CHAPTER
1
Electric Charges and Fields
ELECTROSTATICS
The branch of physics which deals with properties of charges at
rest is called electrostatics.
ELECTRIC CHARGE
Charge is scalar physical quantity associated with matter due to
which it produces and experiences electrical and magnetic effects.
The excess or deficiency of electrons in a body gives the concept
of net charge. A negatively charged body has excess of electrons
while a positively charged body has lost some of its electrons.
Properties of Electric Charge
1. Charges interact with each other i.e., they exert force on each
other. Like point charges repel each other while unlike point
charges attract each other.
2. Charge is of two kind: Positive and negative.
3. Total charge of an isolated system is conserved
(Conservation of charge).
4. Charge is quantised: Charge is an integral multiple of
electronic charge i.e., Q = Ne, where e = 1.6 × 10–19 C and N
is an integer.
5. Charge can be transferred: Charge can be transferred from
one body to an other. This occurs due to transfer of electrons
from one body to another. One of the common example of
transfer of charge is charging by friction.
e–
–
+
Rubbing
Transfer
–
+
A
B
–
+
–
+
Neutral
Neutral
Frictional Electricity: When two bodies are rubbed with
each other, they are found to attract each other. This is so
because, on rubbing, transfer of electrons takes place from
one body to an other. One of them acquires a positive charge
and other acquires a negative charge.
6. Charge is invariant: Charge of a particle is independent of
its speed.
7. Charge cannot exist without mass, while mass can exist
without charge. e.g. Neutron, Neutrino, Antineutrino. All are
Neutral particles having mass.
SI Unit : coulomb.
[1 coulomb = 1 ampere × 1 second]
C.G.S. : stat coulomb or franklin
1 coulomb = 3 × 109 stat coulomb
1
1 coulomb = 3 ×109 esu of charge =
emu of charge [esu
10
= electrostatic unit]
[emu = electro magnetic unit]
1 esu of charge = 1 franklin
Practical Units : 1amp × hr = 3600 coulomb and
1 faraday = 96500 coulomb (charge on 1 mole of electrons)
METHODS OF CHARGING
Charging by Conduction
1. Charging by conduction refers to the technique of charging an
uncharged material by bringing it into touch with some other
charged material.
2. A negatively as well as positively charged item seems to have
an uneven amount of charges.
3. As a consequence, whenever a charged item comes into
interaction with an uncharged conductor, electrons are
transferred from the charged object toward the conductor.
4. Whenever a negative object is being utilized just to charge a
neutral object, both things become negatively charged.
Therefore, charging through conduction involves direct
interaction with charged as well as uncharged bodies, and related
bodies accumulate a very similar type of charge.
There are three types of materials in nature:
(i) Conductor: Conductors are the material in which the outer
most electrons are very loosely bound to the nucleus, so they
are free to move (flow). So in a conductor, there are large
number of free electrons.
Example: Metals like Cu. Ag. Fe. Al.
(ii) Insulator or Dielectric or Non-conductor: Non-conductors
are the materials in which outer most electrons are very
tightly bound to the nucleus, so that they cannot move
(flow). Hence in a non-conductor there are no free electrons.
Example: plastic, rubber, wood etc.
(iii) Semi conductor: Semiconductors are the materials which
have free electrons but very few in number.
Now lets see how the charging is done by conduction. In this
method, we take a charged conductor ‘A’ and an uncharged
conductor ‘B’. When both are connected, some charge will
flow from the charged body to the uncharged body. If both
the conductors are identical & kept at large distance and
connected to each other, then charge will be divided equally
in both the conductors otherwise they will flow till their
electric potential becomes same. Its detailed study will be
done later.
+
+ + ++
++ +
Charged
body
++++
+
+
++
+ ++
+
B
A
Uncharged
body
A
++
++ ++
B
++
++
+
Charging by Friction
In friction when two bodies are rubbed together, electrons are
transferred from one body to the other. This makes one body
become positively charged while the other become negatively
charged, e.g., when a glass rod is rubbed with silk, the rod becomes
positively charged while the silk is negatively charged. Clouds are
also charged by friction. Charging by friction is in accordance with
conservation of charge. The positive and negative charges appear
simultaneously in equal amounts due to transfer of electrons from
one body to the other.
Charging by friction is based on difference in work function (f) of
the bodies we are rubbing. We cannot generate charge by rubbing
two bodies made up of same material.
A body can be charged by induction in the following two ways:
Method-I
Step 1: Take an isolated neutral conductor.
Step 2: Bring a charged rod near it. Due to the charged rod,
charges will induce on the conductor.
++ ++ + +
++ ++ + +
–– +
–– +++
–––
+
–––
+
–––+++
Step 3: Connect another neutral conductor with it. Due to
attraction of the rod, some free electrons will move from the right
conductor to the left conductor and due to deficiency of electrons
positive charges will appear on right conductor and on the left
conductor, there will be excess of electrons due to transfer from
right conductor.
+ ++
++ +
+ ++ + +
+
–– +
–– +++
–––
+
–––
+
–––+++
electrone
transfer
–– +
–– +++
–––
+
–––
+
–––+++
Step 4: Now disconnect the connecting wire and remove the rod.
One body gathers negative charge while second becomes positive
charged.
Method-II
Step 1: Take an isolated neutral conductor.
Charging by Induction
If a charged body is brought near a neutral body, the charged body
will attract opposite charge and repel similar charge present in
the neutral body. One side of the neutral body become positively
charged while the other side negative.
Important Points
™
Inducing body neither gains nor loses charge.
™
The nature of induced charge is always opposite to that of
inducing charge.
™
Induced charge can be lesser or equal to inducing charge (but
never greater) and its maximum value
™
 1
is q′ = −q 1 −  [only when electric field is uniform]
 K
™
q = the inducing charge
™
K = the dielectric constant of the material of the uncharged
body
™
For metals, K = ∞ So q′ = –q
™
i.e., in metals induced charge may be equal or less than the
inducing charge.
2
Step 2: Bring a charged rod near it. Due to the charged rod,
charges will induce on the conductor.
–– +
–– +++
++ ++ + +
–
––
++ ++ + +
+
–––
+
–––+++
Step 3: Connect the conductor to the earth (this process is
called grounding or earthing). Due to attraction of the rod, some
free electrons will move from earth to the conductor, so in the
conductor there will be excess of electrons due to transfer from the
earth, so net charge on conductor will be negative.
––– +++
electron
++
––
+ ++
–
+
––
transfer
+
+
+
––
+ ++ + +
–––+++
+
Step 4: Now disconnect the connecting wire. Conductor becomes
negatively charged.
JEE (XII) Module-1 PW
COULOMB’S LAW
VECTOR FORM OF COULOMB’S LAW
The force of attraction or repulsion between two stationary point
charges is directly proportional to the product of charges and
inversely proportional to the square of distance between them.
This force acts along the line joining the point charges.
If q1 and q2 are charges
r = the distance between them
F = the force acting between them
1
q q
Then F ∝ q1 q2 & F ∝ 2 ∴ F ∝ 1 2
r
r2
1 q1q2
F =
4π ∈0 r 2
q1
q3
r
1
= 9 × 109 Nm²/C²
4πε0
[in electrostatic unit (esu) constant of proportionality = 1]
ε0 = 8.85 × 10–12 C²/Nm² = permittivity of free space or
vacuum
This law is valid only for stationary point charges and cannot
be applied for moving charges.
EFFECT OF MEDIUM
The dielectric constant of a medium is the ratio of the electrostatic
force between two charges separated by a given distance in
vacuum to electrostatic force between same two charges separated
by same distance in that medium.
1 q1q2
1 q1q2
Fvacuum =
and Fmedium =
4πε0 r 2
4πε0ε r r 2
⇒
Fmedium 1
1
=
=
K
Fvacuum ε r
εr or K = dielectric constant or relative permittivity or specific
inductive capacity of medium.
Permittivity: Permittivity is a measure of the ability of the medium
surrounding electric charges to allow electric lines of force to pass
through it. It determines the forces between the charges.
Relative Permittivity: The relative permittivity or the dielectric
constant (εr or K) of a medium is defined as the ratio of the
permittivity ε of the medium to the permittivity ε0 of free space
ε
ε0
Dimension of permittivity
i.e. εr or K =
The dielectric constants of different mediums
Vacuum
1
Teflon
2
Air
1.00059
Glass
5-10
P Electric Charges and Fields
W
PVC
4.5
Another charge q2 is placed at B whose position vector is r2 , such
that AB =| r2 − r1 | = r .
The magnitude of force is given by
F=
k q1 q2
4πε0 r 2
Force on q2 due to q1, in vector form, is given by
F2 =
1 q1q2 AB is a unit vector along line joining
AB , where 4πε0 r 2
A and B, pointing from A to B.
 1  q1q2 ( r2 − r1 )  1  q1q2 =
⇒ F2 
=
 2 
 3 ⋅ ( r2 − r1 )
r2 − r1
 4πε0  r
 4πε0  r
q2
y
q1
AB
F12 A
F21
B
r1
r
2
x
Similarly force on q1 due to q2 is given by,
 1  q1q2 ( r1 − r2 )  1  q1q2 =
F1 
 2 =

