1

ELECTROSTATICS & CAPACITORS

Introduction
Electrostatics deals with interaction between charges at rest. Such an interaction between two charged
particles is experienced in the form of electrostatic forces which they exert on each other. The charge is a
property of matter which when imparted to matter or a body produces an electrical or magnetic effect due to
interaction. There are two types of charges i.e. negative and positive charges.

Basic Properties of Electric Charge

(i) Like charges repel each other and unlike charges attract each other.
(ii) Electric charge is conserved i.e., the algebraic sum of positive and negative charges in an isolated
system remains constant.
(iii) Electric charge is quantised in nature, i.e., electric charges exist in discrete packets rather than in
continuous amount.
(iv) The electric charge is additive in nature, i.e., total electric charge of a body is equal to the algebraic
sum of all the electric charges located any where on the body.
1.1 (i) Coulomb’s Law
The force of attraction or repulsion between two stationary point charges in vacuum or free space is
directly proportional to the product of the two charges and inversely proportional to the square of
distance between these charges.
If q1 and q2 are the magnitudes of two point charges separated by a distance r, the force F due to
interaction is given by :
1  q1 q2 
F  
4 0  r 2 
where 0 is a constant known as permittivity of free space, 0 = 8.85  10–12 coulomb2/newton.metre2.
1
and 4 = 9  109 N.m2.C–2
0

SI unit of measurement of charge is coulomb, its CGS unit of measurement is e.s.u. (electrostatic
unit).
1 coulomb = 3 × 109 e.s.u., another unit is e.m.u., 1 e.m.u. (electromagnetic unit) = 3 × 1010 e.s.u. = 10
coulumb
Relative Permittivity or Dielectric Constant
The relative permittivity or the dielectric constant (r or K) of a medium is the ratio of the absolute
permittivity  of the medium to the permittivity 0 of free space.

 Fv
r  or r = F
0 m

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Electrostatics & Capacitors

Coulomb force F acts on the charged bodies along the line joining the centres of two bodies and is a
vector quantity.
 q1 q2
F12  rˆ , where F12 is force on charge 1 due to charge 2 and r is unit vector from q1 to q2
4 0 r 2

  1  q1 q2 
The electrical forces form an action-reaction pair so that | F12 |  | F21 |    . If one of
4 0  r 2 
the charges is fixed, the free charge will move.
F12 F21 F12 F21
A B
A B
+q1 –q2 +q1 +q2
r r

Super Position Principle


When two or more charges exert forces simultaneously on any charge q1, the resultant force F on
  
charge q1 is the vector sum of forces F12 , F13 , F14 ,...... that the other charges q2, q3, q4, ..... will exert
respectively on this charge.
   
F11  F12  F13  F14  ......

If a force on a point charge q1 is exerted by a large body with charge q, the force dF is found by
dividing charge q into infinitesimally small charge dq located at a distance r from q1 as :
q
1  q1 dq 
dF 
4 0  2  rˆ + +
 r  dq
+ + r +q1 dF
 1 q1 dq +
Total force F  rˆ ++ +
 dF 
4 0  r2 +

(ii) Electric Field Strength or Intensity

 
The electric field strength E at a point in an electric field is defined as the force F experienced by

 F
a unit positive charge placed at that point. If a unit charge q0 is placed at that point, E  Lt
q0  0 q
0


The direction of E at any point is same as the direction of force experienced by a unit positive charge
placed at that point.

 Kq 1
Electric field due to a point charge =  2  , where K = 4
 r  0

Super Position Principle

If a number of charges q1, q2, q3, ..... are present in a space around a point and produce an electric
  
field intensity at that point as E1 , E 2 , E 3 , ....... , respectively the resultant field strength at that point is
equal to vector sum of all these electric fields.
i  n
    
E  E1  E 2  E 3  ........  E n =  Ei
i 1

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Electrostatics & Capacitors

For a continuous distribution of charges,

 1 dq

E  dE   rˆ
4  0 r2

where dE is electric field due to infinitesimally small charge dq.

Electric Lines of Force

A line of force in an electric field is an imaginary line drawn in such a manner that the direction of the
 
field at any point is in the direction of the tangent drawn at that point e.g. E A and E B represent
respective electric fields at A & B.

The lines of force are drawn in such a manner that the number of lines per unit area are proportional

to the magnitude of E . If the lines are crowded with high density at a point P, it indicates large
 
magnitude of E as compared to another point Q where magnitude of E is small due to smaller
number of lines per unit area as shown in the figure (a)

P +
Q –

(a)
(b) (c)
The lines of force for a positively charged particle point radially outwards and for a negatively charged
particle, the lines of force lie along the radii pointing inwards as shown in figures (b) and (c) respectively.
Lines of force start from positive charge and end on negative charge. Lines of force never cross each
other as electric field cannot be in two different directions at the same point. The lines of force meet
a conducting surface normally.

Electric field due to Different Charge Distribution:

(i) An isolated point kq

r P
E 
charge +q E r2

++ +
(ii) A circular ring of radius + +
+ R + 1 qx
R with uniformly + + E
x
+ + P E 4 0 (R  x 2 )3 / 2
2
distributed charge q + +
+
+ + +q

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Electrostatics & Capacitors

(iii) A circular disc of

+
radius R with + R   x 
x E 1  
uniformly distributed + + P E 2 0  2 2 
x R 
+ 
surface charge
density .

(iv) An infinite sheet of + ++

+ ++ 
uniformly distributed + ++ x E
+ ++ P E 2 0
charges with density .

Ey 
Ex  [sin   sin ]
(v) A finite line of charge  4 0 x
x Ex
with constant linear  P 
charge density . Ey  [cos   cos ]
40 x


(vi) An infinite line of 
x 2
charge with linear Ex

charge density . Ex  , Ey = 0
20 x
Sample Problem 1.1:
Two identical conducting small spheres are given charges of opposite sign. These attract each other
with a force of 0.3 Newton when placed at a distance of 30 cm. These spheres are then connected
by a conducting wire. When the conducting wire are removed, the spheres repel each other with a
force of 0.1 Newton at the same distance of 30 cm. Find the initial charges on each sphere.

Sol.: Let the charge on the spheres be +q1 and –q2.

q1 q2
Coulomb force of attraction, F = = 0.3 N
4 0 (0.3) 2

0.3  (0.3) 2
q 1q 2 =
9  109
q1q2 = 3  10–12
When these spheres are connected by a conducting wire, charges will flow such that their potentials
are equal. As the spheres are identical having equal capacitance, these will have equal charges. The
 q1  q2 
net charge will be equal to q1 – q2 and charge on each sphere will be equal to  
 2 
2
 q1  q2 
 
 2 
Coulomb force after disconnection = = 0.1 N
4 0 (0.3) 2

0.1  (0.3) 2  4
(q1 – q2)2 = 9  109 = 4  10–12

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Electrostatics & Capacitors

q1 – q2 = ±2  10–6 ...(i)
q1 + q2 = (q1  q2 ) 2  4 q1 q2 = 4  10 12  4  3  10 12
q1 + q2 = ±4  10–6 ...(ii)
 From (i) and (ii), q1 = ± 3C and q2 = 1 C.
Sample Problem 1.2:
A small sphere of mass 100 mg is given a charge of +2.5  10–10 coulomb. It hangs from an insulating
and inextensible string 0.2 m long. The upper end of the string is connected to a large positively
charged plane vertical conducting plate which has a surface charge density  = 1.5  10–5 C.m–2. Find
the angle which the string makes with the vertical direction at equilibrium.

Sol.: The electric field E due to the large conducting sheet = E = 
0 y
Repulsive force experienced by charge q on the sphere
q
= F = qE =  x
0
The following forces act on the sphere :
(i) Weight = mg 
+
(ii) Electrostatic force of repulsion between sphere and
+
T
metallic plate = F +

(iii) Tension in the string = T +
T cos
The string is deflected through angle  to the vertical. +

When the sphere is in equilibrium state, resultant force +
F
in every direction is zero + T sin
 Fx = 0,  F = T sin, ...(iv) +
mg
Fy = 0, mg = T cos ...(v)
F q
Dividing (iv) by (v), tan = mg   mg
0

( 2.5  10 10 )  (1.5  10 5 )

tan= = 0.432
(8.85  10 12 )  (100  10 6 )  9.8
 = tan–1 (0.432) = 23.4º
Sample Problem 1.3:
A thin vertical wire ring of radius R and a very long uniformly charged straight wire are so placed that
one end of the wire coincides with the centre of the ring. The total charge on the ring is Q and the wire
has a uniform charge per unit length equal to . Find the interaction force between the ring and the
straight wire. A
Sol.: Electric field at P due to ring along axis
1 Qx R
E = 4 (R 2  x 2 ) 3 / 2
0 Ey
O Ex
Force on element dx which is at distance x from P
centre E

1 Q · dq · x
dF = 4 (R 2  x 2 ) 3 / 2
0
B
Qx dx
 dF = 4 ( R 2  x 2 )3 / 2 ( dq – dx)
0

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Electrostatics & Capacitors

 
Q x dx
Total force = F =  dF = 4 
00 0 ( R 2  x 2 ) 3/ 2

Put x = R tan, dx = R sec2  d

dx = R sec2  d, R2 + x2 = R2 + R2 tan2 = R2 sec2
/ 2 / 2
Q R 2 sec 2  tan  d  Q
 F = 4
0
 R 3 sec 3 
= 4  R
0
 sin d 
0 0

Q Q Qλ
= 4  R [– cos ]0 / 2 = 4 [1 –0] =
0 0 4π 0
Sample Problem 1.4:
Rigid insultated wire frame in the form of a right-angled
triangle ABC, is set in a vertical plane as shown. Two
beads of equal masses m and carrying charges q1 and A
q1
q2 are connected by a cord of length l and can slide P
l q2
without friction on the wires. Considering the case when Q

the beads are stationary, determine.

30° 60°
(a) the angle  = APQ, B C

(b) the tension in the cord

(c) the normal reactions on the beads.
(d) If the cord is now cut, what are the values of the charges for which the beads continue to remains
stationary?
Sol. (a) Forces acting on each bead are
K q1q2
(i) mg downward (ii) T, tension (iii) electric force F = (iv) normal reactions.
l2
The directions of the forces are shown in the figure.
Equating the forces along and perpendicular to AB, for equilibrium of the beads
T cos  = F cos  + mg sin 30° ...(i)
F sin  + N1 = mg cos 30° + T sin  ...(ii)
and T sin  = F sin  + mg sin 60° ...(iii)
N2 + F cos  = T cos  + mg cos 60° ...(iv)

From (i) and (iii)

(T – F) cos  = mg sin 30° ...(v)
and (T – F) sin  = mg cos 30° ...(vi) N1 A
Dividing, cot  = tan 30° = cot 60° F
 90° N2
  = 60°
T
T
(b) Substituting the value of  in (1) mg F
mg
1 q1q2 30°
T cos 60° = 4 2 cos 60° + mg sin 30° B C
0 l

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Electrostatics & Capacitors

1 q1q2
or T=  mg
40 l 2
(c) From (ii),
N1 = mg cos 30° + (T – F) sin 60°

3 3
 N1 = mg  mg  3 mg
2 2
1 1
From (iv), N2 = (T – F) cos 60° + mg cos 60° = mg × + mg × = mg
2 2

(d) When the cord is cut, T = 0, then from (i)

0 = F cos 60° + mg sin 30° or F + mg = 0

This result is on the assumption that q1 and q2 are the same sign, so it was taken that there is a force
of repulsion. Since mg is fixed in direction, its sign cannot be reversed but the sign of F can be
reversed because if q1 and q2 are of opposite sign, F will change its sign from + to –
Let q1 and q2 be of the opposite sign, then –F + mg = 0 is the condition for equilibrium
1 q1q2
or  mg or, q1 q2 = 40mgl²
4 0 l 2
Sample Problem 1.5:
–– + ++
A thin conducting ring of radius R has a linear charge ––– + +F
–
density  = 0 cos , where 0 is the maximum linear –– +
+E
–
––
–
+
density and is angle measured anticlockwise from OA. B– O +A
–– +
Calculate the electric field at the centre of the ring. –– dE +
–– +
–– +
Sol. Consider an element EF = Rd ––
++ +
Charge on it = (0 cos ) (Rd)

 0 R cos d  0 cos d

dE = 
4 0 R 2 4  0 R y

 dE  d E cos  iˆ  dE sin  ˆj
x
  cos 2 d ˆ 0 sin  cos d ˆ – + +
 dE   0 i j ––– ++
40 R 40 R – – F Rd
–– E
 0  2  ––
cos 2 d  iˆ – +
 E 
4 0 R  
0 
–
A––
O
+B
– dE +
–– +
 0  2 
=   sin  cos d  ˆj
 ––
–– +
+
4  0 R  0  –– +
++
0 0 ˆ
  iˆ   i
4 0 R 4 0 R
 
 | E | 0
40 R

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Electrostatics & Capacitors

OBJECTIVE QUESTIONS : 1 – I

1. A semi-infinite insulating rod has linear charge density . The electric field at the point P shown in
figure is ++++++++++++++
A
B
r
P

2 2 2 2
(a) at 45° with AB (b) at 45° with AB
(4 0 r ) 2 40 r 2
2 2
(c) at 45° with AB (d) at perpendicular to AB
40 r 40 r
Sol. [c]
Components of electric field intensity at P parallel and perpendicular to wire is same and equal to

40 r .
2. The force between two point charges +Q and –Q placed ‘r’ distance apart is f1 and force between two
spherical conductors, each of radius R placed with their centres ‘r’ distance apart charged with charge
+Q and –Q is f2. If the separation ‘r’ is not much larger than R, then :
(a) f1 > f2 (b) f1 = f2 (c) f1 < f2 (d) f2 = (r/R) f1
Sol. [c]
Conceptual
3. Two identical metallic blocks resting on a frictionless horizontal surface are connected by a light metal-
lic spring having the spring constant 100 N/m and an unstretched length of 0.2m, as shown in figure 1.
A total charge Q is slowly placed on the system, causing the spring to stretch to an equilibrium length of
0.3 m, as shown in figure 2. The value of charge Q, assuming that all the charge resides on the blocks
and that the blocks are like point charges, is
K K
m m m m

Figure 1 Figure 2
(a) 10 µC (b) 15 µC (c) 20 µC (d) 30 µC
Sol. [c]
1 (Q / 2) (Q / 2)
.  (100 N/m) (0.1 m)
40 (0.3m) 2
4  100  (0.1)  (0.3) 2
Q2   Q = 20 × 10–6 C = 20 µC
9  109
4. A particle of charge –q and mass m moves in a circle of radius r around an infinitely long line charge on
1
linear charge density +. Then time period will be [where k  ]
40

m 4 2 m 3
(a) T  2r (b) T2  r
2k q 2k q

1 2k q 1 m
(c) T  (d) T
2r m 2r 2 k q

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Electrostatics & Capacitors

Sol. [a]
We have centripetal force equation
2
 2k   mv
q 
 r  r
2kq 2 r m 1
so v Now, T   2 r where k 
m v 2kq 4 0
5. A ring of radius R is made out of a thin metallic wire of area of cross section A. The ring has a uniform
charge Q distributed on it. A charge q0 is placed at the centre of the ring. If Y is the young’s modulus
for the material of the ring and R is the change in the radius of the ring, then
q0Q q0 Q q0Q q0 Q
(a) R  (b) R  2
(c) R  2 2
(d) R  2
40 RAY 4 0 RAY 8 0 RAY 8 0 RAY
Sol. [d]
Q
dQ  .Rd 
2R

d Q q
2T sin  ( Rd )
2 2R 40 R 2

Qq
Solving, T =
8  0 R 2
2

T R
Using Hooke’s law, Y
A R
6. Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on
the line passing through point O and perpendicular to the plane of the figure at a distance of x ( >>a)
from O is a
+Q –Q

+Q –Q
O

+Q –Q

Qa 2Qa 3Qa
(a) 3 (b) 3 (c) (d) zero
0 x 0 x 0 x3
Sol. [a]
Use concept of electric dipole.

7. A horizontal electric field (E = (mg)/q) exists in space as shown in figure and a mass m is attached at
the end of a light rod. If mass m is released from the position shown in figure find the angular velocity
of the rod when it passes through the bottom most position
g 2g
(a) (b) =45º
  E
mg
 q

3g 5g m
(c) (d) m
l l +q

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Electrostatics & Capacitors

Sol. [b]
Using work energy theorem, W = kf – ki
mgl(1 – cos ) + (mgl sin ) = 1/2 mv2
8. Two point charges +q and –q are held fixed at (–d, 0) and (d, 0) respectively of a (X, Y) coordinate
system. Then

(a) The electric field E at all points on the X-axis has the same direction.

(b) E at all points on the Y-axis is along iˆ
(c) Work has to be done in bringing a test charge from infinity to the origin.
E
(d) The dipole moment is 2qd directed along i ˆ
E  E î
Sol. [b]
 E
The electrical field E at all points on the X-axis will not have the
same direction.

