happen if in the capacitor given in above question,
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.
Soln. C = KC0 = 6 × 18 = 108 pF
(a) If the voltage supply remained connected, then the potential
difference across the capacitor will remain the same i.e. V = 100 V and
hence charge on the capacitor becomes
Q = CV = 108 × 10–12 × 100 or Q = 1.08 × 10–8 C

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18

NCERT Solutions (Class XII)

(b) If the voltage supply was disconnected, then charge on the capacitor
remains the same
i.e. Q = 1.8 × 10–9 C
and hence potential difference across the capacitor becomes
Q 1.8 × 10 −9 C

=
V=
or V = 16.6 V
C 108 × 10 −12 F

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

1
1
CV 2 = × 12 × 10 −12 × 50 2 = 15000 × 10–12 or U = 1.5 × 10–8 J.
2
2
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?
Soln. C1 = 600 pF, V1 = 200 V, C2 = 600 pF, V2 = 0
On connecting charged capacitor to uncharged capacitor, the common
potential V across the capacitors is
Soln. U =

C1V1 + C2V2 600 × 10 −12 × 200 + 0
or V = 100 V
=
C1 + C2
(600 + 600) × 10 −12
Energy stored in capacitors before connection is
V=

1

1
C V 2 + 0 = × 600 × 10 −12 × 200 2 or Ui = 12 µJ
2 1 1
2
and energy stored in capacitors after connection is

Ui =

1
1
(C + C2 )V 2 = (600 + 600) × 10 −12 × 100 2 or Uf = 6 µJ
2 1
2
Hence the energy lost in the process is
∆ = Uf – Ui = (6 – 12) µJ or ∆U = –6 µJ.
Uf =

12. A charge of 8 mC is located at the origin. Calculate the work done in
taking a small charge of – 2 × 10–9 C from a point P(0, 0, 3 cm) to a point
Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).
y
R (0 , 6 , 9)cm
Soln. As electric field is conservative
field, so work done in moving a
charge in electric field is independent
Q
of path chosen to move the charge in (0 , 4 , 0)cm
electric field and depends only on the
z
electric potential difference between

O
P (0 , 0 , 3)cm
q = + 8mc
the two end points. So
 1
q
1 q 
−

WPQ = q0[VQ – VP­] = q0 
 4 πε0 rOQ 4 πε0 rOP 

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Electrostatic Potential and Capacitance

or WPQ =

19

qq0  1
1 
−


4 πε0  rOQ rOP 

1

1 1
= 9 × 109 × 8 × 10–3 C × (–2 × 10–9 C) ×  −  ×
4 3
10 −2
3−4
= 12 × 10–1

= –144 × 10–1 × 
 12 
WPQ = 1.2 J.



13. A cube of side b has a charge q at each of its vertices. Determine the

potential and electric field due to this charge array at the centre of the
cube.
Soln. The length of diagonal of the cube of each side b is

3b2 = b 3
Distance of any of the vertices from the centre of cube,

3
b
2
1 q
1
V=8×
=8×
4 πε0 r

4 πε0


=

q

4q
or V =
b
3 πε0b
2
E = 0, as electric field at centre due to a charge at any corner of cube is
just equal and opposite to that of another charge at diagonally opposite
corner of cube.
3

14. Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm

apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and
passing through the mid-point.
Soln. (a) At mid point C of line joining two charges, electric potential is
EPA sin
EPB sin
VC = VCA + VCB 
or VC =

1  q A qB 

+


4 πε0  rCA rCB 

P

EPB

−6

EPA

EPA cos

EPB cos

 1.5 2.5  10
+
= 9 × 109 
×
 15 15  10 −2

or VC = 2.4 × 105 V
and electric field at point C is
EC = ECB – ECA [As ECB > ECA and
they are directed opposite to
each other]

10 cm

+1.5 C
A

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15 cm
ECB

18 cm

+1 15 cm
C

ECA

+2.5 C
B


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20

or EC =

NCERT Solutions (Class XII)

 2.5 1.5 
10 −6 C
1  qB q A 
 2 − 2  = 9 × 109  2 − 2  ×

4 πε0  rCB rCA 
15 
 15
10 −4 m 2

or EC = 4 × 105 V m–1

directed in direction of ECB i.e. from C to A.
(b) At given point P on perpendicular bisector of
charges, electric potential is
VP = VPA + VPB or VP =

line joining two

1  q A qB 
 1.5 2.5 
10 −6 C
+
 = 9 × 109 

+

×
4 πε0  rPA rPB 
 18 18  10 −2 m

or VP = 2.0 × 105 V
and horizontal component of net electric field at point P is
Ex = EPA cos θ – EPB cos θ = (EPA – EPB) cos θ
 1.5 2.5  10 −6

1  q A qB 
15
or Ex =
×
 2 + 2  cos θ = 9 × 109  2 − 2  ×
−
4
4 πε0  rPA rPB 
18
18  10
 18
5
–1
or Ex = –2.3 × 10 V m
whereas vertical component of net electric field at point p is
Ey = EPA sinθ + EPB sinθ = [EPA + EPB] sinθ
 1.5 2.5  10 −6
1  q A qB 
10
×
or Ey =
 2 + 2  sin θ = 9 × 109  2 + 2  ×
−4
4 πε0  rPA rPB 
18  10
 18
18
or Ey = 6.2 × 105 V m–1
So, magnitude of net electric field at point P is
EP =

Ex2 + Ey2 =

2.32 + 6.2 2 × 10 5

or EP = 6.6 × 105 V m–1 directed at an angle
Ey
6.2 × 10 5
tan θ =
=
= –2.6956 or θ = – 69.6°
Ex
−2.3 × 10 5
with the horizontal in –ve x-direction i.e. at 69.6° with BA.

15. A spherical conducting shell of inner radius r1 and outer radius r2 has a

charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge
density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge)
zero, even if the shell is not spherical, but has any
irregular shape? Explain.
r2
Soln. (a) Surface charge density on the inner surface
r1
of shell is
+q
–q
−q

σin =
Q+q
2
4 πr1

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