 3 ⋅ ( r1 − r2 )
r1 − r2
 4πε0  r
 4πε0  r
Note that F12 = − F21
Both forces are action reaction pair
PRINCIPLE OF SUPERPOSITION
Consider a system of n stationary charges q1, q2, q3, ... qn in
vacuum. What is the force on q1 due to q2, q3 ... qn? Coulomb’s law
is not enough to answer this question. Force on any charge due to a
number of other charges is the vector sum of all the forces on that
charge due to the other charges taken one at a time. The individual
forces are unaffected due to the presence of other charges. This is
termed as the ‘principle of superposition’. (Note that the principle
of superposition is independent of Coulomb’s law and all of
electrostatics is basically a consequence of Coulomb’s law and
superposition principle together).
Now, the force on q1 can be found out by calculating separately the
forces F12 , F13 ……..F1n exerted by q2 , q3 …… qn , respectively
on q1 and then adding these forces vectorially.
[M–1 L–3 T4 A2]
Medium
c
A charge q1 is placed at A whose position vector is r1
Water
80
Metal
∞
Mica
6
Fin
F12
q2
q1
q3
F13
q4
3
So,=
F1 F 12 + F13 +……+ F 1n
Identification of the Nature of the Charge
 q r q3 r

qn r
13
1n
 2 212 + 2 +……. + 2 
r13
r91 
 r12
r −r
q n q rˆ
= 1 ∑ 1 21i where r1i= r1 − r2 and 1 i
| r1 − ri |
4πε0 i =2 r1i
=
∴ F1
q1
4πε0
GOLD LEAF ELECTROSCOPE
Electric charge is the basic physical property of matter that causes it
to experience a force when kept in an electric or magnetic field. An
electric charge is associated with an electric field, and the moving
electric charge generates a magnetic field. Electric charges are of
two types: Positive and Negative, commonly carried by charge
carriers protons and electrons. In this article, let us familiarise
ourselves with a device called Gold Leaf Electroscope, which is
used to detect the charge in a body.
A gold-leaf electroscope is defined as: A type of electroscope that
consists of two gold leaves and is used for detecting the electrical
charge of the body and for the classification of its polarity.
Construction of Gold Leaf Electroscope
The gold leaf electroscope is a sensitive electroscope type that is
used for detecting charges. It consists of a brass rod with a brass
disk at the top, and at the bottom, there are two thin gold leaves
in the form of foils. In order to keep the rod in place, the rod
travels through the insulator. The charges move from the disk to
the leaves through the rod. At the lower portion of the jar, a thin
aluminium foil is connected. The aluminium foil is grounded with
the help of a copper wire so that the leaves are protected from
external electrical disruptions.
To identify the nature of the charge, let’s consider an example. A
positively charged body is brought near the metal cap. Then an
unknown body is brought near the metal cap. If the leaves diverge
further, we can conclude that the unknown body has a positive
charge. If the leaves come closer to each other, then the charge of
the unknown body is negative.
Identification of Body as a Conductor or an Insulator
To identify if a body is a conductor or an insulator, two gold leaf
electroscopes are taken. One gold leaf electroscope is charged so
that the leaves will diverge. Then the other gold leaf electroscope
is connected to the first one. If the leaves of the other electroscope
diverge, then the body is said to be a conductor, and if there is no
change in the leaves, the body is said to be an insulator.
Train Your Brain
Example 1: Two equal point charges (10–3 C) are placed 1 cm
apart in medium of dielectric constant K = 5
(a) Find the interaction force between the point charges.
(b) Net force on any of the charge.
Sol. (a) Interaction force between point charges
(
(
10−3
1 q1q2
9
F =
9
10
=
×
4π ∈0 r 2
10−2
Glass bottle
Brass rod
Brass rod
Metal foil
Earth
(
(
™
™
Detection Charge
For the detection of charge, the object that needs to be tested is
touched with the metal cap. If the leaves diverge, the body is
said to be charged, and if there is no change in the leaves of the
electroscope, then the body is uncharged.
4
9 × 107 N
=
)
)
2
Example 2: Two small balls each of mass m
and charge q on each of them are suspended
through two light insulating string of length
l from a point. Find the expression for
q
angle q made by any of the string with
m
vertical when under static equilibrium.
O
q
m
Sol. Let angle of any string with vertical be q as shown
o
The following are the applications of gold leaf electroscope:
Detect charge
Identification of the nature of the charge
Identification of the body as a conductor or an insulator
2
−3
q1q2 9 × 109 10
1
=
= 18 × 106 N
F='
−2 2
4π ∈0 K r 2
5
10
Applications of Gold Leaf Electroscope
™
2
(b) Net force
Brass disc
Insulator
plug
)
)
T
T
Fe
mg
for horizontal direction
Fe
=
Fe
mg
1
q2
= T sin θ 4π ∈0 ( 2l sin θ )
for vertical direction
T cos θ = mg
...(i)
...(ii)
JEE (XII) Module-1 PW
Dividing (i) by (ii)
tan θ =
Fe
mg
Kq 2
(2a )
2
=
K |Q|q
a2
9
⇒
Q=–



q
4
(i) Ball A repels C and attracts B.
(ii) Ball D attracts B and has no effect on E.
(iii) A negatively charged rod attracts both A and E. For
your information, an electrically neutral styrofoam
ball is very sensitive to charge induction and gets
attracted considerably, if placed nearby a charged
body. What are the charges, if any, on each ball?
From (iii), a –vely charged rod attracts the charged ball
A, so A must be +ve and from exp.
(i) C must also be +ve and B must be –ve.
Example 5: A point charge qA = + 100 mC is placed at point A
(1, 0, 2) m and another point charge qB = + 200 mC is placed at
point B (4, 4, 2) m. Find:
(i) Magnitude of electrostatic interaction force acting
between them
(ii) Find FA (force on A due to B) and FB (force on B

ˆ
ˆ
ˆ

( 4 − 1) i + ( 4 − 0 ) j + ( 2 − 2 ) k 
Sol. F =
µmg (∵ N =
mg )
N
N
q
F
q
F
r
mg
⇒
1 q2
=
µmg
4πε0 r 2
1/2
 1 q2 
⇒r=


 4πε0 µmg 
Example 7: Three equal point charges of charge +q each
are moving along a circle of radius R and a point charge –2q
is also placed at the center of circle (as shown in figure). If
charges are revolving with constant and same speed in the
circle then calculate speed of charges
Sol. +q
v
R
qB = + 200µC
F
–2q
R
B(4, 4, 2)
v
R
+q
+q
qA = + 100 µC
A(1, 0, 2)
v
(i) Value of F:
kq A qB
F
=
=
r2
ˆj  N

Strategy: The separation will be minimum, when
electrostatic force of repulsion is just balanced by limiting
force of friction.
due to A) in vector form
F
3
Example 6: Two particles having mass m and charge q
each are to be placed on a rough horizontal surface with
coefficient of static friction m. What can be the minimum
separation between them so that they remain stationary?
Sol. From (i), as A repels C, so both A and C must be
charged similarly. Either both are +ve or both are –ve. As A
also attract B, so charge on B should be opposite of A or B
may be uncharged conductor.
From (ii) as D has no effect on E, so both D and E
should be uncharged and as B attracts uncharged D, so B
must be charged and D must be an uncharged conductor.
−6
( 4 − 1)2 + ( 4 − 0 )2 + ( 2 − 2 ) 
4 
3
=
7.2  iˆ + iˆ 
5 
5
4
 3
Similarly, FA = 7.2  − iˆ −
5
 5
Example 4: Five syrofoam balls A, B, C, D and E are used in
an experiment. Several experiments are performed on the balls
and the following observations are made :
Sol.
−6
Example 3: For system to be in equilibrium Q = ?
Q
a q
a
q
Sol.
(9 ×10 )(100 ×10 )( 200 ×10 ) ×
kq A qB
(ii)=
FB =
3
F
10 )
(9 ×10 )(100 ×10 )( 200 ×=
9



( 4 − 1)
−6
2
P Electric Charges and Fields
W
+ ( 4 − 0)
F2 − 2 F1cos30° =
−6
2
+ ( 2 − 2 ) 

2
7.2 N
⇒
K ( q )( 2q )
R2
−
mv 2
R
(
2 Kq 2
) cos30° = mv
( 3R ) 2
2
R
5
F1
2. Which one of the following statement regarding
electrostatics is wrong?
(a) Charge is quantized
(b) Charge is conserved
(c) There is an electric field near an isolated charge
at rest
(d) A stationary charge produces both electric and
magnetic fields
F1
30° 30°
+q
R
3R
F2
3R
–2q
R
R
+q
3. When the distance between two charged particle is
halved, the force between them becomes
(a) One fourth
(b) One half
(c) Double
(d) Four times
+q
3R
4. A positively charged body ‘A’ attracts a body ‘B’ then
charge on body ‘B’ may be
(a) Positive
(b) Negative
(c) Zero
(d) Can’t say
kq 2 
1 
2−


Rm 
3
=
⇒v
Example 8: Five point charges, each of value q are placed
on five vertices of a regular hexagon of side L except at A.
What is the magnitude of the force on a point charge of
value – q coulomb placed at the center of the hexagon?
Sol. If there had been a sixth charge + q at the remaining
vertex of hexagon, force due to all the six charges on = q
at O would have been zero (as the forces due to individual
charges will balance each other), i.e., FR = 0
L
E
D
q
q
O
q
q C
F
q
q
A
B
Now if f is the force due to sixth charge and F due to
remaining five charges.
From F + f =
0 i.e., F = − f
or, F=
1 q×q
1 q2
=
=
f
4πε0 L2
4πε0 L2
F=
F=
net
OD
1 q2
along OD
4πε0 L2
Concept Application
1. One quantum of charge should be at least equal to the
charge in coloumb
6
(a) 1.6 ×
10–17
(c) 1.6 ×
10–10
C
C
ELECTRIC FIELD
A region around a charged particle within which a force would
be exerted on other charged particle. An electric field is a vector
quantity and can be visualized as arrows going towards or away
from charges.
Electric Field due to Point Charge
To determine the direction of an electric field, consider a point
charge q as a source charge. This charge creates an electric field at
all points in space surrounding it. A test charge q0 is placed at point
P, a distance r from the source charge, as in figure (a). We imagine
using the test charge to determine the direction of the electric force
and therefore that of the electric field. However, the electric field
does not depend on the existence of the test charge-it is established
solely by the source charge. According to Coulomb’s law. the
qq
force exerted by q on the test charge is Fe = k 20 r
r
where r̂ is a unit vector directed from q toward q0. This force in
figure (a) is directed away from the source charge q, since the electric
field at P, the position of the test charge, is defined by E = F/q, we find
q
that at P, the electric field created by q is, E = k rˆ
r2
When the source charge q is positive as in figure (b) (it shows the
situation with the test charge removed) the source charge sets up
an electric field at point P, directed away from q.
When q is negative, as in figure (c), the force on the test charge
is toward the source charge, so the electric field at P is directed
toward the source charge, as in figure (d)
q0
F
E
+
(b) 1.6 ×
C
q
(d) 4.8 ×
10–10
C
+
+
(a)
(b)
P
r
+
P
P
10–19
q
P
r
q
– r
(c)
(c)
E
P
q r
– P
(d)
JEE (XII) Module-1 PW
Volume charge density: It is the quantity of charge per unit
volume with S.I unit C/m3
Electric Field Due to Continuous
Charge Distribution
Consider the following charged object of irregular shape as shown
in Figure The entire charged object is divided into a large number
of charge elements Dq1, Dq2, Dq3 ..... Dqn, and each charge element
Dq is taken as a point charge.
q1
q3
r3p
r3p
r1p
r1p
E2
p
E1
E3
r2p
q2
r 2p
The electric field at a point P due to a charged object is
approximately given by the sum of the fields at P due to all such
charge elements.
∆qn 
∆q2
1  ∆q1
E≈
 2 rˆ1P + 2 rˆ2 P + ..... + 2  rˆnP
4πε0  r1P
r2 P
rnP 
1 n ∆qi
∑ rˆiP
4πε0 i =1 riP2
To incorporate the continuous distribution of charge, we take
the limit Dq → 0 (= dq). In this limit, the summation in the
equation becomes an integration and takes the following form
1
dq
E=
rˆ
∫
4πε0 r 2
≈
Linear charge density: It is the quantity of charge per unit length
with S.I unit C/m. mathematically:
l=
dq
dl
Mathematically, the volume charge density:
r=
Electric field due to volume
charge distribution of total
charge q is given by:
1
 ρdV 
E=