The electrical field E at all points on the Y-axis will be parallel to the
+q -q
X
X-axis (i.e. iˆ direction). (-d, 0) O (d, 0)
The electric potential at the origin due to both the charge is zero, hence,
no work is done in bringing a test charge from infinity to the origin.
Dipole moment is directed from the –q charge to the +q charge (i.e. –x direction).
9. A particle of charge q and mass m moves rectilinearly under the action of electric field E = A – Bx,
where A and B are positive constants and x is distance from the point where particle was initially at rest
then the distance traveled by the particle before coming to rest and acceleration of particle at that
moment are respectively :
2A qA 2 A qA 2 A qA
(a) ,0 (b) 0,  (c) , (d) ,
B m B m B m
Sol. [c]
F = qE = q (A – Bx)
ma = q(A – Bx)
q
a (A – Bx) ...(1)
m
vdv q q
 ( A  Bx) ; vdv  ( A  Bx) dx
dx m m
0 x
q Bx 2
0 vdv  m 0  A  Bx  dx ; Ax  2  0
2A
x = 0, x = ...(2)
B
From eq. (1) and (2)
q q 2A  q qA
( A  Bx )   A  B    ( A  2 A) 
m m B  m m
10. An electron of mass me initially at rest moves through a certain distance in a uniform electric field in
time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this
uniform electric field. Neglecting the effect of gravity, the ratio of t2/t1 is nearly equal to :
(a) 1 (b) (mp/me)1/2 (c) (me/mp)1/2 (d) 1836
Sol. [b]
Force on a charge particle in a uniform electric field : F = q E
The acceleration imparted to the particle is : a = qE/m

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Electrostatics & Capacitors

1 2 1  qE  2
The distance traveled by the particle in time t is : d  at   t
2 2 m 
For the given problem

t 2p t2 t 2p mp tp mp
 e ;   
m p me te2 me te me

1.2 (i) Electric Dipole

A pair of equal and opposite point charges +q and –q at a distance 2a apart form an electric dipole.

Electric dipole moment is a vector p whose magnitude is the product of either charge and the distance
between the two charges.

p = 2aq r̂ , r̂ is a unit vector from –ve to +ve charge
2a
–q +q

The direction of p for an electric dipole is from negative to positive charge. The unit of electric dipole
moment is coulomb-metre (Cm).

Torque acting on dipole: When dipole is oriented by angle  w.r.t. direction of electric field, F form
couple with arm length 2a sin . qE
+q
  = F × 2 a sin  = qE . 2a sin 
= 2qa E sin 

2a
   –q
i.e.,  = pE
qE E
Force, experienced by dipole in uniform electric field is zero.
Work done in rotating a dipole in uniform field through small angle d will be
dW =  d = pE sin  d
2
W = pE  sin  d = pE(cos 1 – cos 2)
1

if 1 = 0, 2 = 
 W = pE (1 – cos )
If dipole is placed in non-uniform electric field force acting on it is

  E
F p.
a

E 
where is rate of change of E w.r.t a along dipole moment.
a
Potential energy of dipole: Potential energy of dipole is defined as work done in rotating a dipole
from a direction perpendicular to a given direction.
 
i.e. U = (W – W 0) = –PE cos  = – p . E

-1 . 1 1 -
Electrostatics & Capacitors

Net force and net torque on dipole is zero. i.e. dipole is in equilibrium for  = 0 and  = 180.
For  = 0, equilibrium is stable because U is minimum.
For,  = 180, equilibrium is unstable because U is maximum.

Angular S.H.M.: If dipole is displaced by small angle from the direction of electric field and left, it
executes S.H.M.
  = – pE sin 
I = – pE .  ( sin   )

pE
  = – 
I
Comparing with  = – 2
pE I
 = and T = 2 p E
I

(a) Electric field along axial line or end on position : The axial line is the straight line drawn
joining the two charges of a dipole. Consider two point charges +q and –q separated by a distance
2a. Consider a point P at a distance r from centre of dipole along the axial line. The electric field
 
at P will be the resultant of electric field E1 and E 2 due to negative and positive charges
respectively.
+q –q
P
E1 E2
r
2a
    q rˆ q rˆ
E  E1  E 2 , E  2

4  0 ( r  a ) 4  0 (r  a ) 2

where r̂ is unit vector in the direction of dipole moment.

4rqa rˆ 2r p rˆ
 
4 0 ( r 2  a 2 ) 2 4 0 (r 2  a 2 ) 2

 
If r >> a, E 2p .
axial 
4 0 r 3

(b) Electric field along equatorial line or broad on position : The line passing through the
perpendicular bisector of the line joining the charges is called the equatorial line. The electric field
  
E at point P on the equatorial line will be the resultant of electric fields E1 and E 2 due to –q and
+q respectively.
  
E  E1  E 2

-1 . 1 2 -
Electrostatics & Capacitors

 
The Y-components of E1 and E 2 (i.e. components along equatorial lines) being equal in magnitude
and opposite in direction will cancel each other. Therefore the resultant field lies along the direction

E = E1 cos + E2 cos, direction of E is opposite of P E2

q q P E
= 4 (r 2  a 2 ) cos   4 (r 2  a 2 ) cos 
0 0

 2qa r E1
 Eequatorial 
4 0 (r 2  a 2 ) 3 / 2

 
If p
r >> a, Eequatorial  +q p –q
4 0 r 3 2a

Here direction of E is opposite to the direction of dipole moment.

(c) If a point Q lies at a distance r(r >> a) from the mid-point of a small electric dipole such that
radius vector r of point Q (from mid-point of dipole) makes an angle  with the direction of its

dipole moment p , then

1 p 2
E = 4   3 3 cos   1 E1
0 r E II
Q
If  is the angle, which this intensity at Q makes r

with the direction of radius vector of d , then o
–q a a +q
 tan  
tan 
tan  =   = tan–1   


If is the angle, which this intensity at point Q makes with the direction of dipole moment of the
dipole, then
 tan  
 = + = + tan–1  
  

(ii) Gauss’s Law


Electric Flux : Electric flux through small area ds kept in electric field E is defined as

d = E. ds

It represents the no. of lines passing through the given area.

Maximum flux passes through the surface when  = 0. i.e. electric field is perpendicular to surfaces.
for  = 90°,  = 0

For closed body outward flux is taken positive while inward flux is taken negative.

The SI unit of electric flux is Nm2C–1.

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Electrostatics & Capacitors

Gauss’s Law
 1
The net flux of electric field E through a closed surface is equal to  times the charge enclosed by
0
 
the surface. If E is the electric field at an elementary surface area ds , then integral of dot product of
  q
E and ds over the whole surface is equal to 0 . This integral is also called surface integral of E .
 q
s

E = E. ds 
0

Gauss’s law can be used to determine E for symmetric ds

 E
charge distributions. The surface that we choose for the
application of Gauss’s Law is called Gaussian surface.

Electric Potential

If a positive charge is moved towards another positive charge, the work is done against the forces of
repulsion. The electric potential is defined as the work done in moving a unit positive charge from one
point to another point in an electric field. If an external work is done in moving a charge q from one
point to another point, then potential difference between two points V2 – V1 is given by
Wext
V2 – V1 = q

The work done by an external force (against the field forces) in bringing a unit positive charge from
infinity to a point at constant speed is equal to the potential V at that point. If the unit charge moves
from infinity towards a negative charge, the work is negative as it is done by the field itself due to the
forces of attraction. The potential V due to charge +q at a distance r from the charge q is given by :

1 q
V   
40  r 

Similarly, potential due to a negative charge is given by V   1 q.

 
4 0  r 
The unit of electric potential is joule/coulomb or volt. The electrical potential is a scalar quantity. The
potential at any point due to a group of charges is the algebraic sum of potentials due to individual
charges.

V = V1 + V2 + V3 + ..... + Vn

1  q1  1  q2  1  qn  1 qi
= 4   r   4   r   .....  4   r  = 4   ri
0  1  0  2  0  n  0 n

If a body has a continuous charge distribution

1 dq
V  dV  
40 r

-1 . 1 4 -
Electrostatics & Capacitors

Relation between Electric Field and Potential: E and V at a point are related as

dV =  E . dr

or dV = – Edr cos 

dV
or = – E cos 
dr

It means that rate of change of potential w.r.t r along a direction dr is equal to component of electric
dV
field in that direction, for  = 90°, = 0, or V = constant i.e., E is always perpendicular to
dr
equipotential surface.

(iii) Electric potential V due to various charge distributions

1 q
(i) An isolated point +q r P V
charge 40 r

++ +
+ +
(ii) A circular ring of radius + R +
+ + x 1 q
R with uniformly + + P V
+ + 40 R 2  x2
distributed charge q +
++ + q
Note: that unlike electric intensity E, potential at the point P is independent of the charge distribution.

(iii) A circular disc of +

+R +   2
radius R with + + V x  R 2  x
x 
2 0  
++ + P
uniformly distributed +
surface charge at the centre of the disc, x = 0,
R
density . V=
20
+
(iv) A sphere of charge + R+ Inside (0  r  R)
+
with uniform + + r  R2  r2 
P
 + V  3  
volume charge + 6 0  R 2 
density . Outside ( r  R)
 R3 1 Kq
V   
3 0 r r
Trajectory of a Charged Particle in an Electric Field
Consider a charged particle of mass m and of charge q travelling horizontally with velocity v0 and
moving under the influence of an electric field E to reach a point P(x, y) after time t, along its trajectory
as shown in the figure
+ + +
Y
v0 P(x, y )
X
(m, q) E

– – – –

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Electrostatics & Capacitors

x
x = v0t, t  ...(i)
v0

1 1  qE  2 Eqt2
y ay t 2 =  t  ...(ii)
2 2 m  2m

Eqx 2  qE 
Eliminating t from (i) and (ii) y  2
  2 2
2   x or x = ky
2mv0  2mv0 

2m v 20
where k = = a constant
qE

This is the equation of trajectory of the charged particle, which is parabola. Hence, inside the electric
field, the charged particle moves on a parabolic path and on leaving the field, it moves along a straight
line tangent to the curved path.

Kinetic Energy of a Charged Particle Accelerated by a Potential Difference

Let a charged particle (charge q) be released from a point A (potential VA) so that it finally reaches a
point B (potential VB).
Then work done on q by the electric field = Initial P.E. – Final P.E. = 0
WField = qVA – qVB
Also, WField = change in kinetic energy (KE)
 KE = q (VA – VB)
So, if a charged particle q is accelerated by a potential difference V, its kinetic energy increases by
qV.

Electron volt
1 electron volt (eV) is the energy gained by an electron (or a particle having charge equal to that of an
electron) when it is accelerated by a potential difference of 1 volt
1 eV = 1 electrons charge × 1 volt
= 1.6 × 10–19 coulomb × 1 volt
= 1.6 × 10–19 Joule

Conductor inside electric field: When conductor inside electric field is in electrostatic
equilibrium then;
1. Electric field inside conductor is zero.
2. There is no charge inside conductor.
3. Surface of conductor is equipotential surface.
4. Charge density is inversely proportional to the radius of curvature.
5. Electric field just outside conductors
6. Electrostatic lines of force cannot pass through the conductor.
 
E n̂ where n̂ is unit vector perpendicular to surface.
0

-1 . 1 6 -
Electrostatics & Capacitors

Relationship between electric field intensity and electric potential

 
dV   E .dr ...(i)

Also,

  V ˆ V ˆ V 
E   i j kˆ  ...(ii)
 x y z 

To find electric potential at any point if E is given, we use equation (i)

To find E at a point if V is given, we use equation (ii)
Sample Problem 1.6:
Four charges +2C, –4C, +6C and +4C are lying at the corners A, B, C and D respectively of a
square ABCD of side 0.1 m as shown in the figure. Find the work done in assembling these charges.
Sol.: When these four charges are lying at infinity initially, these
A(+2C) B(–4C)
can be brought from infinity to the respective points A, B, C
and D. Thus work has to be done in assembling these
charges because of their interaction with each other. The
total work done will be equal to the sum of potential
energies required for each pair of particles.
D(+4C) C(+6C)
Total potential energy = UAB + UBC + UCD + UDA + UAC + UBD
1  (2  106 )( 4  106 ) ( 4  106 )(6  106 ) ( 6  106 )(4  106 )
= 4 r   
0  0.1 0.1 0.1

( 4  106 )(2  106 ) ( 2  106 )(6  10 6 ) (4  10 6 )( 4  10 6 ) 

   
0.1 2  0.1 2  0.1 
1012  12 16 
=
40  0.1  8  24  24  8   
 2 2

1012  9  109    4   36
=    = = –0.254 J
0.1   2  100 2
Sample Problem 1.7:
A potential difference of 600 volt is applied across two parallel plates, each of length 1.5 cm, separated
by a distance of 5.0 mm. If an electron of energy 1000 eV enters at right angles to the electric field,
find the displacement of the electron in the direction perpendicular to the plates, when it comes out of
space between plates.

V 600
Sol.: Electric field between the plates = E = = = 1.2  105 V/m.
d 5  103
Force experienced by electron = eE
F eE
Acceleration along y-axis = 
m m

-1 . 1 7 -
Electrostatics & Capacitors

Displacement along x-axis in time t = x = v0t = L .....(i)

where v0 is initial velocity of electron.
1 2
displacement along y-axis = y = at .....(ii)
2

2
1 L e EL2 e EL2 e EL2
From (i) & (ii), y = a   = = = , where k is kinetic energy of electron
2  v0  2mv02 1  4k
4  mv02 
2 
when it enters between plates.
+ + + + + y
– y
e
x d
v0
E
x
– – – – –

1
Kinetic energy of electron = mv02 = 1000 eV = 1000  1.6  10–19 J
2

e EL2 (1.6  10 19 ) (1.2  10 5 ) (1.5  10 2 ) 2

 y= = = 6.75  10–3 m = 6.75 mm
4K 4  1000  1.6  10 19
Sample Problem 1.8:
Three concentric spherical shells, made of conducting materials, have radii R1, R2 and R3 and are
given charges Q1, Q2 and Q3 respectively as shown in the figure. Find the potential for points at
distance r when
(a) r < R1 (b) R1  r  R2 (c) R2  r  R3 (d) r > R3
Sol.: The potential at any point within a spherical conducting D
shell is the same as that at its outer surface. The potential
Q3 R3
at any point outside the spherical surface is same as if the C
BR
2
charge on the sphere were concentrated at the centre of
the spherical surface. Q2
O R
1
Q1 A
(a) r < R1.
Q1
Potential at A due to charge Q1 = 4 R
0 1

Q2
Potential at A due to charge Q2 = 4 R
0 2

Q3
Potential at A due to charge Q3 = 4 R
0 3

1 2 Q 3 Q Q
 Total potential at A = 4 R  4 R  4 R
0 1 0 2 0 3

1  Q1 Q2 Q3 
VA = 4π   R  R  R 
0  1 2 3

(b) R1  r  R2

-1 . 1 8 -
Electrostatics & Capacitors

Q1
Potential at B due to Q1 at distance r = 4 r
0

The potential at B due to Q2 and Q3 will be the same as that of surfaces of shells respectively.
1  Q1 Q Q 
 VB =   2  3
4π 0  r R2 R3 
(c) R1  r < R3

1  Q1 Q Q 
Similarly, VC =   2  3
4π 0  r r R3 

(d) r > R3
All the charges can be considered to be concentrated at the common centre of the shells.

1  Q1 Q Q  1
 VD =  2  3 = [Q + Q2 + Q3]
4π 0  r r r  4π 0 r 1
Sample Problem 1.9:
A cone made of an insulating material has a total charge Q spread uniformly over its sloping surface.
Calculate the energy required to bring up a small test charge q from infinity to the apex of the cone.
The cone has a slant length L.
Sol.: If R is the radius of the cone and  be the angle of the cone.
Q A
Surface charge density,  = . As the charge on different
RL l l

elements of the cone will be different, consider an element r
L dl
of cone of radius r lying between slanting lengths of the
cone at distances l and l + dl from the apex A of the cone.
D
B R O C
Area of element dA = 2rdl
Charge on element = dq = dA = (2rdl)
2 r d l 
Potential at the apex A of the cone due to this elementary ring = dVA =  l


L L
2 rd l  2  l sin  dl
Potential at the apex due to all the elements = VA =  dV   40 l

40 l
0 0

  R
= 2 (L sin) = 2  L 
0 0 L

QR Q
=  RL  2 = 2  L
0 0

Qq
Work done to bring charge q from infinite to Apex A = VA q = 2 L
0

Sample Problem 1.10:

A charge Q is distributed over two concentric hollow spheres of radii r and R ( > r) such that the
surface densities are the same. Calculate the potential at the common centre of the two spheres.

-1 . 1 9 -
Electrostatics & Capacitors

Sol.: If q1 and q2 are the charges on two spheres of radii r and R respectively, then the surface charge
q1 q2 q1 r 2
density is given by    or 
4r 2 4R 2 q2 R 2
q1 r2 q q r 2  R2 B
or  1  2  1 or 1 2 
q2 R q2 R2 R
But q1 + q2 = total charge Q o r A

Q r 2  R2 QR r 2  R 2 q2 QR
  2 or  or  2
q2 R q2 R R r  R2
q1 Qr
Similarly,  2
r r  R2
1 q1 1 q2 1  q1 q2 
Potential at the common centre =     
4 0 r 40 R 4 0  r R

1  Qr QR  Q  rR 
=     
4 0  r 2  R 2 r 2  R 2  40  r 2  R 2 

Alternative method : Let  be the surface density of charge on both hollow spheres. Then charges
on spheres will be

q1  4 r 2 and q2  4 R 2 

Q  4 r 2  4 R 2  4 (r 2  R 2 )
Potential at common centre = Potential due to first sphere + Potential due to second sphere.