 rˆ
∫
4πε0 volume  r 2 
SI Unit : N/C or V/m
™
Dimensions: [M L T–3 A–1]
™
In free space Electric field E0 =
Electric field due to surface charge distribution of total charge q
is given by:
1
 σdA 
E ( P) =
∫  r 2  rˆ
4πε0 surface
P Electric Charges and Fields
W
1 q
4πε0 r 2
In a medium of permittivity ε field E =
E ε0 1
So = =
or
ε K
E0
E=
E0
K
1 q
4πε r 2
[as ε = ε0 K]
In presence of a dielectric, electric field decreases and
1
times of its value in free space.
becomes
K
™ The electrostatic field i.e. field produced by charge at rest is
conservative.
of total charge q is
Electric field due to linear charge distribution
given by:
1
 λdl 
E=

 rˆ
∫
4πε0 line  r 2 
dA
™
dl
dV
Important Points
E=
Surface charge density: It is the quantity of charge per unit area
with S.I unit C/m2
Mathematically, the surface charge density:
dq
s=
dA
dq
dV
kq
r
2
kq
E = 2 rˆ
r
kq E= 3r
r
+
–
r = position vector of test point with respect to source charge
r = r test point – r sourcecharge
™
ELECTRIC FIELD AT THE CENTER OF
CHARGED SEMICIRCULAR RING
Let l = linear charge density.
+
+
+
+
dEx
dE
+
d
x
+
y
dEy
7
The arc is a collection of large numbers of point charges.
Consider a part of ring as an element of length Rdθ which
subtends an angle dq at center of ring and it lies between
θ and θ + dθ
=
dE dEx iˆ + dE y ˆj
Ex =
Ey =
∫ dEx = 0
kλ
2k λ
λ
sin θ. =
dθ
=
2πε0 R
R ∫0
R
0
ELECTRIC FIELD DUE TO A
CHARGED RING ON ITS AXIS
z
+
R
+
O
+
P(x,0,0)
dE p
x
dE p
+
∫ dEP cos θ
or
=
EP
(x
k .dQ
2
+R
2
·
)
x
2
x +R
2
=
(x
kx
2
+ R2
)
3/2
∫ dQ
dEP
=0
dx
2




where q = semicone angle subtended by disc at point ‘P’.
π
Case: If R → ∞, then q →
2
Disc becomes infinite sheet i.e., an infinite sheet can be looked
upon as disc of infinite radius.
σ
Einfinite sheet =
2ε0
+ uniformly charged
+ +
+
+ + + + + infinite sheet
+ + + + ++
+
+ + ++ + P
Ep
+ ++ ++ +
++
+ + ++ + +
+ ++ + +
σ
Einfinite sheet =
nˆ
2ε0
R
where n̂ is a unit vector normal to the sheet pointing away
from the sheet.
2
E
- R/ 2 O
X
R/ 2
8
rdr
x
σ 
=
1 −
2 3/2

2
2ε 0 
x + R2
0 (x + r )
This result is easy to remember in following form
σ
E=
(1 − cos θ )
2ε 0
)
The graphical variation of electric field along x-axis is as shown
below:
⇒x= ±
x
dE p
dEP = Electric field due to elementary ring
k .dQx
= 2
( x + r 2 )3/2
∴ EP = ∫ dEP
kQx
Qx
=
3/2
4πε0 ( x 2 + R 2 )3/2
x2 + R2
(
P(x,0,0)
R
Consider an element on the ring, then electric field due to this
element is given by

k .dQ
| dEP |= 2
x + R2
From symmetry of charge distribution, it is clear that component
perpendicular to x-axis will cancel out.
or EP = ∫
dQ = charge on elementary ring
= (2pr) (dr) s = 2ps r dr
or EP = 2πk σx ∫
+
EP =
++
++ +
+ + + ++
+ + ++ +
+ ++
+
+ R +
++
+ ++
+
+ + O+ ++
+
+
+ ++ ++ +
+
++
+ ++
Let us consider an elementary ring of radius ‘r’ and thickness ‘dr’,
then
Let us consider a uniformly charged ring as shown in figure. We
wish to calculate Electric field at a point on its axis.
+
y
z
π
θ
∫ dE sin=
+
Let us consider a uniformly charged disc of radius ‘R’ and Charged
density s as shown in figure. We wish to calculate Electric field
at a point on its axis.
(due to symmetry)
π
∫ dE=y
ELECTRIC FIELD DUE TO CHARGED DISC
ELECTRIC FIELD DUE TO
UNIFORMLY CHARGED WIRE
Electric field due to a finite straight wire carrying uniform linear
charge at a perpendicular distance r from the wire:
JEE (XII) Module-1 PW
dq = dx
\ E=
dx
rsec
x
Parallel to
the line
2
A
En2 + E||2 =
r
2k λ
r
P
Normal to
the line
1
dE
En
r
E||
Consider electric field dE from a small element at a distance x
from point A
kdq
k λdx
=
r 2 sec 2 θ r 2 sec 2 θ
x = rtanq
dE =
dx
= rsec2q
dθ
⇒
E||
dx = xsec2qdq
kλ
dθ
r
kλ
dEn =
cos θd θ
r
r
2
1
En
⇒ dE =
θ
⇒ En =
kλ 2
kλ
=
θd θ
cos
(sin θ2 + sin θ1 ) ∫
r −θ
r
...(i)
Train Your Brain
Example 9: A cube of edge a metres carries a point charge
q at each corner. Calculate the resultant force on any one of
the charges.
C
P
Z
Sol. Let us take one corner
A
Y
of cube as origin O(0, 0, 0) E
and the opposite corner as
X
B
F
P(a, a, a). We will calculate
D
the electric field at P due to O
the other seven charges at corners.
Expressing the field of a point charge in vector form
q E=
r
4πε0 r 3
(i) Field at P due to A, B, C
q
 AP + BP + CP 
=
E1
kλ
3 

dE|| = − sin θd θ
4πε0 a
r
θ
q  kλ 2
kλ
=
a j + ak + ai 
3 

=
θ
θ
θ
−
θ
sin
d
(cos
cos
)
⇒ E|| = −
...(ii)
2
1
4
πε
a
∫
0
r −θ
r
1
(ii) Field at P due to D, E, F
Enet = En2 + E||2
Note that DP = EP = FP = a 2
q
 DP + EP + FP 
1. For infinitely long line charge: In above eq. (i) & (ii) if we
=
E2
3 

put q1 = q2 = 90º we can get required result.
4πε0 a 2
2k λ
λ
q
=
Enet = En =
 a j + ak + ai + a j + ai + ak 
=
r
2πε0 r

4πε0 2 2a 3 
+
q
i + j + k 
=
+

+
4πε0 2a 2 
+
+
(iii) Field at P due to O
+
1
E
+
E
r
r E
+
OP = a 3
+
+
q
E3 =
OP
+
3
+
r
4πε0 a 3
+
+
q
 ai + a j + ak 
=
E3

4πε0 3 3a 3 
2. For Semi-infinite wire
θ1 = 90º and θ2 = 0º so
q
i + j + k 
E3
=
2 

kλ
kλ
4
πε
3
3
a
0
, En =
En =
r
r
1
(
(
)
(
P Electric Charges and Fields
W
)
(
) (
) (
)
)
(
)
(
)
9
Sol.
Resultant Field at P
E = E1 + E2 + E3
q i + j + k 
1
1 
=
E
1+
+

4πε0 a 2 
2 3 3
outward along OP.
(
+
Q
q
Sol. Let us make the force on
F1
F2
q
upper right corner ‘q’ equal Q
to zero. Here both the ‘q’
F2
will have same sign (either
d 2
d
positive or negative), so they
will repel each other, say by
force F1. To balance F1, Q
Q
must apply attractive force on q
q, say F2. That also means Q and q will have opposite signs.
Resultant of both these F2 will be 2F2 (By Pythagoras
theorem) and it will exactly be opposite to F1
or,
F1 = 2F2
From Coulomb’s law
q2
4πε0 d 2
Qq
and F2 =
4πε0 d 2
∵ F1 = 2 F2
q
∴
q 2
(
)
2
+
+
Fe
+
+
+
+
+
+
+ +
+
Electric force acting on the particle, Fe = q0E
)
Example 10: Four charges Q, q, Q Q and q are kept at the four corners of a
square as shown below. What is the d
relation between Q and q so that the
q
net force on a charge q is zero?
F1 =
+
2Qq
=
d2
q
∴Q =
2 2
But as we said Q and q have opposite sign so,
q = −2 2Q.
Example 11: An infinitely large plate of surface charge
density +σ is lying in horizontal xy-plane. A particle having
charge –q0 and mass m is projected from the plate with
velocity u making an angle θ with sheet. Find:
 σ 
Fe = (q0 ) 
 downward
 2ε 0 
Fe
q0 σ
So, acceleration of the particle : =
a =
(uniform)
m 2ε 0 m
This acceleration will act like ‘g’ (acceleration due to
gravity)
So, the particle will perform projectile motion.
2u sin θ 2u sin θ
=
(i) T =
g
 q0 σ 