1 q2 1 q1 1 4R 2  1 4r 2 
V   
40 R 40 r 40 R 40 R

1 1  Q  Q Rr
V 4 ( R  r )   2 2  (R  r) 
4 0 40  r  R  40 R 2  r 2

Sample Problem 1.11:

In nuclear fission, a Uranium-235 nucleus captures a neutron and splits apart into two lighter nuclei.
Sometimes the two fission products are a barium nucleus (charge 56e) and a krypton nucleus. Assume
that these nuclei are positive point charges separated by r = 14.6 × 10–15 m. Calculate the potential
energy of this two charge system in electron volts.
Sol.: The potential energy for two point charges separated by a distance r is U = kq1q2/r. To find this
energy in electron volts we calculate the potential due to one of the charges kq1/r in volts and multiply
by the other charge.
1. The potential energy of the two charges :

kq1q2 k (56e) (36e)

U= =
r r
2. Factor out e and substitute the given values :

k (56e) (36e) ke (56) (36)

U= =e
r r

-1 . 2 0 -
Electrostatics & Capacitors

(8.99  109 N .m 2 / C 2 ) (1.6  1019 C ) (56) (36)

U=e
14.6  1015 m
U = e (1.99 × 108 V) = 199 MeV.
Sample Problem 1.12:
A cube of side b has a charge q at each of its vertices. Determine the potential due to this charge
array at the centre of the cube.
G H

F E D C

O
Sol. D Cb
90°

A B A b B
(i) (ii)
2 2 2
DB 2  b 2  b 2  2b 2 , DE  DB  BE or DE  DB 2  BE 2

or DE  2b2  b 2 or DE  3b
1 3
Now, DO  DE  b
2 2
So, the distance of each vertex from the mid-point of the cube is ( 3 / 2) b . In other words, distance
of each charge from the centre O of the cube is ( 3 / 2) b .
1 q 4q
 Potential at O due to charges at the vertices of the cube = 8  
40 ( 3 / 2) b 30b
Sample Problem 1.13:
Calculate the potential due to a thin charged rod of length  at the points along and perpendicular to the
length. Charge per unit length on the rod is .
kdq k dx
Sol. dV  2 2 1/2

(b  x ) (b  x 2 )1/2
2

a  a 
k dx dx ( a   )  b 2  (  a )2
V  k V  k  ln
 (b 2  x 2 )1/2
 (b 2  x 2 )1/2
;
a  b2  a2
a a

Sample Problem 1.14:

A charge q is distributed uniformly on the surface of a sphere of radius R. It is covered by a concen-
tric hollow conducting sphere of radius 2R. Find the charges on inner and outer surfaces of hollow
sphere if it is earthed.
Sol. The charge on the inner surface should be –q, because if we draw a closed Gaussian

R q

2R

surface through the material of the hollow sphere the total charge enclosed by this
Gaussian surface should be zero. Let q' be the charge on the
outer surface of the hollow sphere.

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Electrostatics & Capacitors

Since, the outer sphere is earthed, its potential should be zero. +q

The potential on it is due to the charges q, –q and q'. –q
q
Hence,
1  q q q 
V   0
4 0  2 R 2 R 2 R 


 q' = 0.
Therefore, there will be no charge on the outer surface of the hollow sphere.
Sample Problem 1.15:
Find the electric field intensity at distance r from the centre of a uniformly charged non-conducting
solid sphere of radius R. (a) r > R (b) r < R.
E
Sol. (a) At an external point (r > R)
+ + dS
+

+
According to Gauss’s law

+
+
Q
E  dS  E (4r 2 )  r Gaussian

+ +
0

+
R surface
+ +
+
1 Q
Therefore, E 
40 r 2
For points outside the charged conducting sphere or the charged spherical shell, the field is same
as that of a point charge at the centre.
(b) At an Internal Point (r < R) :
E
The field still has the same symmetry and so we again pick a
spherical Gaussian surface, but now with radius r less than R. Q
Since the charge enclosed is zero, from Gauss’s law we have 4 0 R 2
E(4r2) = 0  E = 0
Thus, we conclude that E = 0 at all points inside a uniformly
charged conducting sphere or the charged spherical shell. O r=R r

Variation of E with the distance from the centre (r)

OBJECTIVE QUESTIONS : 1 – II

1. In a region of space the electric field is given by E  8iˆ  4 ˆj  3kˆ . The electric flux through a surface
of area of 100 units in x-y plane is :
(a) 800 units (b) 300 units (c) 400 units (d) 1500 units
Sol. [b]
Conceptual
2. A dipole of dipole moment p is kept at the centre of a ring of radius R and charge Q. The dipole moment
has direction along the axis of the ring. The resultant force on the ring due to the dipole is
(a) zero
kpQ
(b)
R3
2kpQ
(c)
R3
kpQ
(d) only if the charge is uniformly distributed on the ring
R3
Sol. [b]

-1 . 2 2 -
Electrostatics & Capacitors
E
Electric field at each point on the surface of ring due to dipole is

kp R
E 3 in direction opposite to the dipole moment.
R
p
kpQ
Hence net force on ring is F  QE 
R3
3. The grid (each square of 1m × 1m), represents a region Y
in space containing a uniform electric field. If potentials
at points O, A, B , C, D, E, F and G, H are respectively D G H
0, –1, –2, 1, 2, 0, –1, 1 and 0 volts, find the electric field
intensity C E F
(a) (iˆ  ˆj ) V/m (b) (iˆ  ˆj ) V/m 1m

(c) (iˆ  ˆj ) V/m X

(d) (iˆ  ˆj ) V/m O A B
1m
Sol. [b]
OEH is an equipotential surface, the uniform electric field must be
Y
perpendicular to it pointing from higher to lower potential
as shown. D G H
v=0
v= 1
 iˆ  ˆj 
Hence, Eˆ   
 2  C E F
v= –1
(V  VB ) 0  (2)
E E   2
EB 2
X
O A B
 ( ˆ  ˆj )
i E
 E  E.Eˆ  2  iˆ  ˆj
2
4. A large uniformly charged (negative) plate is placed in xz plane and a
y
positive point charge is fixed on the y-axis. A dipole is positioned in
+Q
between with its axis along y-axis, as shown. The dipole initially
–
moves in +
(a) negative y-direction
(b) negative x-direction x
– plate
(c) positive x-direction
(d) positive y-direction
Sol. [d]
The electric field due to the plate is uniform, exerting no net force on the dipole. The electric field due
to the point charge Q is non-uniform, exerting net force along the positive y-axis on the dipole, causing
it to move in that direction.

5. When three electric dipoles are near each other, they each experience the electric field of the other
two, and the three dipole system has a certain potential energy. Figure below shows three arrange-
ments (A), (B) and (C) in which three electric dipoles are side by side. All three dipoles have the same
electric dipole moment magnitude and the spacings between adjacent dipoles are identical. If U1, U2
and U3 are potential energies of the arrangements (A) , (B) and (C) respectively, then

-1 . 2 3 -
Electrostatics & Capacitors

(a) (b) (c)

(a) U1 > U3 > U2 (b) U1 > U2 > U3 (c) U1 > U2 = U3 (d) U1 = U2 = U3
Sol. [a]
U1 will be positive and greatest since all forces among dipoles are repulsive, U2 is negative as potential
energy of first and second dipole pair cancels out, leaving only potential energy of interaction of first
and third, that is negative.
6. Four charges are rigidly fixed along the Y-axis as shown. A positive y

charge approaches the system along the X-axis with initial speed just
2 2q
enough to cross the origin. Then its total energy at the origin is
–q
(a) zero
(b) positive v Q x
–1 –q
(c) negative
(d) data insufficient –2 2q

Sol. [b]
There exists a point P on the x-axis (other than the origin), where net electric field is zero. Once the
charge Q reaches point P, attractive forces of the two –ve charge will dominate and automatically
cause the charge Q to cross the origin.
Now if Q is projected with just enough velocity to reach P, its K.E. at P is zero, but while being attracted
towards origin it acquires KE and hence its net energy at the origin is positive. (P.E. at origin = zero)
7. Find the electric flux crossing the wire frame ABCD of length  , width b and whose centre is at a
distance OP = d from an infinite line of charge with linear charge density . Consider that the plane of
frame is perpendicular to the line OP.
A
b B
  b    b 
(a) tan 1   (b) tan 1  
0  2d  2 0  2d  O
d
P l

  b    b 
(c) tan 1   (d) tan 1  
D
C
0  4d  2 0  4d 
Sol. [a]

d   ( dx).l.cos , dx  d (sec 2 ) d 
20 r

 b 
tan1 2d 
 
l
  2  d
20 0

8. A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments
and the Gauss’s law. The flux of the electric field due to this charge through the surface of the hemi-
sphere (figure) is
Q Q Q
(a) 20 (b) 0

2Q 2Q
(c) 0 (d) 30

-1 . 2 4 -
Electrostatics & Capacitors

Sol. [a]
Let us imagine another identical hemispherical surface over given one.
Both being symmetric with respect to Q, hence flux will be same through both the hemisphere
(1 = 2).
Q Q
1  2   1  2  2
0 0
9. A charged particle of mass m and charge q initially at rest is released in an electric field of magnitude
E. Its kinetic energy after time t will be
2E 2 t 2 E 2 q 2t 2 Eq 2 m Eqm
(a) (b) (c) 2 (d)
mq 2m 2t t
Sol. [b]
Conceptual
Y
10. The electric field in the region shown here is given by E  x iˆ volt/m.
Then the total electric flux through the cube of side ‘a’ is

(a) Zero (b) a 2 X

(c) a5/2 (d) a3/2 O
a
Sol. [c]
Conceptual

1.3 Capacitors
If two conductors of any shape and size are placed close to each other and are given equal and
opposite charges, these form a capacitor or condenser. The capacitance C of the capacitor formed by
the two conductors is the ratio of potential difference V between them and the charge Q on either of
them.
Q
C 
V
The unit of capacitance is farad (F). One farad is the capacitance of a capacitor for which a charge
of 1 coulomb raises its potential difference by 1 volt .
1 microfarad (1F) = 10–6F.
The capacitance is also called capacity. It depends upon dimensions of conductors, separation between
conductors and the medium in which conductors are placed.
Parallel Plate Capacitor
If two parallel plates, each of area A, are separated by distance d in air, A t
+ –
 A + –
C 0 + –
d + –
–
+ –
where 0 is permittivity of free space. If a dielectric material of + –
+ –
dielectric constant K is filled between the plates, + –
+ –
K 0 A + –
C d
d
If a conducting slab of thickness t is introduced between the plates with plate separation d, then
C0 0 A
C= ( C0 = )
1 t / d d

-1 . 2 5 -
Electrostatics & Capacitors

If a dielectric slab of thickness t is introduced between the plates,

0 A
C
t
d t 
k
The dielectric constant K of a medium is the ratio of capacitance of a capacitor in the medium to that
of a capacitor with free space (vacuum) as the medium. For air, K 1. For teflon, K 2.1.
Grouping of Capacitors: When capacitor can replace a number of capacitors, it is known as equivalent
capacitor.
1. Series: Capacitor are said to in series between any two points if charge on each individual
capacitor is same i.e., q1 = q2 = q3 and
V = V1 + V2 + V3
1 1 1 1
= C C C
C 1 2 3

if n capacitors are connected in series

1 1 1 1
= C  C C
C 1 2 n
2. Parallel: When potential difference across capacitors is same they are said to be in parallel.
q = q1 + q2 + .....qn

and V1 = V2 = ..... Vn
Ceq = C1 + C2 + ..... Cn
Some more information about capacitors:
1. Force between plates per unit area
1
= 0E2
2
where E is electric field between plates.

2. If any change is done with capacitor (like changing the separation between plates, putting or
taking out dielectric from capacitor etc.), the other parameters related to capacitor changes in
such a way so that charge remains constant if change is done with battery disconnected. Potential
across capacitor remains constant for any change is done when capacitor remains connected to
battery..
(a) If charge is held constant, i.e., battery disconnected and dielectric is inserted between
plates.
(i) Charge remains unchanged, i.e., q = q0, as in an isolated system charge is conserved.
(ii) Capacity increases, i.e., C = KC0, as by presence of a dielectric capacity becomes K times.
(iii) P.D. between the plates decreases, i.e., V = (V0/K)
q q
as, V =  0 [as q = q0 and C = KC0]
C KC0

-1 . 2 6 -
Electrostatics & Capacitors

(iv) E between the plates decreases, i.e., E = (E0/K),

V V E  V0 V0 
as, E =  0  0 as V  K and E 0  d 
d Kd K  
(v) Energy stored in the capacitor decreases, i.e., U = (U0/K),

q2 q02 U
as U =   0 [as q = q0 and C = KC0]
2 C 2 KC0 K

(b) If potential is held constant, i.e., battery remains attached and dielectric in inserted
between plates
(i) PD remains constant, i.e., V = V0, as battery is a source of constant potential difference.
(ii) Capacity increases, i.e., C = KC0, as by presence of a dielectric capacity becomes K times.
(iii) Charge on capacitor increases, i.e., q = Kq0
as q = CV = (KC0)V = Kq0 [as q0 = C0V]
(iv) Electric field remains unchanged, i.e., E = E0,

 V  V0  V0 
as, E =     E0 as V  V0 and d  E 0 
d  d  
(v) Energy stored in the capacitor increases, i.e., U = KU0
1 1
as, U = CV2 = (KC0) (V0)2
2 2
1 1
= KU0 [as C = KC0 and U0 = C0V02]
2 2
3. When two capacitors C1 and C2, charged with battery of V1 and V2 respectively, are connected
together.
C1V1  C 2V2
(a) common potential, V = C1  C 2
C1
(b) final charges on capacitor, q1 = C  C ( q1  q2 )
1 2

C2
q2 = C  C ( q1  q2 )
1 2

where q1 & q2 are initial charges on C1 and C2 i.e., q1 = C1V1, q2 = C2V2

C1C 2 2
(c) Energy loss, U = 2(C  C ) (V1 ~ V2 )
1 2

-1 . 2 7 -
Electrostatics & Capacitors

Energy Stored in a Capacitor

It is equal to the work done in charging a capacitor. Let q be the instantaneous charge on either plate
q
of the capacitor of capacitance C. Then the potential difference between the plates is V = . The
C
external work done dW in transferring an infinitesimal charge dq from the negative to positive plate of
the capacitor is
q
dW = V dq = dq.
C
The total work done in transferring charge Q is the total potential energy stored in the capacitor
Q
q
U = Wext =  dW  C
0
dq

Q2 1 1
 U= = QV = CV2
2C 2 2
Energy density or energy per unit volume in a parallel plate capacitor
(1 / 2) CV 2 (1 / 2)(A  0 / d )(E d ) 2 1
= u = U/volume = = = 0E2.
Ad Ad 2

Wheastone Bridge for Capacitances:

1. If four capacitors, C1, C2, C3 and C4 are arranged B

in a closed loop ABDE and the capacitor C5 rests C1 C2

as a bridge between two diagonally opposite coners

B and E of the bridge, then bridge is said to be A C5 D

balanced if
C3 C4
C1 C3
 E
C2 C4
2. In the balanced state of the bridge, there is no potential difference between the points B and E of
the bridge.

3. There is no charge on C5, therefore it is assumed that C1 and C2 are in series and C3 and C4 wire
in series. The combined capacities of these two branches are in parallel.
To solve the electric circuits involving capacitors, we can use the following :
1. Net charge on plates joined at a junction is zero.
2. In any loop of circuit sum of potential differences across all elements is zero.
Spherical Capacitors

4 0
C
 1 1 
1.   
 R1 R2 

-1 . 2 8 -
Electrostatics & Capacitors

2. For R1 = R, R2 = (Single surface capacitor)

C = 40R

3.

40
C = C1 + C2 =  40 R2
 1 1 
  
 R1 R2 

Cylindrical capacitor

20 L , where L = length of cylinder..

C
R 
log e  2 
 R1 

Dielectrics and Polarisation

1. (a) Dipole moment of a single polar atom  electric field intensity

   
(b) p  E  p = 0  E ...(i)

where  = atomic polarisability and has unit of m3.

 –2
2. P = polarisation vector = dipole moment per unit volume [unit = C m ]

= N.0  E , where N = Number of atoms per unit volume ...(ii)

3. If  qi are induced charges on the surfaces of the dielectric slab of thickness x and area of cross-
section A, then
+++++++++++++++++++
––––––––––––––––

Ei

++ ++ ++ ++ +++ ++ ++ ++ ++
E0 – – – – – – – – – – – – – – – – –

-1 . 2 9 -
Electrostatics & Capacitors

Dipole moment of slab

P = polarisation =
Volume of slab
qi x qi
= A x = A = i ...(iii)

 Polarisation is numerically equal to induced

charge density of dielectric
4. Resultant Electric field intensity across a dielectric
= Applied electric field intensity – induced electric field intensity
i P
E = E0 – Ei = E0 – = E0 – [ P = i]
0 0
But P = 0E , where = electric susceptibility of dielectric
0 E
 E = E0 – = E0 – E
0
E0
 E (1 + ) = E0  1 + = = r ...(iv)
E
5. Where, r = relative permittivity of the medium
(i) For vacuum, r = 1  = 0 for vacuum
(ii) For metals or good conductors, r = 
(iii) (electric susceptibility of medium) is dimensionless.