 2ε0 m 
u 2 sin 2 θ
=
2g
=
(ii) H
=
(iii) R
u 2 sin 2 θ
 qσ 
2 0 
 2ε 0 m 
u 2 sin 2θ u 2 sin 2θ
=
g
 q0 σ 


 2ε 0 m 
Example 12: A block having
mass m and charge Q is resting
Sheet
(–)
on a frictionless plane at a
distance d from fixed large
m
non-conducting infinite sheet
of uniform charge density –σ
d
as shown in Figure. Assuming
that collision of the block with
the sheet is perfectly elastic, find the time period of oscillatory
motion of the block. Is it SHM?
Sol. The situation is shown in Figure. Electric force
produced by sheet will accelerate the block towards the
sheet producing an acceleration. Acceleration will be
uniform because electric field E due to the sheet is uniform.
F QE
a = = , where E =
σ / 2ε0
m
m
As initially the block is at rest and acceleration is constant,
from second equation of motion, time taken by the block to
reach the wall
=
d
1 2
at i.e.,
=
t
2
2L
=
a
2md
=
QE
4md ε0
Qσ
u
+
–
+
+
+
R
+
+
+
+
+
+
+ +
+
(i) The time taken by the particle to return on to the plate.
(ii) Maximum height achieved by the particle.
(iii) At what distance will it strike the plate (Neglect
gravitational force on the particle)
10
m
E
Sheet
(–)
QE
d
mg
JEE (XII) Module-1 PW
As collision with the wall is perfectly elastic, the block will
rebound with same speed and as now its motion is opposite
to the acceleration, it will come to rest after travelling same
distance d in same time t.
After stopping, it will again be accelerated towards the wall
and so the block will execute oscillatory motion with ‘span’
d and time period.
4md ε0
2md
= 2
QE
Qσ
However, as the restoring force F = QE is constant and not
proportional to displacement x, the motion is not simple
harmonic.
Example 13: Positive charge Q is distributed uniformly
over a circular ring of radius a. A point particle having a
mass m and a negative charge –q, is placed on its axis at a
distance y from the center. Find the force on the particle.
Assuming y << a, find the time period of oscillation of the
particle if it is released from there. (Neglect gravity)
Sol. When the negative charge is
E
shifted at a distance x from the center
of the ring along its axis then force
–q
acting on the point charge due to the
F
ring :
e
X
FE = qE (towards center)
Sol. q1 = 90° and q2 = 360° – 37° So
T= 2=
t 2

KQy
= q 
 2
2
 a + y
(
)


3/ 2 

kλ
kλ
En
=
[sin θ1 + sin θ2=
]; E||
[cos θ2 − cos θ1 ]
r
r
Example 15: Three large conducting parallel sheets are
placed at a finite distance from each other as shown in
figure. Find out electric field intensity at points A, B, C & D.
–2Q
Q
A
B
3Q
D
C
y
x
Sol. For point A:
Q
If a >> y then a2 + y2 ≈ a2
–2Q
3Q
E–2Q
1 Qqy
∴ FE =
(Towards center)
4πε0 a 3
Since, restoring force FE ∝ y, therefore motion of charge the
particle will be S.H.M. Time period of SHM
1/2
16π3ε0 ma 3 
m
m
=
=
T=
2π
2π