Sample Problem 1.16:

A slab of thickness t and dielectric constant K is introduced between the plates of a parallel plate
capacitor such that its faces are parallel to the plates and are of same surface area of cross-section as
that of plates of the condenser.
(a) Find the capacitance of the capacitor if the separation between the plates is d.
(b) If the dielectric constant is equal to 5.0, find the value of t/d so that the capacitance of the system
becomes thrice the capacitance with free space in between the plates.
(c) Find the ratio of energy stored in capacitors in both the cases.
Sol.: (a) When the space between the plates is partially filled by a dielectric material, it acts like two
capacitors in series. The area of plates is same. The separation between plates is (d – t) for air
capacitor and t for capacitor having dielectric.
0 A
Capacity of air capacitor, C1 = d  t
A
K 0 A
Capacity of capacitor with dielectric = C2 = d
t K t
The equivalent capacitor C is given by :
1 1 1 d t t 1  t 
     ( d  t )  
C C1 C2 0 A K 0 A 0 A  K

0 A
C=
t
(d  t ) 
K

-1 . 3 0 -
Electrostatics & Capacitors

0 A
(b) Capacitance of air capacitor, C =
d
0 A 3 0 A
As C = 3C , =
t d
(d  t ) 
K
1 3

t d
(d  t ) 
K
3t
or d = 3d – 3t +
K

1  K 1 
2d = 3t 1    3t  
 K  K 
t 2K 25 5
   = 0.83
d 3(K  1) 3 4 6
(c) If U is the energy of capacitor with dielectric and U is energy of capacitor with air.
q2 q2
U/ U = C 1
U= , U = or
C
=
3
2C 2C 
The electric potential energy stored in the capacitor is reduced due to introduction of capacitor.

Sample Problem 1.17:

A capacitor has square plates of side a making small angle  with each other. Find the value of its
capacitance if d is the minimum separation between plates at one end.

Sol.: The spacing between the plates is increasing continuously. Consider an element of strip of width x
lying at a distance x from the left.
The separation between the plates for the strip = d + x tan  d + x ( tan  » )
Area of strip of the plate = a.x
0 . a . x
Capacitance of the strip = C = d  x

The capacitance of the capacitor C is equal to sum of similar elemental strips connected in parallel.
C =  C
In limiting case when x  0, a
x
a
0 a dx 
C=  d  x
0 d

a
 a dx x x
= 0  a
d  x 
0 1  
 d 
a  a 
 1 
0 a x  0 a  d 
= d  log1  d  =  log 1  0 
  0  
 
0 a  a 
C = log1  
  d 

-1 . 3 1 -
Electrostatics & Capacitors

Sample Problem 1.18:

Find out the following if A is connected with C and B is connected with D.
(a) How much charge flows in the circuit.
(b) How much heat is produced in the circuit.
2µF
A + – B

20V
V 0
3µF
+ –
C D
10V
Sol. Let potential of B and D is zero and common potential on capacitors is
V, then at A and C it will be V
3V + 2V = 40 + 30
5V = 70 V = 14 Volt
Final charge on 2 µF = 28µC A
Final charge on 3 µF = 42µC 28µC –28µC
Charge that will flow from 2µ capacitor = 40 – 28 = 12 µC
+12µC +12µC
1 2 1 2 1 2
(b) Heat produced =  2  (20)   3  (10)   5  (14)
2 2 2 C + – D
= 400 + 150 – 490 – 42µC
= 550 – 490 = 60 J 42µC
Sample Problem 1.19:
(a) Find the effective capacitance between A and B of an infinite chain of capacitors joined as
shown in Fig. (A).
(b) For what value of C0 in the circuit shown in Fig. (B) will be net effective capacitance between A
and B be independent of the number of capacitor in the chain ?
C1 C1 C1 C1 C1 C1
A A D

C2 C2 C2 To infinite C2 C2 C2 C0

B (A) B (B) E
Sol. (a) Suppose the effective capacitance between A and B is CR. Since the network is infinite, even if
we remove one repeating unit from the chain remaining network would still have infinite cells,
i.e., effective capacitance between DE would also be CR.

C1 C1
A D A D

CR C2 CR CR C0 C2 C0

B E E
(A) B (B)

-1 . 3 2 -
Electrostatics & Capacitors

In other words the given infinite chain is equivalent to capacity C1 in series with combination of
C2 and CR in parallel as shown in Fig. (A). So
C1 (C2  CR )
CR = C1 S [C2 + CR] i.e., CR 
C1  C2  CR

1 2 
or CR2 + C2CR – C1C2 = 0 i.e., CR   C2  C2  4C1C2 
2 
And as capacitance cannot be negative, only permissible value of CR is :

C2   C  
CR   1  4 1   1 However, if C1 = C2 = C; CR = [5 – 1]C/2
2   C2  
 
(b) Suppose there are n capacitor between A and B and the network is terminated by C0 with
equivalent capacitance CR [Fig. (B)], Now if we add one more capacitor to the network between
D and E, the equivalent capacitance of the network CR will be independent of number of cells if
the capacitance between D and E still remains C0, i.e.,
C1 (C2  C0 )
C1 S [C2 + C0] = C0 or = C0 i.e., C02 + C2C0 – C1C2 = 0
[C1  C2  C0 ]

C2   C  
Which on simplification gives : C0    1  4 1   1
2   C2  
Sample Problem 1.20:  
Find the equivalent capacitance between A and B.
C 2C

A B
2C

2C C

Sol. The given circuit forms a Wheatstone bridge. But the bridge is not balanced. Let us suppose point A
is connected to the positive terminal of a battery and B to the negative terminal of the same battery;
so that a total charge q is stored in the capacitors. Just by seeing input and output symmetry we can
say that charges will be distributed as shown below :
q1 + q2 = q ... (i)
Applying second law, we have
q1 q3 q 2
   0
C 2C 2C
or q2 – q3 – 2q1 = 0 ... (ii)
Plates inside the dotted line form an isolated system. Hence,
q2 + q3 – q1 = 0 ... (iii)
Solving these three equations, we have
2 3 q
q1 = q, q2 = q and q3 = –
5 5 5
Now, let Ceq be the equivalent capacitance between A and B.

-1 . 3 3 -
Electrostatics & Capacitors

Then,
q q1 q 2
V A – VB = = 
Ceq C 2C

q 2q 3q 7q
   
Ceq 5C 10C 10C

10
 Ceq = C
7
Sample Problem 1.21:
The capacitance of all the capacitors shown in figure are in micro farad. What is the equivalent
capacitance between A and B ? If the charges on the 5µF capacitors is 120 µC, what is the potential
difference between A and C ?

Sol. 5 F, 64 Volt

Sample Problem 1.22:

The equivalent capacitance across A and B is.

Sol. 5C/4

OBJECTIVE QUESTIONS : 1 – III

1. The plate of a parallel plate capacitor are separated by d cm. A plate of thickness t cm. with dielectric
constant k1 is inserted and the remaining space is filled with a plate of dielectric constant k2. If Q is the
charge on the capacitor and area of plates is A cm2 each, then potential difference between the plates
is
Q  t d t  4Q  t d  t  4Q  k1 k2  Q  k1 d  t 
(a)    (b)    (c)    (d)   
0 A  k1 k2  A  k1 k2  A  t d t  0 A  t k2 
Sol. [a]
Conceptual

-1 . 3 4 -
Electrostatics & Capacitors

2. The circuit was in the shown state for a long time. Now if the 50V
switch S is closed then the charge that flows through the switch
S, will be
4µF 2µF
400
(a) C (b) 100 µC
3
100 S
(c) C (d) 50 µC
3 2µF 4µF
Sol. [d]
Initial and final charges are marked on 4µf and 2µf capacitor as shown.
50V

q1=200/3µC
q1=100µC
4µF 2µF
1 q 1=100/3µC

S
1 q2=50/3µC
2µF 4µF
q 1=200/3µC q1=50µC

Hence charge passing through segment 1 and 2 are

100 50
q1  C , q2  C
3 3
 charge through switch = q1 + q2 = 50 µC
3. A capacitor of 1µF withstands a maximum voltage of 6 kilovolt while another capacitor of 2µF
withstands a maximum voltage of 4 kilovolt. If the two capacitors are connected in series, the system
will withstand a maximum voltage of
(a) 2 kV (b) 4 kV (c) 6 kV (d) 9 kV
Sol. [d]
For series combination
CC 2
Cs  1 2  Cs  F
C1  C2 3
When connected in series the maximum charge that can flow through the combination equals the lower
value of charge accommodated by the first capacitor i.e. 6000 µC
 Q1 = 6000 µC and 8000 µC
Q 6000C
Vs  1 
Cs (2 / 3)C  Vs = 9kV
4. Three uncharged capacitors of capacities C1, C2 and C3 are connected to one another as shown in the
figure. Points A, B and D are at potential V1, V2 and V3 then the potential at O will be
V1C1  V2C2  V3C3 V1  V2  V3 A
(a) C1  C2  C3 (b) C  C  C
1 2 3 C1
V1 V2  V3  V1V2V3
(c) C C  C (d) C C C O
1  2 3  1 2 3 C2 C3
Sol. [a]
Using Kirchhoff’s Law B D

-1 . 3 5 -
Electrostatics & Capacitors

5. A parallel plate capacitor of area ‘A’ plate separation ‘d’ is filled with two dielectrics as shown. What
is the capacitance of the arrangement ?
A/2 A/2

K d
d
K
2

3K 0 A 4 K 0 A  K  1 0 A K  K  3  0 A
(a) (b) (c) (d) 2  K  1 d .
4d 3d 2d
Sol. [d]
( A / 2) 0 A 0 A A c1
c1   , c2  K 0 , c3  K 0 c3
(d / 2) d d 2d K d
d K c2
2
ceq. 
c1  c2
 c3 
 3  K  KA0
 c1  c2 2d  K  1
( C1 and C2 are in series and resultant of these two in parallel with C3)
6. Two conducting spheres of radii a and b are separated by a large distance. The capacity of this
system between points A and B is

4o
A B b
(a) 1 1 (b) 4o (a  b)
 a
a b

(c) 2o (a  b) (d) none of the above

Sol. [a]
Two spherical capacitors are in series.
7. In an isolated parallel plate capacitor of capacitance C the four surfaces have charges Q1 , Q2, Q3 and
Q4 as shown in the figure. The potential difference between the plates is
Q1 Q3
Q1  Q2 Q2  Q3
(a) (b)
C C Q2 Q4
Q3 1
(c) (d)  Q1  Q2    Q3  Q4 
C C
Sol. [c]
|Q2| = |Q3| and having opposite polarity.
8. Three capacitors each having capacitance C = 2F are connected to battery of emf 30 V as shown in
the figure. When the switch is closed. Which of the following is incorrect. S
(a) The amount of charge that flows through the battery is 20C
C
(b) Heat generated in the circuit is 0.6 mJ C
C
(c) Work done by the battery is 0.6 mJ
(d) The charge flowing through the switch is 60C
30V
Sol. [b]
4
When the switch is open, the equivalent capacitance C1    F
3

-1 . 3 6 -
Electrostatics & Capacitors

4
Charge flown through the battery =  30  40C
3
When switch is closed
C1  2F
q1  2  30  60C
Hence , 20C extra charge flows when S is closed
Work done by the battery = 20 × 10–6 × 30 = 0.6 mJ
When switch is closed, it is short circuited. Hence all the charge flows through S.
9. In the circuit shown, some potential difference is applied between points A and B. If C is joined to
D.
(a) No charge will flow between C and D.
(b) Some charge will flow between C and D.
(c) Equivalent capacitance between A and B will decrease.
(d) Equivalent capacitance between A and B will change.
Sol. [a]
C and D are at same potential
10. A dielectric slab of thickness d is inserted in a parallel plate capacitor where negative plate is at
x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is given
some charge. As x goes from 0 to 3d
(a) the magnitude of electric field remains the same.
(b) the direction of electric field remain the same.
(c) the electric potential doesn’t change.
(d) the electric potential increases at first, then decrease and again increases.
Sol. [b]
The potential is a continuous function of x.

-1 . 3 7 -
 CHAPTER ASSIGNMENT
STRAIGHT OBJECTIVE TYPE
1. Two identical rings, each of radius R, are co-axially placed. The distance between their centres is R. A
charge Q is placed on each ring. The work done in moving a small charge q from the center of one ring to
that of the other is
1
(a) 4   R 
 2q Q 
(b) 4  
1  

 2 1 q Q
(c) 4  
1 

2q Q 
(d) zero

0   
0  2 R   ( 2  1) R 
0 

2. Two concentric hollow spheres of radii R and r (R > r) have positive charge of equal surface density .
The electrical potential at their common centre is
  (R  r )  (R  r )  (R  r )
(a)  (b) 0 (c) 0 (d) 0 r
0

3. Four charges, each equal to q, are placed at the corners of a square of side l. The electrical potential at
the centre of the square is

q 1  4q  1 
2q  2q
(a)   l (b)    2l  (c)    l  (d)   l
0 0   0   0
20 F
4. The charge and potential difference across 4 F capacitors
+
in the given circuit is 310 V 4 F 4 F
–
(a) 1200 C, 310 V (b) 600 C, 310 V
(c) 1200 C, 150 V (d) 600 C, 150 V 12 F

5. Five capacitors are connected as shown. The equivalent 12 F 10 F 9 F 8 F

capacitance between terminals A and B and charge on A
5 F
5F capacitor will be respectively +
(a) 8 F, 100 C (b) 4 F, 50 C 60 V
–
(c) 12 F, 150 C (d) 16 F, 200 C
B
6. Three charges +q, +q and Q are placed at the vertices of a right-angled
isosceles triangle as shown in the figure. The net electrostatic potential +q
energy of the configuration is zero if Q is equal to
q  2q a 90º a
(a) 2 1
(b)
2 1
q
(c) 2 2 (d) –2q +q Q

7. Two equal negative charges –q are fixed at points with coordinates (0, a) and (0, –a) on y-axis. A
positive charge Q is released from rest at the point with coordinates (2a, 0) on x-axis. The charge Q will
(a) move to origin and remain at rest
(b) execute simple harmonic motion about the origin
(c) move to infinity
(d) execute oscillatory motion but not simple harmonic motion

-1 . 3 8 -
Electrostatics & Capacitors

8. The charge on the capacitor of 5 F in the circuit given in the 3 F

figure is 2 F

5 F
(a) 7 C (b) 9 C
4 F
(c) 12.6 C (d) 21 C + –
6V
9. Two concentric hollow metal spheres have radii R1 and R2. The outer sphere of radius R2 is given a
positive charge q and the inner is earthed. The charge on the inner sphere is
(a) zero
+q
(b) –q
(c) –R1q(R1+R2)
(d) –R1q/R2
10. Two slabs of the same dimensions having dielectric constants
K1 and K2 completely fill the space between plates of a parallel
plate capacitor as shown in the figure. If C is original K1 K2
capacitance of the capacitor, the new capacitance is
 K1  K2   2 K 1K 2 
(a)  C (b)   C
 2   K1  K 2 
 K 1K 2  d/2 d/2
(c) (K1 + K2) C (d)   C
 K1  K 2 
11. A solid sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical
shell. The potential difference between the surface of the solid sphere and that of the outer surface of
the hollow shell is V. If the shell is now given a charge of –3Q, the new potential difference between the
two surfaces is
(a) V (b) 2V (c) 4V (d) –2V

12. Seven capacitors, each of capacitance 2 F are to be connected in a configuration to obtain an effective
capacitance of (10/11) F. Which of the combination(s), as shown in the figure, will achieve the desired
result?

(a) (b)

(c) (d)

13. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due
to the sphere at a distance r from its centre
(a) increases as r increases for r < R (b) decreases as r increases for 0 < r < 
(c) increases as r increases for R < r <  (d) is discontinuous at r = R

-1 . 3 9 -
Electrostatics & Capacitors

14. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 ........... up to infinity on the x-axis and
a charge –q is fixed at each of the points x = 2x0, x = 4x0 and x = 6x0 ........... where x is a positive
Q
constant. Take the electric potential at a point due to a charge +Q at a distance r from it to be 4 r .
0
Then the potential at the origin due to the above system of charges is .........
q q ln 2
(a) 0 (b) (c)  (d) 4   x
8 0 ln 2 0 0

15. A charge Q is distributed over two concentric hollow spheres of radii R and 2R such that the surface
charge densities are equal. The potential at the common centre is

2 Q 3 Q 2 Q 3 Q
(a) 3 . 4  R (b) 4 . 4  R (c) 5 . 4  R (d) 5 . 4  R
0 0 0 0

MULTIPLE CORRECT ANSWERS TYPE

16. Mark the correct options
(a) Gauss’s law is valid only for uniform charge distributions
(b) Gauss’s law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss’s law is the field due to all the charges.
(d) The flux of electric field through a closed surface due to all the charges is equal to the flux due to the
charges enclosed by the surface.

17. Select the correct alternative

(a) The charge gained by the uncharged body from a charged body due to conduction is equal to half of
the total charge initially present.
(b) The magnitude of charge increases with the increase in velocity of charge.
(c) Charge cannot exist without matter although matter can exist without charge
(d) Between two non-magnetic substances repulsion is the true test of electrification (electrification
means body has net charge)

18. The electric field intensity at a point in space is equal to-

(a) Magnitude of the potential gradient there
(b) The electric charge there
(c) The magnitude of the electric force, a unit charge would experience there
(d) The force, an electron would experience there

B D
19. Figure shows a charge q placed at the centre of a hemisphere. A
second charge Q is placed at one of the positions A, B, C and D.
In which position(s) of this second charge, the flux of the electric
field through the hemisphere remains unchanged ? C q A

(a) A (b) B (c) C (d) D

-1 . 4 0 -
Electrostatics & Capacitors

20. An electric dipole is placed at the centre of a sphere, mark the correct options –
(a) The flux of the electric field through the surface is zero.
(b) The electric field is zero at every point of the sphere.
(c) The electric field is not zero anywhere on the sphere.
(d) The electric field is zero on a circle on the sphere.
21. A small uncharged metallic sphere is positioned exactly at a point midway between two equal and
opposite point charges separated by very small distance. If the sphere is slightly displaced towards the
positive charge and released, then
(a) it will oscillate about it’s original position.
(b) it will move further towards the positive charge.
(c) its potential energy will decrease and kinetic energy will increase.
(d) the total energy remains constant but is non-zero.
22. A, B and C are three large, parallel conducting plates, placed horizontally. A and C are rigidly fixed and
earthed. B is given some charge. Under electrostatic and gravitational forces, B may be:
(a) in equilibrium exactly midway between A & C A
(b) in equilibrium if it is closer to A than to C B
(c) in equilibrium if it is closer to C than to A
(d) B can never be in stable equilibrium
C
23. Two concentric shells have radii R and 2R, charges qA and qB and potentials 2V and (3/2) V respectively.
Now shell B is earthed and let charges on them become q ' A and q 'B . Then
qA q
B
(a) qA / qB = 1/2
B
A
(b) | q ' A | / | q 'B |  1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
24. In the circuit shown, the potential difference across the 3F capacitor is V, and the equivalent
3F 6F
capacitance between A and B is CAB f
e
(a) CAB = 4F V

18 F
(b) C AB  F c d
11
(c) V = 20 volt
B
(d) V = 40 volt A
60V
25. In the circuit shown, each capacitor has a capacitance C. The emf of the cell is . If the switch S is
closed
C S
(a) some charge will flow out of the positive terminal of the cell
(b) some charge will enter the positive terminal of the cell
(c) the amount of charge flowing through the cell will be C C
4 C
(d) the amount of charge flowing though the cell will be C +

3

-1 . 4 1 -
Electrostatics & Capacitors

LINKED COMPREHENSION TYPE

Comprehension : 1
Van de Graaff generators like the one shown in figure are used to produce very high voltage. In the
figure, the + signs represent positive charge and the –ve signs represent negative charge. In this
common Van de Graaff generator, charge is separated by the frictional contact of the belt and the
lower pulley shown. Positive charge collects on the lower pulley and an equal amount of negative
charge spreads out along the inside of the belt. Electrons from the ground are attracted to the outside
of the belt by the net positive charge on the lower portion of the belt-pulley system. These electrons
travel up the belt and are transferred to the dome, which is a hollow metal sphere. A high negative
charge density can be built up on the dome, because the electrons from the outside of the belt do not
experience a repulsive force from the charge built up on the outside of the sphere.
The electric potential of the dome is V = Er where E is the –
– –
electric field just outside the dome and r is the radius. The – –
– –
charges on the surface of the dome do not affect the electric-
– –
field inside the cavity. The potential that can build up on the – – Upper pulley
dome is limited by the dielectric strength of the air, which is – –
– –
about 30,000 V/cm for dry air. Beyond this air molecules are Belt
– –
– – Lower pulley
ionized. This enables the air to conduct electricity. – –
Van de Graaff generators are routinely used in college physics – ++ –
++ –
laboratories. When a student gets within a few inches of a Ground
Van de Graaff generator, she may draw a spark with an
instantaneous current of 10A and remain uninjured. An
instantaneous current is the transfer of charge within 1 s.

26. The 660 volt rails on a subway can kill a person upon contact. A 10,000 volt Van de Graaff generator,
however, will only give a mild shock. Which of the following best explains this paradox?
(a) The generator provides more energy per charge, but since it has few charges it transfers a lesser
amount of energy
(b) The generator provides more energy, but since there is little energy per charge the current is small
(c) Most of the energy provided by the generator is dissipated in the air because air presents a smaller
resistance than the human body
(d) Most of the energy flows directly to the ground without going through the human body since the
generator is grounded

27. What is the maximum potential the dome, with a radius of 10 cm, can sustain in dry air?
(a) 3 kV (b) 5 kV (c) 300 kV (d) 500 kV

28. What is the work required to move a charge q from the top of the belt to the surface of the dome, if the
amount of charge on the dome is Q and q is the only charge on the belt?
(a) 0 (b) kQq/2r (c) kQq/r (d) kq/r

-1 . 4 2 -
Electrostatics & Capacitors

Comprehension : 2
Two capacitors A and B with capacities 3 F and 2 F are charged to a potential difference of 100V
and 180 V respectively. The plates of the capacitors are connected as shown in the figure. With one
wire from one each capacitor free. The upper plate of A is positive and that of B is negative. An
uncharged capacitor C with lead wires falls on the free ends to complete the circuit.
2F C

3 F 2F
A B

2
29. Final charge on capacitor having capacitance 3 F will be
(a) 90C (b) 150C (c) 120C (d) 300C
30. Charge flow through section 2 in the direction as shown in figure.
(a) 210C (b) 150C (c) –210C (d) –150C
31. Energy loss in redistribution of charges on capacitors will be
(a) 47.5 mJ (b) 45.6 mJ (c) 1.8 mJ (d) 29.4 mJ

MATRIX MATCH TYPE

32. An electric dipole is placed in an electric field. The column I gives the description of electric field and
 
the angle between the dipole moment p and the electric field intensity E and the column II gives the
effect of the electric field on the dipole. Match the description in Column I with the statements in
column II.
Column I Column II
A. Uniform electric field, = 0 (p) Force = 0
B. Electric field due to a point charge, = 0 (q) Torque = 0

C. Electric field between the two oppositely charged large plates, = 90° (r) p.E  0
D. Dipole moment parallel to uniformly charged long wire. (s) Force 0
33. Match the following :
Column I Column II
A. Change in Electric field (p) Is in the case of a parallel plate capacitor.
B. Uniform electric field (q) The equipotential surfaces are parallel planes
which are perpendicular to the field lines.

 0A 
C. Capacitance  C   (r) Travels with speed of light.
 d 
D. Electric potential Energy (s) Is stored in a uniformly charged thin spherical
shell from the surface upto infinity.
(t) Is stored in spherical capacitor

-1 . 4 3 -
Electrostatics & Capacitors

INTEGER TYPE QUESTIONS

34. Two identical charged spheres are suspended by strings of equal length. The strings make an angle of
30° with each other. When suspended in a liquid of density 0.8 gm/cc, the angle remains the same.
What is the dielectric constant of the liquid ? (Density of the material of sphere is 1.6 gm/cc.)

35. Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a mass m
and a negative charge q, is placed on its axis at a distance x from the centre. Assuming x < < R, find the
time period of oscillations of the particle if it is released from there. Given [m2R3 = 4kQq]. All the
values are in S.I. system.
C
36. Three concentric conducting shells, A, B and C of radii a, b and c B
A
are as shown in figure. Find the capacitance of the assembly a
between A and C. Given [c = a + 40ac]. All the values are b
in S.I system. c

2
37. A system consists of a ball of radius (where ‘a’ is constant) carrying a spherically symmetric
a
charge and the surrounding space filled with a charge of volume density  = a/r, where r is the
distance from the centre of ball. Find the ball’s charge at which the magnitude of the electric field is
independent of r outside the ball . All the values are in S.I. system.

38. In the figure shown there are three thin large metallic plates.
The middle plate carries a total charge q = 4C. Plates
1 and 2 are connected by a wire. Find the charge induced
on the outer surface of plate 1 and 2 in micro coulumb.
Given l1 = 5 cm, l = 10 cm.

39. When the switch S in the figure is thrown to the left, the plates
of capacitors C1 acquire a potential difference V = 2.5 V.
Initially the capcitors C2 and C3 are uncharged. The switch is
now thrown to the right. What are the final charges on C2 and
C3 capacitors. Given C1 = 5C, C2 = 5C, C3 = 10C.

40. The connections shown in figure are established with the switch S open. A charge Q = 6KC where K
is constant will flow through the switch if it is closed. Then find K value ?

41. How much heat will be generated in the circuit shown in the
C C0 C
adjoining figure after the switch is shifted from position 1 to 2
position 2 ? All the values are in S.I.system.
1 2
10 5
Given : V   .
C0 C
V

-1 . 4 4 -
Electrostatics & Capacitors

42. Two capacitors C1 = 1F and C2 = 4F are charged to a potential difference of 100 volts and 200 volts
respectively. The charged capacitors are now connected to each other with terminals of opposite sign
connected together. If final charge on capacitor C1 = 1F in steady state is 20xC, then find the value
of x .
43. Figure shows two conducting thin concentric shells of radii
r and 3r. The outer shell carries charge q = 6C. Inner shell
is neutral. Find the charge (in C) that will flow from inner
shell to earth after the switch S is closed.

-1 . 4 5 -
 PREVIOUS YEAR QUESTIONS
IIT-JEE/JEE-ADVANCE QUESTIONS

1. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown
in the figure as [IIT]

(a) 1 (b) 2 (c) 3 (d) 4

2. The magnitude of electric field E in the annular region of a charged cylindrical capacitor.
(a) is same throughout
(b) is higher near the outer cylinder than near the inner cylinder
(c) varies as 1/r, where r is the distance from the axis
(d) varies as 1/r2 where r is the distance from the axis [IIT]

3. Seven capacitors each of capacitance 2 µF are to be connected to obtain a capacitance of (10/11) µF.
Which of the following combination is possible [IIT]
(a) 5 in parallel 2 in series (b) 4 in parallel 3 in series
(c) 3 in parallel 4 in series (d) 2 in parallel 5 in series
4. Two identical metal plates are given positive charges Q1 and Q2 (<Q1) respectively. If they are now
brought close together to form a parallel plate capacitor with capacitance C, the potential difference
between them is [IIT]
(a) (Q1 + Q2)/(2C) (b) (Q1 + Q2)/C (c) (Q1 – Q2)/C (d) (Q1 – Q2)/(2C)

5. Four the circuit shown, which of the following statements is true? [IIT]

(a) With S1 closed, V1 = 15 V, V2 = 20 V (b) With S3 closed, V1 = V2 = 25 V

(c) With S1 and S2 closed, V1 = V2 = 0 (d) With S1 and S3 closed, V1 = 30 V, V2 = 20 V

6. Three charges Q, + q and +q are placed at the vertices

of a right-angled isosceles triangle as shown. The net
electrostatic energy of the configuration is zero, if Q is
equal to
q 2 q
(a) (b) A
1 2 2 2
(c) –2 q (d) + q [IIT]

-1 . 4 6 -
Electrostatics & Capacitors

7. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different
dielectric materials having dielectric constants k1, k2 and k3 as shown. If a single dielectric materials is
to be used to have the same capacitance C in this capacitor, then its dielectric constant k is
given by [IIT]

1 1 1 1 1 1 1 k1 k 2
(a)    (b)   (c) k = k  k + 2 k3 (d) k = k1 + k2 + 2k3
k k1 k 2 2k 3 k k1  k 2 2k 3 1 2

8. A uniform electric field pointing in positive X-direction exists in a region. Let A be the origin, and B be
the point on X-axis at x = 1 cm and C be the point on Y-axis at y = 1 cm. Then the potential at points A,
B and C satisfy [IIT]
(a) VA < VB (b) VA > VB (c) VA < VC (d) VA > VC

9. Three point charges of equal value q are placed at the vertices of an equilateral triangle. The resulting
lines of force should be sketch as in [IIT]

(a) (b) (c) (d)

10. A capacitor A with charge Q0 is connected through a resistance to another identical uncharged capacitor
B at t = 0. The charges on A and B at t = t are QA and QB respectively. Indicate the correct curves.

(a) (b) (c) (d)

11. Plates of a parallel plate air capacitor of capacitance 0.01 F are held vertical. The capacitor is first
charged by a battery of emf 20 V and then the battery is disconnected. The capacitor is reconnected
across the same battery but with reversed polarity. Pick the correct statements
(a) After reconnection of the battery, heat generated in the circuit is equal to 8 J.
(b) No energy is supplied by the battery after reconnection because final energy of the capacitor is
equal to energy just before reconnection.
(c) Energy is supplied by the battery after reconnection and it is equal to 2J
(d) No charged mass can remain in equilibrium in the space between plates of the capacitor.

-1 . 4 7 -
Electrostatics & Capacitors

12. For spherical symmetrical charge distribution, variation of electric potential with distance from centre
is given in diagram. Given that

q q
V= for r  r0 and V= for r  r0
40 r0 4 0 r
Then which option(s) are correct V [IIT]
(a) Total charge within 2r0 is q
(b) Total electrostatic energy for r  r0 is zero
(c) At r = r0 electric field is discontinuous
r0 r
(d) There will be no charge anywhere except at r = r0

13. A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of
larger radius. Both the cylinders are initially electrically neutral. [IIT]
(a) A potential difference appears between the two cylinders when a charge density is given to the
inner cylinder
(b) A potential difference appears between the two cylinders when a charge density is given to the
outer cylinder
(c) No potential difference appears between the two cylinders when a uniform line charge is kept
along the axis of the cylinders
(d) No potential difference appears between the two cylinders when same charge density is given to
both the cylinders

14. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its
volume as shown in figure. The electric field inside the emptied space is [IIT]
(a) zero everywhere
(b) non-zero and uniform
(c) non-uniform
(d) zero only at its center

 a  a
15. Positive and negative point charges of equal magnitude are kept at  0, 0,  and  0, 0,  ,
 2  2

respectively. The work done by the electric field when another positive point charge is moved from
(–a, 0, 0) to (0, a, 0) is [IIT]
(a) positive (b) negative
(c) zero (d) depends on the path connecting the initial and final positions
16. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric
constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed
V, the time constant as a function of time t is [IIT]

6 0 R (15d  9Vt ) 0 R
(a) (b) C
5d  3Vt 2d 2  3dVt  9V 2 t 2
d d/3 R
6 0 R (15d  9Vt ) 0 R
(c) (d)
5d  3Vt 2d 2  3dVt  9V 2 t 2

-1 . 4 8 -
Electrostatics & Capacitors

q q 2q
17. Consider a system of three charges , and  placed at points A, B and C, respectively, as shown
3 3 3
in figure. Take O to be the centre of the circle of radius R and angle CAB = 60º [IIT]
y
q
(a) The electric field at point O is 8 R 2 directed along the negative x-axis B
0

C
(b) The potential energy of the system is zero x
O
q2 60º
(c) The magnitude of the force between the charges at C and B is
54 0 R 2 A
q
(d) The potential at point O is 12 R
0

18. STATEMENT-1: For practical purposes, the earth is used as a reference at zero potential in electrical
circuits.
STATEMENT-2: The electrical potential of a sphere of radius R with charge Q uniformly distributed
Q
on the surface is given by . [IIT]
4 0 R
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 and 2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

Paragraph for Question Nos. 19 to 21

The nuclear charge (Ze) is non-uniformly distributed within
a nucleus of radius R. The charge density (r) [charge per
(r)
unit volume] is dependent only on the radial distance r from
d
the centre of the nucleus as shown in figure. The electric
field is only along the radial direction. [IIT]

19. The electric field at r = R is a R r

(a) independent of a (b) directly proportional to a
(c) directly proportional to a2 (d) inversely proportional to a

20. For a = 0, the value of d (maximum value of  as shown in the figure) is

3Ze 3Ze 4Ze Ze

(a) (b) (c) (d)
4 R 3 R 3 3R3 3R3

21. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies

R 2R
(a) a = 0 (b) a (c) a = R (d) a
2 3

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Electrostatics & Capacitors

22. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre
at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from
x = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0),
respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The
electric flux through this cubical surface is [IIT]
y
 2C 2C
(a) o (b) o
x
10C 12C
(c) o (d) o

23. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively.
It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio
of the charges given to the shells, Q1 : Q2 : Q3, is [IIT]
(a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 1 : 4 : 9 (d) 1 : 8 : 18

24. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical
orbit. Find out the correct statement (s) [IIT]

(a) The angular momentum of the charge –q is constant

(b) The linear momentum of the charge –q is constant
(c) The angular velocity of the charge –q is constant
(d) The linear speed of the charge –q is constant

25. Column I gives certain situations in which a straight metallic wire of resistance R is used and Column
II gives some resulting effects. Match the statements in Column I with the statements in Column II and
indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.
Column I Column II
(a) A charged capacitor is connected to (p) A constant current flows through the wire
the ends of the wire
(b) The wire is moved perpendicular to its (q) Thermal energy is generated in the wire
length with a constant velocity in a
uniform magnetic field perpendicular to
the plane of motion
(c) The wire is placed in a constant electric (r) A constant potential difference develops
field that has a direction along the length between the ends of the wire
of the wire
(d) A battery of constant emf is connected (s) Charges of constant magnitude appear at the
to the ends of the wire ends of the wire [IIT]

-1 . 5 0 -
Electrostatics & Capacitors

26. Column II gives certain systems undergoing a process. Column I suggests changes in some of the
parameters related to the system. Match the systems in Column I to the appropriate process (es) from
Column II. [IIT]
Column-I Column-II
(A) The energy of the system is increased (p) System: A capacitor, initially uncharged.
Process: It is connected to a battery
(B) Mechanical energy is provided to the system, (q) System: A gas in an adiabatic container
which is converted into energy of random fitted with an adiabatic piston
motion of its parts
Process: The gas is compressed by pushing
the piston
(C) Internal energy of the system is converted into (r) System: A gas in a rigid container
its mechanical energy Process: The gas gets cooled due to colder
atmosphere surrounding it.
(D) Mass of the system is decreased (s) System: A heavy nucleus, initially at rest
Process: The nucleus fissions into two
fragments of nearly equal masses
and some neutrons are emitted
(t) System: A resistive wire loop.
Process: The loop is placed in a time varying
magnetic field perpendicular to its
plane.
27. A solid sphere of radius R has a charge Q distributed in its volume with a charge density   r a ,
R 1
where  and a are constants and r is the distance from its centre. If the electric field at r  is
2 8
times that at r = R, find the value of a. [IIT]

28. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air.
When two dielectrics of different relative primitivities (1 = 2 and 2 = 4) are introduced between the
C2
two plates as shown in the figure, the capacitance becomes C2. The ratio C is [JEE-Adv]
1
d/2

+ –
S/2

d
(a) 6/5 (b) 5/3 (c) 7/5 (d) 7/3

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Electrostatics & Capacitors

29. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the
x-axis are shown in the figure. These lines suggest that [IIT]

(a) |Q1| > |Q2|

(b) |Q1| < |Q2|
Q1 Q2
(c) at a finite distance to the left of Q1 the electric field is zero
(d) at a finite distance to the right of Q2 the electric field is zero

30. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of  per
unit area. It is made of two hemispherical shells, held together by pressing them with force F (see
figure). F is proportional to [IIT]

1 2 2 1 2
(a) R (b) R
0 0
F F
1 2 1  2

(c) (d)
0 R 0 R 2

31. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric

81
field of strength  105 Vm –1. When the field is switched off, the drop is observed to fall with
7
terminal velocity 2 × 10–3 ms–1. Given g = 9.8 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the
density of oil = 900 kg m–3, the magnitude of q is [IIT]

(a) 1.6 × 10–19 C (b) 3.2 × 10–19 C (c) 4.8 × 10–19 C (d) 8.0 × 10–19 C

32. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors
have no charge initially, at what time (in second) does the voltage across them become 4 V?