k
Qq
 Qq  


3 

 4πε0 a 
EQ E3Q A
Q ˆ 3Q ˆ 2Q ˆ
Q ˆ
Enet =+
EQ E3Q + E−2Q =
−
−
i−
i+
i=
i
2 Aε0
2 Aε0
2 Aε0
Aε0
For point B:
Q
Example 14: Figure shows a long wire having uniform
charge density λ as shown in figure. Calculate electric field
intensity at point P.
–2Q
3Q
E–2Q
E3Q B
EQ
3Q ˆ 2Q ˆ
Q ˆ
0
Enet =
E3Q + E2Q + EQ =
−
i+
i+
i=
2 Aε0
2 Aε0
2 Aε0
For point C:
Q
–2Q
3Q
E3Q EQ
37°
E–2Q C
P
r
P Electric Charges and Fields
W
11
™
2Q ˆ
Q ˆ 3Q ˆ 2Q ˆ
Enet =+
EQ E3Q + E−2Q =
+
i−
i−
i=
−
i
Aε0
2 Aε0
2 Aε0
2 Aε0
For point D:
Q
™
–2Q
3Q
EQ E3Q
E–2Q D
™
Q ˆ 3Q ˆ 2Q ˆ
Q
Enet =+
EQ E3Q + E−2Q =
+
i+
i−
i = iˆ
Aε0
2 Aε0
2 Aε0
2 Aε0
Lines of force never cross each other because if they cross then
electric field intensity at that point will have two directions
which is not possible.
In electrostatics the electric lines of force can never be
closed loops, as a line can never start and end on the same
charge. If a line of force is a closed curve, work done round a
closed path will not be zero and electric field will not remain
conservative.
Lines of force have tendency to contract longitudinally like a
stretched elastic string producing attraction between opposite
charges and repel each other laterally resulting in, repulsion
between similar charges and ‘edge-effect’ (curving of lines of
force near the edges of a charged conductor)
+
–
+
+
Concept Application
5. Four equal but like charges are placed at four corners
of a square. The electric field intensity at the center of
the square due to any one charge is E, then the resultant
electric field intensity at center of square will be
6.
(a) Zero
(b) 4E
(c) E
(d) 1/2E
™
A charge of 4 ×10–9C is distributed uniformly over the
circumference of a conducting ring of radius 0.3 m.
Calculate the field intensity at a point on the axis of
the ring at 0.4m from its center and also at the center?
(a) 112N/C, 2N/C
(b) 112N/C, 3N/C
(b) 115.2N/C, Zero
(d) 113.2N/C, Zero
edge effect
+
+
+
E= 0
+q
ELECTRIC LINES OF FORCE
Important Points
™ Electric lines of force usually start or diverge out from positive
charge and end or converge on negative charge.
™
12
The number of lines originating or terminating on a charge is
proportional to the magnitude of charge.
electric lines of force for a system
of two positive charges
Electric lines of force end or start normally on the surface of
a conductor. If a line of force is not normal to the surface of a
conductor, electric intensity will have a component along the
surface of the conductor and hence charge an the surface will
start flowing. But in electrostatics charge is at rest. So, charge
continues to flow until tangential component of electric field
vanishes.
d
The idea of lines of force was introduced by Michael Faraday.
A line of force is an imaginary curve, the tangent to which at a
point gives the direction of electric field intensity at that point and
the number of lines of force per unit area normal to the surface
surrounding that point gives the magnitude of electric field
intensity at that point.
repulsion
attraction
electric lines of force for a dipole
™
™
™
+
–
+
–
E=
0 –
+
+ Uniform–
+ field –
E=
E= 0
+
0
+
+
+
E= 0
+
E=
0
+
+
+
+
+
–
+
+
+
–
+
fixed point charge near
infinite metal plate
parallel metal plates
having dissimilar charges
parallel metal plates
having dissimilar charges
(A)
(B)
(C)
If in a region of space, there is no electric field there will
be no lines of force. This is why inside a conductor or at a
neutral point where resultant intensity is zero there is no lines
of force.
The number of lines of force per unit normal area at a point
represents magnitude of electric field intensity. The crowded
lines represent strong field while distant lines show a weak
field.
The tangent to the line of force at a point in an electric
field gives the direction of intensity. It gives direction of
force and hence acceleration which a positive charge will
experience there (and not the direction of motion). A positive
point charge free to move may or may not follow the line
of force. It will follow the line of force if it is a straight line
(as direction of velocity and acceleration will be same) and
will not follow the line if it is curved as the direction of motion
will be different from that of acceleration.
JEE (XII) Module-1 PW
ELECTRIC FLUX
ds
The mathematical quantity related to number of lines passing
through a surface is called the electric flux f. The electric flux
through a surface which is perpendicular to a uniform electric field
E is defined as the product of electric field E and surface area S :
f = ES
Since the electric field is proportional to density of lines of force,
the electric flux is proportional to number of lines of force passing
through the surface area : f a N.
If the surface area is not perpendicular to the electric field, then the
electric flux is given by
ˆ
=
φ E=
E cos=
θ S En S
. nS
where n̂ is a unit vector perpendicular to the surface and En is the
component of electric field perpendicular to the surface (normal
component).
n̂
S
E
The electric flux over a curved surface over which electric field
may vary in direction and magnitude can be computed by dividing
the surface into large number of very small area elements. Let
nˆi be the unit vector perpendicular to such an area element and
DSi be its area. The flux of the electric field through such an area
element is
=
∆φ
Ei . ni ∆Si
i
The total flux through the surface area is found by adding flux
through each area element. In the limit, when number of area
elements approaches infinity and area of each element approaches
zero, the sum becomes an integral.
=
φ lim ∑=
Ei . nˆi ∆ Si ∫ E . nˆ dS
n→∞
i
Sometimes we are interested in finding out the flux through a
closed surface. The unit vector n̂ in such a case is defined to be
directed outward from each point. Note that when an electric line
comes out of the closed surface, then E . nˆ is positive and if it
enters the surface is E . nˆ negative.
The net flux through a closed surface is written as
=
φnet
. nˆ dS ∫ En dS .
∫ E=
Please note that f is proportional to the net lines of force going out
of the surface, i.e., number of lines going out of the surface minus
the number of lines going into the surface.
Important Points
™ It is a scalar physical quantity.
™ SI Units: volt × m
Dimensions: [ML3T–3A–1]
P Electric Charges and Fields
W
P
E
d Eds cos
ds
P
E
º
(dmax Eds
ds
º
P
E
(dmin 0
It will be maximum when cos q is max = 1, i.e.,
q = 0°, i.e., electric field is normal to the area with
(dfE)max = EdS
™
™
™
It will be minimum when cosq is min = 0, i.e., q = 90°,
i.e. field is parallel to the area with (dfE) min = 0
For a closed body outward flux is taken as positive while
inward flux is taken as negative.
GAUSS’S LAW
If a closed surface encloses a positive change the net flux is positive,
and if it encloses a negative charge the net flux is negative. A
charge residing outside the closed surface cannot contribute to the
net flux because every line of force that enters the surface at one
point leaves the surface at some other point.
Consider a closed spherical surface of radius r with a point
charge q at the center. The electric field at the surface due to this
q
charge is normal to the surface and its magnitude will be
.
4πεo r 2
Therefore, the flux through the surface is
q =
q ×=
=
φ ∫ Eˆ =
.nˆ d S
dS
4πr 2 q
ε0
4πε0r 2 ∫
4πε0r 2
Notice that the radius of the spherical surface cancels out,
for while the surface area goes up as r2 the field goes down as
1/r2, and so the product is constant. In terms of electric lines of
force, this makes sense since the same number of lines of force
pass through any spherical surface centered at the point charge
irrespective of its size. In fact it didn’t have to be a spherical
surface - any surface whatever its shape, size would have the same
number of lines of force passing through it. Thus, the flux through
any closed surface enclosing the charge is q/e0.
Now suppose we have number of point charges spread over a
region. From the superposition principle, the field at a point is
n given by E = ∑ Ei
i =1
The flux through any surface due to all of the charges enclosed by
the surface will be
n n q
i
=
=
∫ E.nˆ dS ∑
∫ Ei .nˆ dS ∑
=i 1 =i 1 ε 0
On the other hand, a charge residing outside the surface would
contribute nothing to the net flux because every line of force that
enters the surface at one point leaves the surface at another point.
It follows then that for any closed surface.
qenclosed
=
φnet
E .nˆ dS
∫ =
ε0
surface
where qenclosed is the total charge enclosed inside the surface. This
important result is called Gauss’s Law and can be stated as follows:
For a system of charges, the net flux through any closed surface ‘S’
is equal to 1/e0 times the net charge inside the surface.
13
FLUX THROUGH OPEN SURFACES
USING GAUSS’S THEOREM
Consider a point charge +q is
placed at the center of curvature of
a hemisphere. Then flux through
the hemispherical surface can be
calculated as follow:
Lets put an upper half hemisphere.
Gaussian Surface
™
+q
2. Cylindrical Symmetry: When, by symmetry, electric field
is same at all points at a perpendicular distance r from an
infinite lime and electric field is directed away from the line.
Choose gaussian surface to a coaxial cylinder with the line
F
q
Now flux passing through the entire sphere =
ε0
+q
E
As the charge q is symmetrical to the upper half and lower
half hemispheres, so half-half flux will emit from both the
surfaces.
Gaussian Surface
3. Planar Symmetry: When due to symmetry, electric field
is same at all points at a distance ‘d’ from an infinite plane
and electric field is directed away from the plane.
Here, one can choose our gaussian surface to be a pill boxshaped cylinder of length ‘2d’.
F
Flux cutting
q
lower half surface =
20
Flux cutting
q
upper half surface =
20
Let us assume a charge Q is placed at
a distance a/2 above the center of a
horizontal square surface of edge a as a
shown in figure. Then the flux of the
electric field through the square surface
can be calculated as follow:
We can consider imaginary faces of
cube such that the charge lies at the
center of the cube. Due to symmetry we
can say that flux through the given area
(which is one face of cube).
Q
f=
6ε0
d
™
a/2
d
a
E
Q
APPLICATIONS OF GAUSS‘S LAW
Gauss law can be used to calculate the electric field E when there
is a high degree of symmetry due to which the direction of E is
known at every point and magnitude of E is constant over some
simple surface. In such situation, Gauss’s law provides the easiest
ways of calculating the electric fields
There are three kinds of symmetries where gauss’s law can be
used to find the electric field:
1. Spherical Symmetry: When due to symmetry, electric field
is same at a distance ‘r’ from a fixed point and is directed
away from that point. Choose gaussian surface to be a
spherical surface concentric with that point
14
SPHERICAL SYMMETRY
Finding E due to a Spherical Shell or Sphere
Electric field outside the Sphere: Since, electric field due
to a shell will be radially outwards. So lets choose a spherical
Gaussian surface Applying Gauss`s theorem for this spherical
Gaussian surface,
+
+
+
+
+
qin
∫ E.ds =φnet =∈0
q,R
+ +
+
+
+
+
+
r
+
q
=
∈0
JEE (XII) Module-1 PW
⇒ E(4pr2) =
⇒ Eout =
Electric field inside the solid sphere: For this choose a spherical
Gaussian surface inside the solid sphere Applying gauss’s theorem
for this surface
qin
ε0
q
4πεο r 2
+ Q,R +
+ ++
+
r
++
+ +
+ + +
+ +
+
+ +
+
+
+ +
+
+ +
+ ++
+ ++
+
+
+ +
++
+ + + ++ + +
+ + +
+
+
+ +
+
+
+
+ +
+ +
Electric field inside a sphere: Lets choose a spherical Gaussian
surface inside the shell.
Applying Gauss`s theorem for this surface
+
+
+
+
+
q,R
+ +
r +
+
4
× πr 3
4 3 3
πR
→ →
q
Qr 3
3
in
E . dS =
φ
=
=
= 3
net
∫
εο
εο
εο R
Q
+
+ + +
Qr 3
⇒ E(4pr2) =
qin
q
=0 =0
2 = fnet =
ε0
4πεο r
E =
⇒ E (4pr2) = 0
εο R3
Qr
4πεο R
E = (Qr)/(40
E
⇒ Ein = 0
kQ
⇒ Ein =
3
R
3
r=
ρr
3εo
E=
1
40
r3)
Q
R2
E = Q/(40r2)
E
0
r<R
r>R
r=R
r
Electric Field due to Uniformly
Charged Solid Sphere
r
R
CYLINDRICAL SYMMETRY
Electric Field due to Infinite Uniform Line Charge (l)
+
+
+
Electric field outside the solid sphere: Direction of electric
field is radially outwards, so we will choose a spherical Gaussian
surface Applying Gauss`s theorem
surface(1)
++
cylindrical
gaussian
surface
+
Q,R
+
++ +
+ + +
+ + +
+ +
r
+
surface(3)
+
+ surface(2)
+
→ →
∫ E . dS = fnet =
⇒ E(4pr2) =
Q
εο
Electric field due to infinite wire is radial so we will choose
cylindrical Gaussian surface as shown in figure.
qin
Q
=
εο
εο
⇒ Eout =
P Electric Charges and Fields
W
net
Q
4πεο r 2
1
=0
2
=0
3
=0
15
q
λl
φ3= in=
∈0 ∈0
R
⇒ E ∫ =
ds E ( 2πrl )
=
+
+
r
λ
εο
⇒ E =
λ
= 2k λ
2πεο r
r
Electric Field due to Infinitely Long Charged Tube
(having uniform surface charge density s)
R
+
+
+
+
+
+
+
+
+
+
+
r
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ S
+
+
+
+
Applying Gauss`s Theorem
q
ρ× πR 2 E × 2prl = in =
⇒
εο
εο
Eout =
ρR 2
2r ε 0
(ii) E at inside point: Lets choose a cylindrical gaussian
surface inside the solid cylinder.
Applying gauss`s theorem
r
(i) E outside the tube: Lets choose a cylindrical gaussian
surface
qin
σ2πR
=
εο
εο
fnet =
σ2πRl
⇒ Eout × 2πrl =
∈0
⇒ E =
σR
r ε0
(ii) E inside the tube: Lets choose a cylindrical gaussian
surface inside the tube.
fnet =
qin
=0
εο
qin
ρ× πr 2 E × 2prl =
=
εο
εο
Ein =
ρr
2ε0
E
So, Ein = 0
E
Einµr
Eout 1
r
r=R
Ein = 0
r
r=R
E due to Infinitely Long Solid Cylinder of Radius R
(having uniform volume charge density ρ)
(i) E at outside point: Lets choose a cylindrical gaussian
surface.
16
Eout 1
r
r
PLANAR SYMMETRY
Electric field Strength due to a Non-conducting Uniformly
Charged Sheet: To find the electric field strength at a point P
infront of the charged sheet we consider a cylindrical Gaussian
surface as shown in figure of face area S. If we apply Gauss law
for this surface, we have
JEE (XII) Module-1 PW
→
→
∫ E . dS =
qencl
or
∈0
→ →
→ →
+
E
.
dS
∫
∫ E . dS +
I
II
→ → σS
∫ E . dS =
∈0
III
[As here qencl = sS]
Field at O due to this infinitesimal element = dE
( dq )( R sin θ )
dE =
3/ 2
2
4πε0 ( R sin 2 θ + R 2 cos 2 θ )
coul/m
II
E
I
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
area S
E
III
II
As the lateral surface of cylinder is parallel to the direction of
electric field strength, no flux is coming out from the lateral
surface, hence we have
I
σS
or
∫ EdS =
∈0
III
(using the result for the filed at axis of a ring)
P
x
→ →
In this case ∫ E . dS = 0
∫ EdS +
Charge on the element = dq = s (area)
dq = s (2pR sin q) Rdq
2 ES =
σS
∈0
[As electric field is uniform on both sides]
σ
or E =
2 ∈0
∫ dE=
⇒ E=
σ
=
2
ε0
σ
=
4ε0
π/ 2
∫
π/ 2
∫
(
)
R sin θ 2πσR 2 cos θ
4πε0 R
0
2
dθ
π/ 2
sin θ=
cos θdθ
0
σ sin 2 θ
2ε0 2 0
Example 17: A system consists of a ball of radius R carrying
a spherically symmetric charge and the surrounding spaced
filled with a charge of volume density r = a/r where a is a
constant, r is the distance from the center of ball. Find the
ball’s charge at which the magnitude of the electric field is
independent of r outside the ball. How high is this strength?
Sol. Let us consider a spherical surface of radius r(r > R)
concentric with the ball and apply Gauss’s Law.
q
∫ E .dA = ε0 Let Q = Total charge on the ball
Variation of E with r Variation of E with s
E
E
(
)
r
ε0 E 4πr 2 = Q + ∫ ρ 4πx 2 dx
R
O
r O
Train Your Brain
Example 16: Find by direct integration the electric field
at the center of a hemispherical surface of charge having
uniform surface density s.
Sol.
Rd B
d
A
R
P Electric Charges and Fields
W
(
)
(
)
(
ε0 E 4πr 2 = Q + 2πa r 2 − R 2
 Q − 2πaR 2
=
⇒E 
 4πε0
)
R
O
 1 2πa
 2+
4πε0
r
r
For E to be independent of r,
Q = 2paR2
a
E=
2ε0
Example 18: Find out flux through the given Gaussian
surface.
Sol. f =
O
Consider a ring-shaped element on the surface of
the hemisphere at an angle q above the base as
shown.
AB = the radius of the ring = R sin q
Distance of center of ring B from O = R cos q
r
a
ε0 E 4πr 2 = Q + 4π∫ x 2 dx
x
R
=
Qin
ε0
2µC − 3µC + 4µC
ε0
−6
= 3 × 10 Nm2/C
ε0
q4=–6C
q1=2C
q2=–3C
q3=4C
q5=2C
q6=3C
Gaussian surface
17
Example 19: If a point charge q is placed at the center of
a cube then find out flux through any one surface of cube.
q
Sol. Flux through 6 surfaces =
ε0
Since all the surfaces are symmetrical
1 q
so, flux through one surfaces =
6 ∈0
Example 20: The electric field in a region is given by
3
4
=
E
E0 iˆ + E0 ˆj with E0 = 2.0 × 103 N/C. Find the flux
5
5
of this field through a rectangular surface of area 0.2 m2
parallel to the Y – Z plane.
3
4

Sol. φE = E ⋅ S =  E0 iˆ + E0 ˆj  ⋅ 0.2iˆ
5
5

(
= 240
+Q
F
q
F
F
x
Sol. We can divide the circular region into small rings.
E
dr
+Q
ds
r
X
Lets take a ring of radius r and width dr. Flux through this
small element dφ = E ds cos θ


r =R
kQ
x
∴ φnet = ∫ Edscosθ = ∫
( 2πrdr )  2 2 
2
2
r =0
x +r
 x +r 
(
)

 Q
x
[1 − cos β]
1 −
=

x 2 + R 2  2e0
Example 22: Consider a point charge q = 1mC placed at a
corner of a cube of sides 10 cm. Determine the electric flux
through each face of the cube.
Strategy: We will learn about the utility of symmetry in
solving problems with the help of Gauss’ law. Here we’ll
use the symmetry of the situation, which involves the faces
joining at the corner at which the charge resides.
18
(a) A charge q is placed at the comer of a cube
N − m2
C
R
Q
2ε 0
q +
)
Example 21: Find the electric flux due to a point charge
‘Q’ through the circular region of radius R if the charge is
placed on the axis of ring at a distance x.
=
E
(b) By surrounding the charge with a series of cubes such
that the charge is at the center of a larger cube, we have
created an arrangement sufficiently symmetric to be able to
solve for desired flux values.
You can see from figure (a) that for these faces E ⋅ n =
0.
Since the normal is perpendicular to the surfaces while the
electric field goes off in a spherically symmetric pattern
and lies in the sides. In other words, the electric field that
originates at the charge is tangential to the surface of these
three sides. This means there is no flux through these sides.
The electric flux through each of the remaining three faces
of the cube must be equal by symmetry. We’ll refer to these
faces with the label F.
To find the flux through each of the sides F, we can use
a technique that puts the single charge in the middle of a
larger cube. It takes seven other similarly placed cubes to
surrounds the points charge q completely in fig.(b). The
charge is at the dead center of the new larger cube. So the
flux through each of the six sides of the large cube will
now have an electric flux of one sixth of the total flux. The
large sides of the cube consist of four smaller square, one
of which is in fact one of the sides F. So given that the total
structure is completely symmetric, the flux through a side F
is one fourth of the flux through the large side
q
. So that the flux through each of
Sol. The total flux is
ε0
the sides of the large cube is
q
and one quarter of that
6ε0
q
goes through each of the far sides of the small cube.
24ε0
Numerical evaluation gives.
φ=
q
1× 10−3 C
=
= 5 × 106 Nm 2 /C
24ε0 24 8.85 × 10−12 C2 / Nm 2
(
)
JEE (XII) Module-1 PW
Concept Application
7. If an insulated non-conducting sphere of radius R has
charge density r. The electric field at a distance r from
the center of sphere (r < R) will be
ρR
ρr
(a)
(b)
3ε0
ε0
8. The given figure gives electric lines of force due to
two charges q1 and q2. What are the signs of the two
charges?
–q
+q
(–a, 0)
(a, 0)
r
Here we wish to find the electric field at point P having
coordinates (r, 0) (where r >> 2a). Due to positive charge
of dipole electric field at P is in outward direction and due
to negative charge it is in inward direction.
kq
Enet at P =
(r – a)
2
–
kq
(r + a)
2
=
4kqar
2
(r – a 2 )2
2kpr
(r 2 – a 2 )2
As r >> 2a
∴we can neglect a w.r.t. r
2kpr
Enet at P =
9. If electric field is uniform, then the electric lines of
forces are
(a) Divergent
(b) Convergent
(c) Circular
(d) Parallel
10. 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 center.
(a) Increases as r increases, for r ≤ R
(b) Decreases as r increases, for 0 < r < ∞.
(c) Decreases as r increases, for R < r < ∞.
(d) Is discontinuous at r = R
(r 2 – a 2 )2
As we can observe that for axial point direction of field is
in direction of dipole moment
2kp
∴ Vectorially, E =
r3
2. At an equatorial point: Again we consider the dipole
placed along the x-axis and we wish to find, electric field
at point P which is situated equatorially at a distance
r (where r >> 2a) from origin.
Vertical component of the electric field vectors cancel out
each other.
kq
Enet at P = 2E cos θ [where E = 2
]
r + a2


2kq
a
a
.
Enet at P = 2
∵cos θ =

2
2
2
2
2
r +a
r + a 
r + a 
ELECTRIC DIPOLE
A system of two equal and opposite charges separated by a small
distance is called electric dipole, shown in figure. Every dipole has
a characteristic property called dipole moment. It is defined as the
product of magnitude of either charge and the separation between
the charges. It is given as.
p = qd
d
p
kq
(r 2 + a 2 )
Enet
P
kq
2
(r + a 2 )
r
–q
+q
Dipole moment is a vector quantity and its direction is given from
negative pole to positive pole.
P Electric Charges and Fields
W
P
O
∴ Enet at P =
Both are negative
Both are positive
q1 is positive but q2 is negative
q1 is negative but q2 is positive
–q
kq
(r - a) 2
kq
(r + a) 2
As P = 2aq
q1 q2
(a)
(b)
(c)
(d)
1. At an axial point: Figure shows an electric dipole placed
on x-axis at origin
2a
3ρR
(d)
ε0
ρr
(c)
3ε0
ELECTRIC FIELD DUE TO A DIPOLE
(–a, 0)
O
q
(a, 0)
2kqa
kp
E=
=
( As p 2aq )
net =
2
2
2 3/2
(r + a 2 )3/ 2
(r + a )
19
As we have already stated that r > > 2a
∴ Enet at P =
DIPOLE IN UNIFORM ELECTRIC FIELD
kp
r3
We can observe that the direction of dipole moment and
electric field due to dipole at P are in opposite direction.
∴ Vectorially
→
→ –k p
E = 3
r
Figure shows a dipole of dipole moment p placed at an angle θ to
the direction of electric field. Here the charges of dipole experience
forces qE in opposite direction as shown.
E
+q
qE
ELECTRIC FIELD AT A GENERAL
POINT DUE TO A DIPOLE
d Figure shows a electric dipole placed on x-axis at origin and we
wish to find out the electric field at point P with coordinate (r, q)
Enet =
2
 2kp cos θ   kp sin θ 

 +

r3

  r3 
qE
–q
2
Thus we can state that when a dipole is placed in a uniform electric
field, net forces on the dipole is zero. But as equal and opposite
forces act with a separation in their line of action, they produce
a couple which tend to align the dipole along the direction of
electric field. The torque due to this couple can be given as
kp
1 + 3cos 2 θ
r3
kp sin θ
r3
And tan a =
2kp cos θ
r3
=
t = Force × separation between lines of action of forces
= qE × d sin q
= pE sin q
E
Eeq
P
s r = OP
O
P
P
sin
tan θ
2
tan θ 
α =tan 

 2 
–1 
Basic Torque Concept
τ= r×F
⇒ If the net translational force on the body is zero then the torque
of the forces may or may not be zero but net torque of the
forces about each point of universe is same.
⇒ If we have to prove that a body is in equilibrium then first
we will prove Fnet is equal to zero and after that we will
show tnet about any point is equal to zero.
⇒ If the body is free to rotate then it will rotate about the axis
passing through center of mass and parallel to torque vector
direction and if the body is hinged then it will rotate about
hinged axis.
20
or vectorially we can write the torque on dipole is
τ= p × E
Eaxial
o
Pc
tan a =
dsin
Angular SHM of Dipole: When a dipole is suspended in a
uniform electric field, it will align itself parallel to the field.
Now if it is given a small angular displacement q about its
equilibrium, the (restoring) couple will be
τ = – pE sin q
or τ = – pE q [as sin q ≈ q, for small q]
™
or I
d 2θ
dt 2
=
− pE θ
2
pE
or d θ =
−
θ
2
I
dt
2
or d θ = −ω2 θ
dt 2
pE
where ω2 =
I
This is standard equation of angular simple harmonic motion
2π 
with time-period T  =
 . So the dipole will execute
 ω
angular SHM with time-period
T = 2π
I
pE
JEE (XII) Module-1 PW
DIPOLE IN NON UNIFORM ELECTRIC FIELD
If the dipole is placed along E , (shown in figure)
E(x)
dx
q
qE(x)
+q
E(x)
(q) E(x + dx)
E(x + dx)
Then, net force on the dipole : Enet = qE(x + dx) – qE(x)
E ( x + dx) − E ( x)
(dx); here (q(dx) = P)
Enet = q
dx
 dE 
\ Fnet = P 