[Take : ln 5 = 1.6, ln 3 = 1.1] [IIT]

2MQ 2µF

A 2MQ B
2µF 1 2
33. A 2 µF capacitor is charged as shown in figure. The
s
percentage of its stored energy dissipated after the switch S
is turned to position 2 is [IIT]
(a) 0 % (b) 20 %
V
(c) 75% (d) 80%

34. Consider an electric field E  E0 xˆ , where E0 is a constant. z
The flux through the shaded area (as shown in the figure)
(a,0,a) (a,a,a)
due to this field is [IIT]
(a) 2E0a 2 (b) 2E0 a 2
E0 a 2 y
(c) E0 a 2 (d) (0,0,0) (0,a,0)
2 x

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Electrostatics & Capacitors

35. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart
and each is given charge +Q. Now they are connected by a thin metal wire. Then [IIT]
(a) E Ainside  0 (b) QA  QB
 A RB
(c)  (d) E Aon surface  EBon surface
 B RA

36. A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a

charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of
the block will be [IIT]

(a) of the same frequency and with shifted mean position E

(b) of the same frequency and with the same mean position
+Q
(c) of changed frequency and with shifted mean position

(d) of changed frequency and with the same mean position

37. Which of the field patterns given below is valid for electric field as well as for magnetic field?
[IIT]

(a) (b) (c) (d)

38. Which of the following statement(s) is/are correct? [IIT]

(a) If the electric field due to a point charge varies as r–2.5 instead of r–2, then the Gauss law will still
be valid

(b) The Gauss law can be used to calculate the field distribution around an electric dipole

(c) If the electric field between two point charges is zero somewhere, then the sign of the two
charges is the same

(d) The work done by the external force in moving a unit positive charge from point A at potential VA
to point B at potential VB is (VB – VA)

39. Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side
‘a’. The surface tension of the soap film is . The system of charges and planar film are in equilibrium,
1/ N
 q2 
and a  k   , where k is a constant. Then N is [IIT]
 

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Electrostatics & Capacitors

40. Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage
source of potential difference X. A proton is released at rest midway between the two plates. It is
found to move at 45° to the vertical JUST after release. Then X is nearly [IIT]
(a) 1 × 10–5 V (b) 1 × 10–7 V (c) 1 × 10–9 V (d) 1 × 10–10 V

41. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface

charge density. The variation of the magnitude of the electric field | E ( r ) | and the electric potential
V(r) with the distance r from the centre, is best represented by which graph? [IIT]
V(r) V(r)

| E (r ) | | E (r) |
(a) (b)

O R r O R r
V(r) V(r)

| E (r) | | E (r) |
(c) (d)

O R r O R r
42. A cubical region of side a has its centre at the origin. It
encloses three fixed point charges, –q at (0, – a/4, 0) +3q at
z
(0, 0, 0) and –q at (0, + a/4, 0).
Choose the correct option (s). a
(a) The net electric flux crossing the plane x = + a/2 is
equal to the net electric flux crossing the plane
–q
x = – a/2. y
–q 3q
(b) The net electric flux crossing the plane y = + a/2 is
more than the net electric flux crossing the plane
y = – a/2. x
(c) The net electric flux crossing the entire region is q/0
(d) The net electric flux crossing the plane z = + a/2 is equal to the net electric flux crossing the plane
x = +a/2 z [IIT]
43. An infinitely long solid cylinder of radius R has a uniform
volume charge density . It has a spherical cavity of radius
R
R/2 with its centre on the axis of the cylinder, as shown in R/2
the figure. The magnitude of the electric field at the point P, P
y
which is at a distance 2R from the axis of the cylinder, is 2R
23R x
given by the expression 16k  . The value of k is [IIT]
o

-1 . 5 4 -
Electrostatics & Capacitors

44. In the given circuit, a charge of +80 C is given to the upper

plate of the 4 F capacitor. Then in the steady state, the
charge on the upper plate of the 3 F capacitor is
(a) + 32 C (b) + 40 C
(c) + 48 C (d) + 80 C [IIT]

45. Six point charges are kept at the vertices of a regular hexagon
of side L and centre O, as shown in the figure. Given that L
F E
+q
1 q
K P
4o L2 , which of the following statement (s) is (are)
correct?
A S T D
(a) The electric field at O is 6K along OD +2q O –2q
(b) The potential at O is zero
(c) The potential at all points on the line PR is same R

B
(d) The potential at all points on the line ST is same +q –q C [IIT]

46. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2
respectively, touch each other. The net electric field at a distance 2R from the centre of the
1
smaller sphere, along the line joining the centres of the spheres, is zero. The ratio  can
2

be [JEE-Adv.]
32 32
(a) – 4 (b)  (c) (d) 4
25 25
47. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The
switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then
pressed to charge the capacitor C2. After some time, S2 is released and then S3 is pressed. After some
time. [JEE-Adv.]
S1 S2 S3
(a) the charge on the upper plate of C1 is 2CV0
(b) the charge on the upper plate of C1 is CV0
(c) the charge on the upper plate of C2 is 0 2V0 C1 V0
C2
(d) the charge on the upper plate of C2 is –CV0

48. Two non-conducting spheres of radii R1 and R2 and

carrying uniform volume charge densities + and –,
respectively, are placed which that they partially
overlap, as shown in the figure. At all points in the R1 R2
overlapping region [JEE-Adv.]
(a) the electrostatic field is zero (b) the electrostatics potential is constant
(c) the electrostatic field is constant in magnitude (d) the electrostatic field has same direction

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Electrostatics & Capacitors

49. A parallel plate capacitor has a dielectric slab of dielectric constant K between
its plates that covers 1/3 of the area of its plates, as shown in the figure.
Q1 E1
The total capacitance of the capacitor is C while that of the portion with
dielectric in between is C1. When the capacitor is charged, the plate area
covered by the dielectric gets charge Q1 and the rest of the area gets Q2 E2
charge Q2. The electric field in the dielectric is E1 and that in the other
portion is E2. Choose the correct option/options, ignoring edge effects.
[JEE-Adv.]

E1 E1 1 Q1 3 C 2K
(a) 1 (b)  (c)  (d) 
E2 E2 K Q2 K C1 K

50. Let E1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charges Q, an
infinitely long wire with constant linear charge density , and an infinite plane with uniform surface
charge density . If E1(r0) = E2(r0) = E3(r0) at a given distance r0, then [JEE-Adv.]


(a) Q  4r02 (b) r0 
2
(c) E1 (r0 / 2)  2 E2 (r0 / 2) (d) E2 (r0 / 2)  4 E3 ( r0 / 2)

51. Charge Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2,
R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R
from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then [JEE-Adv.]
P P P
R R
Q 2Q 4Q
R/2
2R
Sphere 1 Sphere 2 Sphere 3
(a) E1 > E2 > E3 (b) E3 > E1 > E2 (c) E2 > E1 > E3 (d) E3 > E2 > E1

52. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x-axis at x = – 2a, –a, +a and
+2a, respectively. A positive charge q is placed on the positive y-axis at a distance b > 0. Four options
of the signs of these charges are given in List I. The direction of the forces on the charge q is given in
List II. Match List I with List II and select the correct answer using the code given below the lists.
q(0, b)
List I List II
P. Q1, Q2, Q3 and Q4 all positive 1. + x
Q. Q1, Q2 positive, Q3, Q4 negative 2. – x
R. Q1, Q4 positive, Q2, Q3 negative 3. + y
Q1 Q2 Q3 Q4
S. Q1, Q3 positive, Q2, Q4 negative 4. – y (–2 a, 0) (–a, 0) (+a, 0) (+2a, 0)

(a) P-3; Q-1; R-4; S-2 (b) P-4; Q-2; R-3; S-1 [JEE-Adv.]
(c) P-3; Q-1; R-2; S-4 (d) P-4; Q-2; R-1; S-3

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Electrostatics & Capacitors

z
53. An infinitely long uniform line charge distribution of
charge per unit length  lies parallel to the y-axis in
3
the y-z plane at z  a (see figure). If the D L
2 C
magnitude of the flux of the electric field through the
rectangular surface ABCD lying in the x-y plane with a
y
L O
its centre at the origin is n (0 = permittivity of
o

free space), then the value of n is A B

z [JEE-Adv.]
54. The figures below depict two situations in which two infinitely long static line charges of constant
positive line charge density  are kept parallel to each other. In their resulting electric field, point
charges q and –q are kept in equilibrium between them. The point charges are confined to move in the
x direction only. If they are given a small displacement about their equilibrium positions, then the correct
statement(s) is(are) [JEE-Adv.]

x x
+q –q

(a) Both charges execute simple harmonic motion.

(b) Both charges will continue moving in the direction of their displacement
(c) Charge +q executes simple harmonic motion while charge –q continues moving in the direction of
its displacement.
(d) Charge –q executes simple harmonic motion while charge +q continues moving in the direction of
its displacement.
R2
55. Consider a uniform spherical charge distribution of radius P
R1 centred at the origin O. In this distribution, a spherical a
R1
cavity of radius R2, centred at P with distance OP = a = R1
O
– R2 (see figure) is made. If the electric field inside the
 
cavity at position r is E ( r ) , then the correct statement (s)
is (are)
[JEE-Adv.]

(a) E is uniform, its magnitude is independent of R2 but its direction depends on r

(b) E is uniform, its magnitude depends on R2 and its direction depends on r

(c) E is uniform, its magnitude is independent of a but its direction depends on a

(d) E is uniform and both its magnitude and direction depend on a

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Electrostatics & Capacitors

PARAGRAPH for Question 56 and 57

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and
an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight
and soft material and coated with a conducting material are placed on the bottom plate. The balls have
a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that
the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface, the balls will get
charged, will becomes equipotential with the plate and are repelled by it. The balls will eventually
collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft
nature of the material of the balls. The electric field in the chamber can be considered to be that of a
parallel plate capacitor. Assume that there are no collisions between that balls and the interaction
between them is negligible. (Ignore gravity). [JEE-Adv.]

56. Which one of the following statements is correct?

(A) The balls will execute simple harmonic motion between the two plates
(B) The balls will stick to the top plate and remains there
(C) The balls will bounce back to the bottom plate carrying the same charge they went up with
(D) The balls will bounce back to the bottom plate carrying the opposite charge they went up with
57. The average current in the steady state registered by the ammeter in the circuit will be
(A) proportional to V02 (B) zero
(C) proportional to the potential V0 (D) proportional to V01/2
+Q
58. A point charge +Q is placed just outside an imaginary
hemispherical surface of radius R as shown in the
R
figure. Which of the following statements is/are
correct?

[JEE-Adv.]

Q  1 
(a) The electric flux passing through the curved surface of the hemisphere is  1  
2o  2
(b) The component of the electric field normal to the flat surface is constant over the
surface
Q
(c) Total flux through the curved and the flat surfaces is 
o
(d) The circumference of the flat surface is an equipotential

-1 . 5 8 -
Electrostatics & Capacitors

59. Three identical capacitors C1, C2 and C3 have a capacitance of 1.0 F each and they are
uncharged initially. They are connected in a circuit as shown in the figure and C1 is then
filled completely with a dielectric material of relative permittivity r . The cell electromotive
force (emf) V0 = 8V. First the switch S1 is closed while the switch S2 is kept open. When the
capacitor C3 is fully charged, S1 is opened and S2 is closed simultaneously. When all the
capacitors reach equilibrium, the charge on C3 is found to be 5 C. The value of r  .
[JEE-Adv.]

60. An infinite long thin non-conducting wire is parallel to the

z-axis and carries a uniform line charge density . It pierces
a thin non-conducting spherical shell of radius R in such a
way that the arc PQ subtends an angle 120° at the centre
O of the spherical shell, as shown in the figure. The
permittivity of free space is 0. Which of the following
statements is (are) true?
[JEE-Adv]
(a) The electric flux through the shell is 3R / 0
(b) The z-component of the electric field is zero at all the points on the surface of the shell
(c) The electric flux through the shell is 2 R / 0
(d) The electric field is normal to the surface of the shell at all points

61. A particle, of mass 10–3 kg and charge 1.0 C, is initially at rest. At time t = 0, the particle

comes under the influence of an electric field E (t )  E0 sin t iˆ , where E0 = 1.0 NC–1 and
  10 3 rad s–1. Consider the effect of only the electrical force on the particle. Then the
maximum speed, in ms–1, attained by the particle at subsequent times is ______
[JEE-Adv.]

62. The electron field E is measured at a poitn P(0, 0, d) generated due to various charge distributions and
the dependence of E on d is found to be different for different charge distributions. List-I contains
different relations between E and d. List-II describes different electric charge distributions, along with
their locations. Match the fucntions in List-I with the related charge distributions in List-II.

List-I List-II
P. E is independent of d 1. A point charge Q at the origin.
1
Q. E 2. A small dipole with point charges Q at
d
(0, 0, l) and –Q at (0, 0, –l). Take 2l << d.

-1 . 5 9 -
Electrostatics & Capacitors

1
R. E 3. An infinite line charge coincident with the
d2
x-axis, with uniform linear charge density 
1
S. E 4. Two infinite wires carrying uniform linear
d3
charge density parallel to the x-axis. This one
along (y = 0, z = l) has a charge density +
and the one along (y = 0, z = – l) has a charge
densty –. Take 2l << d.
5. Infinite plane charge coincident with the xy-
plane with uniform surface charge density
(a) (P)-(5); (Q)-(3,4); (R)-(1); (S)-(2) (b) (P)-(5); (Q)-(3); (R)-(1,4); (S)-(2)
(c) (P)-(5); (Q)-(3); (R)-(1,2); (S)-(4) (d) (P)-(4); (Q)-(2,3); (R)-(1): (S)-(5)

DCE QUESTIONS

1. Electric flux linked with a hemisphere placed in a uniform electric field parallel to its axis, [DCE]
(a) r 2 E (b) 3/2 r2E (c) 2 r2E (d) zero

2. Where will be the electric field be zero [DCE]

(a) towards left of (–8Q)
x
(b) towards right of (+2Q)
–8Q +2Q
(c) in between but nearer to (–8Q)
(d) in between but nearer to (+2Q)

3. Equivalent capacitance between A and B is 4 F 4 F [DCE]

(a) 8 F
(b) 6 F 4 F
(c) 268 F
A B
(d) 10/38 F 4 F 4 F
4. According to Gauss Theorem, electric field of an infinitely long straight wire is proportional to
1 1 1
(a) r (b) (c) (d) [DCE]
r2 r3 r

5. Equivalent capacitance is [DCE]

(a) 15 F 15 F
(b) 20 F 15 F 15 F
15 F
(c) 25 F
(d) 30 F

-1 . 6 0 -
Electrostatics & Capacitors

6. What is the potential at the centre c? [DCE]

(a) 0 +q a –q
(b) Kq/a 2
a c a
(c) 2 Kq/a
(d) none of these –q a +q

7. A capacitor of capacitance 1 µF is filled with two dielectrics

of dielectric constant 4 and 6. What is new capacitance
(a) 10 µF (b) 5 µF

(c) 4 µF (d) 7 µF [DCE]

8. What is angle between electric field and equipotential surface [DCE]
(a) 90° always (b) 0° always (c) 0° to 90° (d) 0° to 180°
9. Eight identically charged drops are joined to form bigger drop. By what factor the charge and potential
changes [DCE]
(a) 8, 4 (b) 8, 8 (c) 6, 8 (d) 8, 10
10. Three charges 1C, 2C, 3C are kept at vertices of an equilateral triangle of side 1 m. If they are
brought nearer so that they now form equilateral triangle of side 0.5 m, then work done is
[DCE]
(a) 11 J (b) 1.1 J (c) 0.011 J (d) 0.11 J
11. Two capacitors of capacitance C are connected in series. If one of them is filled with dielectric
substance k, what is the effective capacitance ? [DCE]

kC 2kC
(a) 1  k  (b) C(k + 1) (c) (d) none of these
1 k
12. A charge Q is divided in two parts q, and Q – q. What is value of q for maximum force between
them ? [DCE]