 dx 
Electric dipole placed at distance r from a point charge.
Q
p
dE
1
d −2
1 2 pQ
. pQ
r
.
=
= −
dr 4πε0
dr
4πε0 r 3
( )
Negative sign indicates that the force is in the direction of
decreasing r.
\ Force is attractive.
If the same dipole is placed at an angle q(q ≠ 0, 180°) then it will
experience force as well as torque.
Q
r
dF
2kP cos θQ PQ cos θ
F = Pcosq
=
−
=
dr
2πε0 r 3
r3
Train Your Brain
Example 23: What is the dipole moment of the system
shown in figure
–q
a
a
A
2q
a
–q
Sol. There are two dipoles of P = q (a )
so =
Pnet
=
3p
or
r
= 21/3
r1
Concept Application
Q
4πε0 r 2
\F= P
2
1
r3
=
or
=2
r 3 r13
r13
r
Here τ = p × E = pE sin 0°
E=
Example 24: An electron and a proton are placed at distance
of 1Å. What will be dipole moment of so formed dipole?
Sol.
p = qd = 1.6 × 10–19 × 1 × 10–10
= 1.6 × 10–29 coulomb metre
Example 25: The electric field due to a short dipole at a
distance r, on the axial line, from its mid point is the same as
that of electric field at a distance r1, on the equatorial line, from
r
its mid-point. Determine the ratio .
r1
1 2p
1 p
Sol.
=
3
4πε0 r
4πε0 r13
3qa
P Electric Charges and Fields
W
P
60º
P
11. The torque acting
on a dipole of moment P in an
electric field E is
(a) P · E
(b) P × E
(c) Zero
(d) E × P
12. If an electric dipole is kept in a uniform electric field,
Then it may experience
(a) A force
(b) A couple and move
(c) A couple and rotates
(d) A force and moves
13. An electric dipole consists of two opposite charges
each of magnitude 1 × 10–6 C separated by a distance
2 cm. The dipole is placed in an external field of
10 × 105N/C. The maximum torque on the dipole is
(a) 0.2 × 10–3 N-m
(b) 1.0 × 10–3 N-m
(c) 2 × 10–2 N-m
(d) 4 × 10–3 N-m
MOTION OF A CHARGED PARTICLE
IN A UNIFORM ELECTRIC FIELD
A charged body of mass ‘m’ and charge ‘q’ is initially at rest in
a uniform electric field of intensity E. The force acting on it is
given by F = Eq.
™ The direction of F is in the direction of field if ‘q’ is +ve and
opposite to the field if ‘q’ is –ve.
™ The body travels in a straight line path with uniform
F Eq
acceleration, =
a =
m m
21
At an instant of time t.
 Eq 
Its final velocity, v =u + at =
t  m 
1 2 1  Eq  2
Displacement s =+
ut
at =
t 2
2 m 
Momentum, P = mv = (Eq)t
Kinetic energy,
(⸪ u = 0)
(⸪ u = 0)
1 2 1  E 2q2  2
mv
=

t
2
2  m 
When a charged particle enters perpendicularly into a uniform
electric field of intensity E with a velocity u then it describes
parabolic path as shown in the figure.
K .E
=
™
2Eq
(b)
m
(a)
2E
(c)
m
u 0,=
a
Sol. (a) Given =
2Em
(d)
q
Eq
2m
qE
,=
s m
2
⸪ v=
u 2 + 2as
2qE 2qE \ v2 = 0 +
⇒v=
m
m
Example 27: An oil drop having a mass 4.8 × 10–10g
and charge of 30 ×10–18C is at rest between two charged
horizontal plates separated by a distance of 1cm . If now
polarity of the plates is reversed, instantaneous acceleration
of the drop is (g = 10 m/s2)
(a) 10 m/s2 (b) 15 m/s2 (c) 25 m/s2 (d) 20 m/s2
Sol. (d) Case I: When oil drop is in equilibrium
Negative plate
F
b
F = qE
q mg
Positive plate
mg = qE
Case II: when the polarity of the plates are reversed
Positive plate
Along the horizontal direction, acceleration is zero and hence
x = ut.
™
Along the vertical direction,
Force Eq
acceleration
= =
(here gravitational force is not
mass
m
considered)
F
F = qE
q mg
Negative plate
mg + qE = ma
2mg = ma [⸪ qE = mg]
2 × 10 = a
\ a = 20m/s2
Thus, vertical displacement,
y=
=
y
1  Eq  2

t
2 m 
Concept Application
2
 qE  2
1  qE   x 
=

x

 
2  m  u 
 2mu 2 
At any instant of time t, horizontal component of velocity,
vx = u
Vertical component of velocity is
™
™
 Eq 
v y= at= 
t
 m 
∴v = v =
v x2 + v y2 =
u2 +
E 2 q 2t 2
m2
Train Your Brain
Example 26: An electron having mass m and charge q
is accelerated from rest in a uniform electric field E. The
velocity acquired by it as it travels a distance l is
22
14. A proton of mass ‘m’ charge ‘e’ is released from rest
in a uniform electric field of strength ‘E’. The time
taken by it to travel a distance ‘d’ in the field is
(a)
2dE
me
(b)
2dm
Ee
(c)
2dE
me
(d)
2Ee
dm
15. A particle having mass m and charge q is thrown with
a speed u against a uniform electric field E. How much
distance will it travel before coming to rest?
(a) mu2/2qE (b) 3mu2/qE (c) mqE/2u2 (d) mu/qE
16. A charged particle of mass 3 × 10–3 Kg and charge
6 × 10–6 C enters in an electric field of 7 V/m. Then
its kinetic energy after 1m is
(a) 10–4 J
(b) 19 × 10–2 J
(c) 1.05 × 10–3 J
(d) 4.2 × 10–5 J
JEE (XII) Module-1 PW
Short Notes
(i) Two charges must be of like nature.
Electric Charge
Charge of a material body is that property due to which it
(ii) Third charge should be of unlike nature.
interacts with other charges. There are two kinds of chargesQ1
−Q1Q2
positive and negative. SI unit is coulomb. Charge is quantized
=
x =
r and q
Q1 + Q2
and additive.
Q1 + Q2
(
Coulomb’s Law
q1q2 1
Force between two charges F =
r , ∈r = dielectric
4π ∈0 ∈r r 2
constant
q1
q2
r
™
Coulomb’s Law is applicable only for static point charges.
Electric Dipole
™
™
Electric field at a general point due to a dipole.
F
E = unit is N/C or V/m.
q
–q
F
1 Q
E L=
im
r
=
q0 → 0 q
4π ∈0 r 2
0
Q1 r
Q1 ± Q2
™
x → distance of null point from Q1 charge
(+) for like charges
(–) for unlike charges
Direction tan =
α
Eθ 1
=
tan θ
Er 2
P Electric Charges and Fields
W
Electric field at an axial point (or End-on) E =
dipole
™
1 2p
of
4π ∈0 r 3
Electric field at an equatorial position (Broad-on) of dipole
1 (− p)
4π ∈0 r 3
Electric Lines of Force
Electric lines of electrostatic field have following properties.
(i) Imaginary
(iii) Can never be closed loops
X
q
r
x
(ii) Can never cross each other
Equilibrium of Three Point Charges
Q1
+q
1 p 1 + 3cos 2 θ
4π ∈0
r3
E=
;
Er
P
Electric field E =
Electric Field due to Point Charge Q
Q2
r
Electric Field or Electric Field Intensity
x=
E
E
The force due to one charge is not affected by the presence
of other charges.
If Q1 > Q2
Torque on dipole placed in uniform electric field τ= p × E
F = F 1 + F 2 + F 3 + ........
Q1
⇒ Null point near Q2
™
Force on a point charge due to many charges is given by
r
negative to positive charge.
y
Null Point for Two Charges
2
Electric dipole moment p = qd , where d is, distance from
Principle of Superposition
™
)
Q2
(iv) The number of lines originating or terminating on a charge
is proportional to the magnitude of charge.
23
+ A
(v) Lines of force ends or starts normally at the surface of a
conductor.
(vi) If there is no electric field there will be no lines of force.
(vii) Lines of force per unit area normal to the area at a
point represents magnitude of intensity, crowded lines
represent strong field while distant lines weak field.
(viii) Tangent to the line of force at a point in an electric field gives
the direction of Electric Field.
B–
qA > qB
Name/Type
Formula
Point charge
kq
E = 2 rˆ
|r |
Note
™
™
Infinitely long line charge
Infinite non-conducting thin sheet
q is source charge.
r is vector drawn from source charge
to the test point.
™
λ
2k λ
rˆ =
rˆ
2πε0 r
r
σ
nˆ
2ε 0
E
Outwards due to +charges and inwards
due to -charges.
™
l is linear charge density (assumed
uniform)
™
r is perpendicular distance of point
from line charge.
™
r is radial unit vector drawn from the
charge to test point.
™
s is surface charge density.
(assumed uniform)
™
Graph
n̂ is unit normal vector
r
E
r
E
/20
r
Uniformly charged ring
E=
(R
kQx
2
+x
Ecentre = 0
Infinitely large charged
conducting sheet
)
2 3/ 2
σ
n̂
ε0
™
Q is total charge of the ring
™
x = distance of point on the axis from
center of the ring.
E
Emax.
r
R
™
Electric field is always along the axis.
™
s is the surface charge. density
(assumed uniform)
E
n̂ is the unit vector perpendicular to the
surface.
/0
™
2
r
Uniformly charged hollow (i) for r ≥ R
conducting/non-conducting/
kQ
E = 2 rˆ
solid conducting sphere
|r |
(ii) For < R
E=0
™
™
™
™
™
24
R is radius of the sphere.
r is vector drawn from center of
sphere to the point.
Sphere acts like a point charge placed
at center for points outside the sphere.
E is always along radial direction.
E
KQ/R2
r
R
Q is total charge (= s4pR2).
(s = surface charge density)
JEE (XII) Module-1 PW
Uniformly charged solid non- (i) for r ≥ R
conducting sphere (insulating
kQ
E = 2 rˆ
material)
|r |
(ii) for r ≤ R
™
r is vector drawn from center of
sphere to the point
E
KQ/R2
Sphere acts like a point charge placed at
the center of points outside the sphere
kQr
E = 3 rˆ
R
™
™
E is always along radial direction
r
R
 4