3Q Q Q
(a) (b) (c) Q (d)
4 3 2
AIEEE/JEE-MAINS QUESTIONS
1. A charged particle ‘q’ is shot towards another charged particle ‘Q’, which is fixed, with a speed ‘v’.
It approaches ‘Q’ upto a closest distance r and then returns. If q were given a speed of ‘2v;, the
closest distances of approach would be [AIEEE]

(a) r/2 (b) 2 r (c) r (d) r/4

2. A charged oil drop is suspended in a uniform field of 3 × 104 v/m so that it neither falls nor rises. The
charge on the drop will be (Take the mass of the charge = 9.9 × 10–15 kg and g = 10 m/s2)
[AIEEE]
(a) 1.6 × 10–18 C (b) 3.2 × 10–18 C (c) 3.3 × 10–18 C (d) 4.8 × 10–18 C

-1 . 6 1 -
Electrostatics & Capacitors

3. A charged ball B hangs from a silk thread S, which makes an angle  with
+
a large charged conducting sheet P, as shown in the figure. The surface P+
charge density  of the sheet is proportional to [AIEEE] +
+ 
S
(a) sin  (b) tan  +
+ B
(c) cos  (d) cot  +
4. Two point charges +8q and –2q are located at x = 0 and x = L respectively. The location of a point
on the x-axis at which the net electric field due to these two point charges is zero is [AIEEE]
(a) 8 L (b) 4 L (c) 2 L (d) L/4
5. Two thin rings each having a radius R are placed at a distance d apart with their axes coinciding.
The charges on the two rings are +q and –q. The potential difference between the centres of the
two rings is [AIEEE]
 
Q 1 1 
(a) zero (b) 
40  R R2  d 2 
 
 
Q Q 1 1 
(c) (d) 
4 0 d 2 2  0  R R  d2
2 
 
6. A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the
capacitance between any two adjacent plates is C then the resultant capacitance is [AIEEE]
(a) C (b) nC (c) (n – 1)C (d) (n + 1)C
7. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance
wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature
of the block is raised by T, the potential difference V across the capacitance is [AIEEE]

ms T 2ms T 2m C  T mCT

(a) (b) (c) (d)
C C s s

8. An electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will experience
(a) a translational force only in the direction of the field
(b) a translational force only in a direction normal to the direction of the field
(c) a torque as well as a translational force
(d) a torque only [AIEEE]
9. Two insulating plates are both uniformly charged in such a way that the potential difference between
them is V2 – V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m
and can be treated as infinitely large. An electron is released from rest on the inner surface of
plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10–19 C, me = 9.11 × 10–31 kg) [AIEEE]
(a) 2.65 × 106 m/s
Y
(b) 7.02 × 1012 m/s 0.1 m
(c) 1.87 × 106 m/s X

(d) 32 × 10–19 m/s

1 2

-1 . 6 2 -
Electrostatics & Capacitors

10. Two spherical conductors A and B of a radii 1 mm and 2 mm are separated by a distance of 5 cm an
are uniformly charged. If the spheres are connected by a conducting wire then in
equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A
and B is [AIEEE]
(a) 4 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4

11. An electric charge 10–3 C is placed at the origin (0, 0) of X – Y co-ordinate system. Two points A
and B are situated at  2, 2  and (2, 0) respectively. The potential difference between the points
A and B will be [AIEEE]
(a) 2 volt (b) 4.5 volt (c) 9 volt (d) zero

12. A battery is used to charge a parallel plate capacitor till the potential difference between the plates
becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor
and the work done by the battery will be [AIEEE]
(a) 1/4 (b) 1/2 (c) 1 (d) 2

13. Charges are placed on the vertices of a square as shown. Let E be the electric field and V the
potential at the centre. If the charges on A and B are interchanged with those on D and C respectively,
then

(a) E and V remain unchanged

(b) E changes, V remains unchanged

(c) E remains unchanged, V changes

(d) Both E and V change [AIEEE]
14. The potential at a point x (measured in m) due to some charges situated on the x-axis is given by
V(x) = 20/(x2 – 4) Volts
The electric field E at x = 4 m is given by [AIEEE]
(a) 10/9 Volt/m and in the –ve x direction (b) 10/9 Volt/m and in the +ve x direction
(c) 5/3 Volt/m and in the –ve x direction (d) 5/3 Volt/m and in the +ve x direction
15. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the
following graphs most closely represents the electric field E(r) produced by the shell in the range
0  r < , where r is the distance from the centre of the shell ? [AIEEE]

(a) (b) (c) (d)

16. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation
between its plates is d. The space between the plates is now filled with two dielectrics. One of the
dielectrics has dielectric constant k1 = 3 and thickness d/3 while the other one has dielectric constant
2d
k2 = 6 and thickness . Capacitance of the capacitor is now [AIEEE]
3
(a) 45 pF (b) 40.5 pF (c) 20.25 pF (d) 1.8 pF
17. Two points P and Q are maintained at the potentials of 10V and –4V, respectively. The work done in
moving 100 electrons from P to Q is [AIEEE]
(a) 9.60 ×10–17 J (b) –2.24 ×10–16 J (c) 2.24 × 10 –16
J (d) –9.60 ×10 –17
J

-1 . 6 3 -
Electrostatics & Capacitors

18. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of
the two corners. If the net electrical force on Q is zero, then Q/q equals. [AIEEE]
1
(a) –1 (b) 1 (c)  (d) – 22
2
19. Statement-1: For a charged particle moving from point P to point Q, the net work done by an
electrostatic field on the particle is independent of the path connecting point P to point
Q. [AIEEE]
Statement-2: The net work done by a conservative force on an object moving along a closed loop
is zero.
(a) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1
(b) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1
(c) Statement-1 is false, Statement-2 is true
(d) Statement-1 is true, Statement-2 is false
Q
20. Let P( r )  r be the charge density distribution for a solid sphere of radius R and total charge
R 4
Q. For a point p inside the sphere at distance r1 from the centre of the sphere, the magnitude of
electric field is [AIEEE]

Q Qr12 Qr12
(a) (b) (c) (d) 0
4 o r12 4 o R 4 3 o R 4
21. Let there be a spherically symmetric charge distribution with charge density varying as
5 r 
(r )  o    upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The
4 R
electric field at a distance r (r < R) from the origin is given by [AIEEE]

4o r  5 r  o r  5 r  4o r  5 r  o r  5 r 
(a)  (b)  (c)  (d) 
3 o  4 R  3 o  4 R  3 o  3 R  4o  3 R 

22. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time
taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken
for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2 will be [AIEEE]
(a) 1/4 (b) 2 (c) 1 (d) 1/2

23. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net
 j
field E at the centre O is [AIEEE]
q ˆj q ˆj
(a)  (b)
2 2 0 r 2 2 2 0 r 2
q ˆj q
(c) (d)  2 2 ˆj
4 0 r 2
2
4  0 r i
O
24. The electrostatic potential inside a charged spherical ball is given by  = a r2 + b where r is the
distance from the centre, a, b are constants. Then the charge density inside the ball is [AIEEE]
(a) 24  a o r (b) 6 a  o r (c) 24 a  o (d) 6 a  o

-1 . 6 4 -
Electrostatics & Capacitors

25. Two identical charged spheres suspended from a common point by two massless strings of length l
are initially a distance d(d << l) apart because of their mutual repulsion. The charge begins to leak
from both the spheres at a constant rate. As a result the charges approach each other with a velocity
v. Then as a function of distance x between them, [AIEEE]
(a) v  x 1/ 2 (b) v  x 1 (c) v  x1/ 2 (d) vx

26. In a uniformly, charged sphere of total charge Q and radius R, the electric field E is plotted as a
function of distance from the centre. The graph which would correspond to the above will be:
[AIEEE]
E E E E

(a) (b) (c) (d)

R r R r R r R r

27. This question has statement 1 and statement 2. Of the four choices given after the statements, choose
the one that best describes the two statements.
An insulating solid sphere of radius R has a uniformly positive charge density . As a result of this
uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the
surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero.
Statement-1: When a charge q is taken from the centre to the surface of the sphere, its potential
q
energy changes by 3  [AIEEE]
0

r
Statement-2: The electric field at a distance r(r < R) from the centre of the sphere is 3 
0

(a) Statement 1 and is true Statement 2 is not the correct explanation for Statement 1
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1

28. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric
potential at the point O lying at a distance L from the end A is [JEE-Mains]

A B
O
L L
Q Q ln 2 Q 3Q
(a) 4o L ln 2 (b) 4o L (c) 8o L (d) 4o L

29. Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass m and
q
charge qo  is placed at the origin. If charge q0 is given a small displacement (y << a) along the y-
2
axis, the net force acting on the particle is proportional to [JEE-Mains]
1 1
(a) y (b)  y (c) y (d) – y

-1 . 6 5 -
Electrostatics & Capacitors

30. Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting
them together the potential on each one can be made zero. Then [JEE-Mains]
(a) 3C1 + 5C2 = 0 (b) 9C1 = 4C2 (c) 5C1 = 3C2 (d) 3C1 = 5C2

31. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a
dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is
3 × 104 V/m, the charge density of the positive plate will be close to [JEE-Mains]
4 2 4 2 –7 2
(a) 3 × 10 C/m (b) 6 × 10 C/m (c) 6 × 10 C/m (d) 3 × 10 C/m2
–7


32. Assume that an electric field E  30 x 2iˆ exists in space. Then the potential difference VA – VO, where
VO is the potential at the origin and VA the potential at x = 2 m is [JEE-Mains]
(a) –80 J (b) 80 J (c) 120 J (d) –120 J
33. A long cylindrical shell carries positive surface charge  in the upper half and negative surface charge
–  in the lower half. The electric field lines around the cylinder will look like figure given in: (figures
are schematic and not drawn to scale). [JEE-Mains]

+
++ ++ + + ++
+ + ++ +

(a) (b)

++ ++
++ + +
++ ++
+ +
(c) (d)

34. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ) on its
3V0 5V0 3V0 V0
surface. For this sphere the equipotential surfaces with potentials , , and have radius
2 4 4 4
R1, R2, R3 and R4 respectively. Then [JEE-Mains]

(a) R1 = 0 and R2 > (R4 – R3) (b) R1  0 and (R2 – R1) > (R4 – R3)
(c) R1 = 0 and R2 < (R4 – R3) (d) 2R < R4

35. In the given circuit, charge Q2 on the 2 F capacitor changes

C
as C is varied from 1 F to 3 F. Q2 as a function of ‘C’ is
given properly by: (figures are drawn schematically and are
not to scale)

E
[JEE-Mains]

-1 . 6 6 -
Electrostatics & Capacitors

charge charge
Q2 Q2

(a) (b)
C C

charge charge
Q2 Q2

(c) (d)
C C

36. A combination of capacitors is set up as shown in the figure.

The magnitude of the electric field, due to a point charge Q
(having a charge equal to the sum of the charges on the
4 F and 9 F capacitors), at a point distant 30 m from it,
would equal. [JEE-Mains]
(a) 480 N/C (b) 240 N/C + –
8V
(c) 360 N/C (d) 420 N/C

37. The region between two concentric spheres of radii a and b respectively
A
(see figure), has volume charge density   where A is a constant and
r a
r is the distance from the centre. At the centre of the sphere is Q a point
charge Q. The value of A such that the electric field in the region between b
the spheres will be constant, is : [JEE-Mains]

2Q Q Q 2Q
(a) (b) 2 a 2 (c) 2  (b 2  a 2 ) (d)  (a 2  b 2 )
a 2

38. An electric dipole has a fixed dipole moment P , which makes angle  with respect to x-axis. When
 
subjected to an electric field E1  Eiˆ, it experiences a torque T1  kˆ. When subjected to another
  
electric field E2  3E1 ˆj it experiences a torque T2  T1 . The angle  is [JEE-Mains]

(a) 60º (b) 90º (c) 30º (d) 45º

39. A capacitance of 2F is required in an electrical circuit across a potential difference of 1.0 kV. A
large number of 1 F capacitors are available which can withstand a potential difference of not more
than 300 V. The minimum number of capacitors required to achieve this is [JEE-Mains]
(a) 24 (b) 32 (c) 2 (d) 16

40. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have
surface charge densities ,   and  respectively. The potential of shell B is
[JEE-Mains]

  b2  c 2    a 2  b2    a 2  b2    b2  c 2 
(a)   a (b)   c (c)   c (d)   a
o  c  o  a  o  b  o  b 

-1 . 6 7 -
Electrostatics & Capacitors

41. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a

5
deelectric material of dielectric constant K  is inserted between the plates, the magnitude
3
of the induced charge will be [JEE-Mains ]

(a) 0.9 n C (b) 1.2 n C (c) 0.3 n C (d) 2.4 n C

-1 . 6 8 -
 CHAPTER TEST
SECTION-I: STRAIGHT OBJECTIVE TYPE
This section contains 5 multiple choice questions numbered 1 to 5. Each question has 4
choices (A), (B), (C) and (D), out of which ONLY-ONE is correct.

1. An uncharged parallel plate condenser, having a dielectric constant K, is connected to a similar parallel
plate condenser with air as medium in between plates and charged to a potential V. The two condensers
are connected so that their common potential is V. The dielectric constant K is
V  V V  V V  V V V 
(a) (b) (c) (d)
V  V V V V

2. A system of three parallel plates, each of area A, are separated

by distances d1 and d2. The space between these plates is 1 2 3
filled with dielectrics of relative permittivity 1 and 2 as shown
in the figure. The permittivity of free space is 0. The 1 2
equivalent capacitance of the system is
012 A 120 A
(a)  d   d (b)  d   d
2 1 1 2 1 1 2 2 d1 d2

0 A 0 A
(c)  d   d (d)  d   d
1 1 2 2 1 2 2 1

3. A parallel plate capacitor of plate area A and separation between the plates d is charged to a potential
difference V by a battery. The battery is then disconnected and a slab of dielectric constant K is then
inserted between the plates of the capacitor so as to fill the space between the plates completely. The
work done to insert the slab is
0 AV 2  1   AV 2 ( K  1) 0 AV 2 0 AV 2 K
(a)   1 (b) 0 (c)
d ( K  1)
(d)
2d  K  2d d
4. Two conducting balls of same mass M and radius r are suspended from insulating threads of length L
each and are given charge Q each as shown in the figure. If the angular displacement  from the vertical
is small, the distance d between the balls is
1/ 3 3/ 2
 Q2 L   Q2L   
(a)  
 (b)  

 4 0 M g   4 0 M g  L L
1/ 3 1/ 2
 Q2 L   Q2L 
(c)  
 (d)  
 M
Q Q
M
 2 0 M g   2 0 M g  d

5. A parallel plate capacitor with air as medium has a capacitance

of 14 F. The capacitor is filled with three media of dielectric d
K2 2
constant K1 = 2, K2 = 3 and K3 = 4. The capacitance of the d K1
system will be d
K3
(a) 19 F (b) 21 F 2

(c) 25 F (d) 38 F

-1 . 6 9 -
Electrostatics & Capacitors

SECTION-II: MULTIPLE CORRECT ANSWERS TYPE

This section contains 5 multiple choice questions numbered 6 to 10. Each question has 4
choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

6. A thin walled spherical conducting shell S of radius R is given charge Q. The same amount of charge is
also placed at its centre C. Which of the following are correct –
Q
(a) On the outer surface of S charge density =
2R 2
(b) The electric field is zero at all points inside S
(c) At a point just outside S, the electric field is double, the field at a point just inside S.
(d) At any point inside S, the electric field is inversely proportional to the square of distance from C.

7. Electric field, due to an infinite line of charge, as shown in figure at a point P at a distance r from the line
is E. If one half of the line of charge is removed from either side of point A, then :
(a) Electric field at P will have magnitude E/2

(b) Electric field at P in x direction will be E/2.

(c) Electric field at P in y direction will be E/2.

(d) none of these

8. An electric dipole moment p  (2.0iˆ  3.0ˆj) C . m is placed in a uniform electric field

E  (3.0iˆ  2.0k)
ˆ  105 NC 1

 
(a) The torque that E exerts on p is (0.6iˆ  0.4ˆj  0.9k)Nm
ˆ

(b) The potential energy of the dipole is –0.6 J

(c) The potential energy of the dipole is 0.6 J
(d) If the dipole is rotated in the electric field, the maximum potential energy of the dipole is 1.3 J

9. An electric field converges at the origin whose magnitude is given by the expression E = 100 r N/C.,
where r is the distance measured from the origin.
(a) Total charge contained in any spherical volume with its centre at origin is negative.
(b) Total charge contained at any spherical volume, irrespective of the location of its centre, is negative.
(c) Total charge contained in a spherical volume of radius 3cm with its centre at the origin has magnitude
3×10–13 C.
(d) Total charge contained in a spherical volume of radius 3 cm with its centre at the origin has magnitude
3 × 10–9 C.

-1 . 7 0 -
Electrostatics & Capacitors

10. A uniform electric field of strength Ejˆ exists in a region. An electron (charge –e, mass m) enters a point
A with velocity v ˆj . It moves through the electric field and exits at point B. Then –

2amv 2 ˆ y V
(a) E   j
ed 2
B(2a, d)
4ma 2 v3 V
(b) Rate of work done by the electric field at B is
d3
(c) Rate of work by the electric field at A is zero x
(0, 0) A(a, 0)
2av ˆ ˆ
(d) Velocity at B is i  vj
d

SECTION- III: LINKED COMPREHENSION TYPE

This section contains 1 Paragraph. Based upon the paragraph, 3 multiple choice questions
have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY
ONE is correct.

Comprehension
Each plate of a parallel plate capacitor is a square of side a. The plates are kept at a separation d
(d << a), air being the dielectric medium between the plates. The capacitor is connected to a battery.
A block of length a, width a and
Square of side a
thickness y  d is inserted a distance x x

into the capacitor as shown. Assuming

dielectric constant of material is K. ydy
d
Answer the following questions.

11. Electric energy stored in the capacitor,

as a function of x, can be expressed as

0a v2 0a v2
(a) (a  x(K  1)) (b) (a  x(K  1))
2d 2d

0a v2 0a v2
(c) ( x  a(K  1)) (d) ( x  a(K  1))
2d 2d

12. Magnitude of force that acts on the block is

 0 a v 2 (K  1) 0 a v 2  0 a v 2 (K  1) 0a v2
(a) (b) (c) (d)
2d 2d (K  1) 2d 2d (K  1)

13. Which of the following is correct?

(a) Energy density is maximum when x = 0 (b) Energy density is maximum when x = d
(c) Energy density is minimum when x = 0 (d) Energy density is independent of x

-1 . 7 1 -
Electrostatics & Capacitors

SECTION- IV: MATRIX MATCH TYPE

This Section contains 4 multiple choice questions. Each question has matching lists. The codes for
lists have choice (A), (B), (C) and (D) out of which ONLY ONE may be correct.

14. Column II describe graph for charge distribution given in column I. Match the description.
Column I Column II
E (Electric
Field Intensities)

A. Uniformly charged ring (p) r

V (Electric
Potential)

B. Infinitely large charge conducting sheet (q) r

C. Infinite non conducting thin sheet. (r)

r
V

D. Hollow non conducting sphere. (s)

r

SECTION- V: INTEGER TYPE QUESTIONS

This section contains 8 questions. The answer to each of the question is a single digit integer,
ranging from 0 to 9. The bubble corresponding to the correct anInteger Type Questions

15. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting
hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of
the outer of hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference
between the same two surface is nV. Find the value of n.
16. In the given figure, find the value of equivalent capacitance between A and B in µF.

17. A parallel plate capacitor has an electric field of 105 V/m between the plates . If the charge on the
capacitor plate is 1 mC , then the force on each capacitor plate is n/100 Newton find the value of n.

-1 . 7 2 -
Electrostatics & Capacitors

18. Amount of heat that generated in the circuit shown in the figure after the switch Sw is shifted from
position 1 to position 2 is 1/n (CE22) find the value of n ?

19. 12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. The
potential difference VB – VA is 300n volts. Find the value of n

20. The electric field in a region is given by E  (A / x)3 iˆ . An expression for the potential in the region is
A
assuming the potential at infinity to be zero. Find the value of n
nx 2

21. The capacitance of all the capacitors is shown in the given figure. If the charges on the 5µF capacitors
is 120 µC, then the potential difference between A and C is 8n volts. Find the value of n?

22. A spherical shell of radius R carries uniformly distributed

charge of density  on its surface. Find potential difference
between point A on the edge and centre B; (VA – VB).

-1 . 7 3 -
 ANSWERS
Chapter Assignment Answer Keys

1. (d) 2. (c) 3. (c) 4. (c) 5. (b)

6. (b) 7. (d) 8. (b) 9. (d) 10. (b)
11. (a) 12. (a) 13. (a) 14. (d) 15. (d)
16. (c,d) 17. (c,d) 18. (a,c) 19. (a,d) 20. (a,c)
21. (b,c,d) 22. (b,d) 23. (a,b,d) 24. (a,d) 25. (a,d)
26. (a) 27. (c) 28. (a) 29. (a) 30. (c)
31. (d) 32. A-(p),(q); B-(q),(s); C-(p),(r); D-(p),(r)
33. A-(r); B-(p),(q); C-(p); D-(p),(s),(t) 34. (2) 35. (4)
36. (1) 37. (8) 38. (2) 39. (5) 40. (2)
41. (5) 42. (7) 43. (2)

Chapter Assignment Clues

1. [d] Work done = charge to be moved × potential difference = 0

1   4r 2 1   4R 2
2. [c] V    
40 r 40 R

1 q
3. [c] V  4   4
0 l / 2

1 1 1 1
4. [c] C  20  12  8
e

120
Ce  F
31

120
q  310   1200C
31

1200
P.D. across 4F capacitors =  150V
8

-1 . 7 4 -
Electrostatics & Capacitors

1 1 1 1
5. [b] C  12  24  8 , Ce  4F
e

Total charge = 4 × 60 = 240C

Charge on 5F capacitor = 50C

k  q  q kq  Q k ( q  Q )
6. [b]   0
a a 9 2

2q
Q
2 1

7. [d] Conceptual

3
8. [b] P.D. across 5F =  6  1.8 V
10
Charge = 1.8 × 5 = 9C

kq kq
9. [d] R  R  0
1 2

R1
q   q
R2

0 A
10. [b] C  
d /2 d /2

K1 K2

11. [a] Potential at each point inside shell will change by same value.
12. [a] Find equivalent capacitance in each case.
13. [a] Use formula.

kq  1 1 1 
14. [d] v  1     ... 
x0  2 3 4 

15. [d] Let charges on outer and inner shell are q1 and q2

q1 q2
2

40 R 4 0 (2 R) 2

Also q1 + q2 = Q

 kq kq 
V  1  2 
 R 2R 

-1 . 7 5 -
Electrostatics & Capacitors

16. [c,d] Conceptual

17. [c,d] Conceptual

18. [a,c] Conceptual

19. [a,c] Conceptual

20. [a,c] Conceptual

21. [b,c,d] + -- +
+
centre

22. [b,d] Due to induction - Attractive forces between the pairs AB and BC are equal to F1 and F2.
Also F1 = mg + F2.
23. [a,b,d] Before earthing

q q 
VA  2V  K  A  B  ...(1) q
A
q
B
 R 2R 
A
B
3  q  qB  R
VB  V  K  A  ...(2)
2  2R  2R

1 4 2q A  qB 1
   q A : qB 
 2  3 q A  qB 2

 qA  qB 
After earthing VB  0  K    qA   qB
 2R 
24. [a,d] Ceq = 2 + 2 = 4F
Qeq = Ceq.V = 240C
 Charge flowing through ef = 120C

 V  120  40V .
3
25. [a,d] When switch S is closed then the corresponding capacitor will be short circuited.
Initially when the switch is open.
2Cε ε
qi = ,VA = VB =
3 3
after the switch is closed, capacitor C is short circuited and qf = 2C

4C
Therefore more charge will flow from battery extra charge flown q  q f  qi 
3
So (a and d) are correct.

26. [a] Conceptual

kq
27. [c] Use  E  Breaking strength of medium.
R2

-1 . 7 6 -
Electrostatics & Capacitors

 kQ kq 
28. [a] W    0
 R r 
29. [a]
30. [c]
31. [d]
For Q. 29, 30 and 31
By Applying Kirchhoff’s voltage law and conservation of charge.

q1'  q3'  360 ...........(1)

q1'  q3'  360 ...........(2)

q1'  q3'  540 ...........(3)

32. A-(p),(q); B-(q),(s); C-(p),(r); D-(p),(r)

33. A-(r); B-(p),(q); C-(p); D-(p),(s),(t)
Electrostatic energy inside a capacitor is stored between the plates. When battery remains connected
with a capacitor then potential difference between the plates remains constant.
34. 2
Initially as the forces on each ball are tension T, weight mg and electric force F, for its equilibrium along
vertical,
Tcos = mg ..... (1)
and along horizontal
Tsinθ = F ..... (2)
Dividing Equation (2) by (1), we have

F
tanθ = ..... (3)
mg

When the balls are suspended in a liquid of density  and dielectric constant K, the electric force will
become (1/K) times, i.e., F  = (F/K) while weight

mg = mg  Th = mg  Vσg [as Th = Vg]

 σ  m
i.e., mg  = mg 1    as V = 
 ρ  ρ

-1 . 7 7 -
Electrostatics & Capacitors

F F
So for equilibrium of ball, tanθ = =
mg  Kmg 1  (σ / ρ)

According to given problem  = ; so from Equation (4) and (3), we have

ρ 1.6
K= = =2
(ρ  σ) (1.6  0.8)

35. 4
We know that the electric field due to uniformly charge ring of radius R at distance x from its centre on
its axis is given by

KQx
E
( r  x 2 )3/2
2

 Force on the negative charge q will be

KQqx
F 
( R  x 2 )3/ 2
2

In this case x << R

KQq  KQq
F= x; a= x
R3 mR 3
The motion is S.H.M. in nature we get

KQq KQq
ω2 = ω=
mR 3 mR3

mR 3
T = 2π
KQq

Put the given condition to get the answer

36. 1
To calculate the capacitance of capacitor let charge on inner surface a and outer surface c is +q and
–q

Kq K ( q)  Kq K ( q ) 
 Va  Vc    
a c  c c 
a

c  a  4 0 ac q c
 Va  Vc  Kq    C b
 ca Va  Vc
 ac 

put the give condition in the final answer then Capacitance turns to be 1.

-1 . 7 8 -
Electrostatics & Capacitors

37. 8
Let us consider a spherical surface of radius r(r > R) concentric with the ball and apply Gauss’s Law.
  q
 E.dS  0
Let Q = total charge of the ball
r R
2 2 O
0 E (4r )  Q   4x dx
R

r
r
a
0 E (4r )  Q  4  x 2 dx
2

R x

0E(4r2) = Q + 2a(r2 – R2)

 Q  2aR 2  1 2a
 E   2
 40  r 4 0

For E to be independent of r, Q = 2aR2 by putting the value of the R then it leads to charge is equal to
8 coulumb.
38. 2
Let the charge distribution in all the six face’s be as shown in figure. we have used the fact that two
opposite faces have equal and opposite charges on them. Net charge on the outer’s plate are zero

2z – x – y = 0
2z = x + y ...(i)
further A and C are at same potential Hence VB – VA = VB – VC
x y
 l1   l  l1 
A0 A0

 l  l1 
x y ...(ii)
 l1 
Charge on the middle plate will conserved. Hence
x+y=q ...(iii)
from equation (ii) and (iii)

-1 . 7 9 -
Electrostatics & Capacitors

l
y  q ; y  l1 q
l1 l

 l  l1 
and x   q
 l 
put the values of l, l1 and q value then we will be getting x = 2
39. 5
1 2

1F 2F
24V S

2F 1F

4 3

(a)

Charge on the capacitor C1 before switch is thrown to the right is C1V = Q. Let charge on the capacitor
after switch is thrown to the right are x, y and z as shown in figure. For isolated plate between C2 and
C3
y–x=0 ...(1) S
and between C1 and C2 x
-x C2
C1 and C2 ...(2) V z
C1 -z y
-y C3
Applying Kirchhoff’s second law in loop,

x y 2
+ = ...(3)
C2 C3 C1
1 2
from equation (1), (2) and (3),
1F 2F
24V S
C1C2C3V 2F 1F
x=y=
C1C2 + C2 C3 + C3C1 4 3

putting the values of C1 C2 and C3 and V then x = 5 (a)

40. 2

2
When the switch is open, capacitors (2) and (3) are in series. Their equivalent capacitance is F .
3
2
The charge appearing on each of these capacitors is, therefore, 24V  F  16C .
3

2
The equivalent capacitance of (1) and (4), which are also connected in series, is alsoF and the
3
charge on each of these capacitors is also 16C. The total charge on the two plates of (1) and (4)
connected to the switch is, therefore, zero.
The situation when the switch S is closed is shown in figure. Let the charges be distributed as shown in
the figure. Q1 and Q 2 are arbitarily chosen for the positive plates of (1) and (2). The same magnitude
of charges will appear at the negative plates (3) and (4).

-1 . 8 0 -
Electrostatics & Capacitors

-Q1 Q1 Q2 -Q2
1 2
1F 2F
24V S
0 V0
2F 1F
4 3
Q2 -Q2 -Q1 Q1

(b)
Take the potential at the negative terminal to the zero and at the switch to be V0.
Writing equations for the capacitors (1), (2), (3) and (4).
Q1 = (24V – V0) × 1F ...(i)
Q2 = (24V – V0) × 2F ...(ii)
Q1 = V0 × 1 F ...(iii)
Q2 = V0 × 2F ...(iv)
From (i) and (iii), V0 = 12 V.
Thus, from (iii) and (iv),
Q1 = 12C and Q2 = 24C
The charge on the two plates of (1) and (4) which are connected to the switch is, therefore Q2 – Q1 =
12C. When the switch was open, this charge was zero. Thus, 12C of charge has passed through the
switch after it was closed.then K = 2.
41. 5
When the switch is in the position 1, C and C0 are in parallel and C is in series with the combination

C (C  C0 )
Hence Ceq 
(2C  C0 )

C (C  C0 )
Charge on Ceq  V
2C  C0

VC (C  C0 )C0 VCC0
By the charge distribution principle q2  
(2C  C0 )(C  C0 ) (2C  C0 )

VC (C  C0 )C C 2V
q1  
(2C  C0 )(C  C0 ) (2C  C0 )
In position 2

C 2V VC (C  C0 )
q3  , q2  q2 , q1 
2C  C0 2C  C0
Heat produced = loss in stored energy + extra energy drawn from the battery.
Here loss in stored energy is zero because Ceq is the same in both the position of the key.
Energy drawn from the battery = V q  V (q1  q1 ) or V (q3  q3 )

 VC (C  C0 ) VC 2  V 2 CC0
 Heat produced = V   
 2C  C0 2C  C0  2C  C0

-1 . 8 1 -
Electrostatics & Capacitors

42. 7
V1 V

+ + – – – – – –
C1 C2 C1 C2
– – + + + + + +

V2

Initial charge on C1 = C1V1 = 100C; Initial charge on C2 = C2V2 = 800C

C1V1 < C2V2
when the terminals of opposite polarity are connected together, the magnitude of net charge finally is
equal to the difference of magnitude of charges before connection.
(charge on C2)i – (charge on C1)i = (charge on C2)f – (charge on C1)f
Let V be the final common potential difference across each.
The charges will be redistributed and the system attains a steady state when potential difference
across each capacitor becomes same.
C2V2 – C1V1 = C2V + C1V
C1V2  C1V1 800  100
V   140 volts
C2  C1 5
43. 2
Kq kq1
Potential at the inner sphere, V  
3r r
Now, v = 0
q
 q1 
3

Previous Year Questions

IIT-JEE/JEE-ADVANCE QUESTIONS

1. (d) 2. (c) 3. (a) 4. (d) 5. (d)

6. (b) 7. (b) 8. (b) 9. (c) 10. (a)
11. (a,d) 12. (a,b,d) 13. (a) 14. (b) 15. (c)
16. (a) 17. (c) 18. (b) 19. (a) 20. (b)
21. (c) 22. (a) 23. (b) 24. (a)
25. A-(q); B-(r),(s); C-(s); D-(p),(q),(r),(s) 26.A-(p),(q),(s),(t); B-(q); C-(s); D-(s)
27. (2) 28. (d) 29. (a,d) 30. (a) 31. (d)
32. 2 33. (d) 34. (d) 35. (a,b,c,d) 36. (a)
37. (c) 38. (a,c,d) 39. (3) 40. (c) 41. (d)

-1 . 8 2 -
Electrostatics & Capacitors

42. (a,c,d) 43. (7) 44. (c) 45. (a,b,c) 46. (b,d)
47. (b,d) 48. (c,d) 49. (d) 50. (c) 51. (c)
52. (a) 53. (6) 54. (c) 55. (d) 56. (d)
57. (a) 58. (a, d) 59. (1.50) 60. (a,b) 61. (2.00)
62. (b)

DCE QUESTIONS
1. (d) 2. (a) 3. (a) 4. (d) 5. (b)
6. (b) 7. (b) 8. (a) 9. (a) 10. (c)
11. (a) 12. (d)
AIEEE/JEE-MAINS QUESTIONS
1. (d) 2. (c) 3. (b) 4. (c) 5. (d)
6. (c) 7. (b) 8. (c) 9. (a) 10. (c)
11. (d) 12. (b) 13. (b) 14. (b) 15. (d)
16. (b) 17. (c) 18. (d) 19. (a) 20. (b)
21. (d) 22. (a) 23. (a) 24. (d) 25. (a)
26. (c) 27. (c) 28. (b) 29. (c) 30. (d)
31. (c) 32. (a) 33. (a) 34. (c,d) 35. (b)
36. (d) 37. (b) 38. (a) 39. (b) 40. (c)
41. (b)

Chapter Test Answer Keys

1. (d) 2. (a) 3. (a) 4. (c) 5. (d)

6. (a,c,d) 7. (a,b,c) 8. (a,b,d) 9. (a,b,c) 10. (a,b,c,d)

11. (b) 12. (c) 13. (d) 14. A-(p),(q); B-(r); C-(r); D-(s)

15. (1) 16. (4) 17. (5) 18. (2) 19. (4)

20. (2) 21. (8) 22. (0)

-1 . 8 3 -