Q is total charge  ρ ⋅ πR 3  .
 3

(r = volume charge density)
™
Inside the sphere E ∝ r .
™
Outside the sphere E ∝ 1/r .
Solved Examples
1. Four point charges are fixed at the corners of a square of side
length a. (in horizontal x – y plane). A positive charge q0 is
placed at a distance a from center of square perpendicular
to the plane of square. If point charge q0 is in equilibrium

3 
then its mass m is  a =
m .
2 

(a)
2qq0
9πε0 g
3
2
(b)
4qq0
9πε0 g
3
2
(c)
5qq0
9πε0 g
3
2
(d)
8 qq0
9 πε0 g
3
2
a
a2
a +
2
2
+
2kqq0
 2 a2
 a +
2

a

a2
 a 2 +
2

=
dF
 2 × 8  ( qq0 ) 3
=
m = 

9g
2
 4πε0  9 g
3
2
2. A thin straight rod of length l carrying a uniformly distributed
charge q is located in vacuum. Find the magnitude of the
electric force on a point charge ‘Q’ kept as shown in the
figure.
Q
a
P Electric Charges and Fields
W
a
k .dq.Q k .Q.q.dy
=
y2
y 2l
=
mg
\ F = ∑ dF
( 2kqq0 ) ( a ) 8
4qq0
9πε0 g
P
All forces are along the same direction,
2kqq a
03/2 ( 3) = mg
 3a 2 


 2 
m=
Electric force on ‘Q’ due to element
dy
y
Sol. Freal = mg
2 ( k )( 2q ) q0
 2 a2 
 a + 
2 

Sol. As the charge on the rod is not point charge, therefore, first
we have to find force on charge Q due to charge over a very
small part on the length of the rod. This part called element
of length, dy can be considered as point charge.
q
Charge on element, dq = ldy = dy
l
therefore F =
This sum can be calculated using integration,
a +l
∫
y =a
=
KQqdy
y 2l
a +l
KqQ  1 
=
= − 
l
 y a
KQ.q  1
KQq
1 
−
=


l  a a + l  a(a + l )
NOTE: (1) The total charge of the rod cannot be
considered to be placed at the center of the rod as we
do in mechanics for mass in many problems.
(2) If a >> l then, F =
KQq
a2
i.e., Behavior of the rod is just like a point charge.
3. Two charges q and 4q are placed at x = 0 and x = L along
x-axis. Find where another particle of charge q, is to be placed
on x-axis, so that net force on charge at x = 0 becomes zero.
Sol. The charge 4q repels the charge q towards x-axis. So the
charge q has to be placed on the left of charge q. Let it be
placed at x = –r.
25
In liquid this force is F' and is given as:
q2
F'=
where k is dielectric constant
4π ε 0 k a 2
Let F1 , be the force on q (at x = 0) due to charge at
x = –r, given as
 1  q2
F1 = 
 2 iˆ
 4πε0  r
Let F2 be the force on q (at x = 0) due to charge 4q at
x = L, given as
 1
F2 = − 
 4πε0
 q × 4q
 2 iˆ
 L
According to the condition given,
q 2 4q 2
Net force F1 + F2 ⇒
=2
r2
L
L
⇒ r=
2
−L
.
2
4. Two identical charged spheres are suspended by strings of
equal length. The strings makes an angle of 30° with each
other. When suspended in a liquid of density 800 kg/m3,
the angle remains the same. What is the dielectric constant
of the liquid? The density of the material of the sphere is
1600 kg/m3.
\ The charge q should be placed at r =
Sol. Let T, T' be the tensions in the strings when spheres are in
air and in liquid respectively.
T
F
mg
IN AIR
T sin q = F
T cos q = mg
⇒ F = mg tan q
1
⇒ F ' =F
k
From force diagrams:
F
mg
=
F ' mg − B
Vdg
d
1600
=
⇒k
=
=
⇒k
= 2
Vdg − V ρg d − ρ
1600 − 800
5. In which of the following figures, electric field at pint O is
non-zero?
+
++ ++
+
+
+
(a) ++
+
O
+
+
+
+
++ ++
Uniform charge/length
= 0 cos = Linear charge
density
(b)
O
Change per unit length varies with angular position
= 0 sin = Linear charge
density
(c)
O
Charge per unit length varies with angular position
–– – +++++
––
++
–
–
++
–
O
–
+
(d) +
–
++
––
++
––
++
+++ – –––
Uniform charge/length for different quadrants
Sol. Symmetrical configuration will lead to zero field.
6. A dipole with electric dipole moment p1 , another dipole
with dipole moment p2 are placed in a region as shown. If
| P1 |=| P 2 | then find out the electric field at the origin.
y
B
T
F
x
mg
IN LIQUID
T' sin q = F'
B + T' cos q = mg, (where, B is buoyant force)
⇒ F' = (mg – B) tan q
F = electrostatic repulsion in air
q2
2
4πε0 a
F=
26
60º
r
(a) –
(c)
3kP j
r3
– 2kP j
r3
60º
r
(b) – 3kP j
2r 3
3kP
(d) + 3 j
r
JEE (XII) Module-1 PW
Sol. (a) By resolving the dipole moments along x and y-axis,
we can get net electric field.
After resolving dipole moments along x and y axis,
y
P
3
2
8. A charge Q is uniformly distributed over a hollow
hemispherical shell. Let E is the electric field at point
A whose co-ordinates are (–x0, 0, 0). Find out the electric
field at point B whose co-ordinates are (x0, 0, 0).
(Note, x0 < R, where, R is the radius of hollow shell)
y
3
2
P
x
O
P
2
A
P
2
Electric field at center due to horizontal components of
both dipoles cancels out.
Hence, net electric field at center will be:
P 3
P
E=
−k 23 × 2iˆ =
− 3k 3 ˆj
r
r
7. A long uniformly charged thread having linear charge density
l is placed along y-axis. At a distance ‘r’ from it, there a
short dipole of dipole moment P . Find the force exerted
on electric dipole if it is along the thread.
y
''
r
x
P
(a)
λP
2πε 0 r 2
(b) zero
(c)
2λP
2πε 0 r 2
(d)
3λP
2πε 0 r 2
+
Eq
z
(a) − E
kQ
(c) + E − 2 kˆ
x0
r
Eq
Sol. (a) Electric field inside a conducting sphere is zero.
If we complete the sphere, then E.F. at point
A⇒ E=0
\ at point B ⇒ E =
+E
9. A proton is revolving with speed v = x × 106 ms–1 in a
circular orbit of radius of 2 cm, in a region between two
hollow large cylindrical shells having uniform charge
densities, –2 × 10 –4 cm –1 on the inner cylinder and
2 × 10–4 cm–1 on the outer cylinder.
The radius of inner shell is 1 cm and outer shell is 3 cm.
Find x.
2k λ
E=
r
\ force on proton
2k λ
r
For circular motion,
F= q ×
v=
2 λq
4πε0 m
v=
2 × 9 × 109 × 2 × 10−4 × 1.6 × 10−19
m/s
1.67 × 10−27
–
E=
λ
2πε0 r
Net force on dipole = Eq – Eq = 0
P Electric Charges and Fields
W
kQ
(b) − E + 2 kˆ
x0
(d) + E
2k λq mv 2
=
r
r
–q
+q
x
Sol. [18.5]
Force due to linear charge will be used as centripetal
force on the proton.
E.F. between the cylinders,
Sol. (b) The dipole is kept inside the electric field produced by
the line charge
B
o
v = 18.5 × 106 m/s
27
10. Two free ends of a uniformly charged wire makes
60° and 48° at a point on x-axis. If wire is along y-axis, find
out the angle (in degrees) of electric field strength vector at
that point on the x-axis
Sol. [6.00]
Find out the electric field components and determine
the angle from x-axis.
Sol. (c) Electric field will be radially outwards, so Gaussian
surface must be cylindrical.
qin
∫ E.dA = ∈0
At A:
E ( 2πr1=
)
E
⇒=
60°
48°
At B:
ρr1
ρR
=
2 ∈0 4 ∈0
E ( 2πr2=
)
to wire)
E⊥ (perpendicular=
Kλ
(sin 60° + sin 48°)
R
Kλ
(cos 60° − cos 48°)
R
E cos 60° − cos 48°
tan=
θ
=
E⊥ sin 60° + sin 48°
E|| (parallel=
to wire)
q from x-axis
E cos 60° − cos 48°
tan=
θ
=
E⊥ sin 60° + sin 48°
−2sin 54° sin 6°
=
2sin 54° cos 6°
tan q = tan 6°
6° from x-axis
11. In the given figure, we have
Co-axial solid and hollow
cylinders of charge density
s (per unit area) and r (per
unit volume) respectively.
Radius of outer cylinder
is 2R. Use Gauss law to
find out the electric field
(Column-II) at points given
in Column-I.
B.
A
(a)
(b)
(c)
(d)
28
C
ρR 2
ρR
=
2 ∈0 r2 3 ∈0
At C:
1
(ρπR 2 + σ2π2 R)
E (2πr3=
)
∈0
=
E
=
E
ρR 2
2σR
+
2 ∈0 r3 ∈0 r3
=
E
5ρR 2
ρR
=
2 ∈0 r3 ∈0
Electric field strength at point B,
q.
at a distance 3R/2 from axis
Electric field strength at any point
s.
on the axis
12. In a uniform sphere of charge (charge density r), a small
cavity is created. The center of cavity is at a distance a from
the center of sphere. Taking center of sphere as origin and
as a unit vector along the line joining the center of cavity
and origin, show that the electric field at any point inside
ρa
aˆ .
the cavity is
3ε0
Strategy: Consider any point P inside cavity. The position
vector of P is r . The field at P can be assumed to be a
superposition of field due to a uniform sphere of charge
E1 and field due to a negative charged sphere of the size
of cavity E2 .
( )
R
Electric field strength at point A,
p.
at a distance R/2 from axis
A-(p); B-(p); C-(p); D-(p)
A-(p); B-(q); C-(p); D-(q)
A-(q); B-(p); C-(s); D-(r)
A-(p); B-(q); C-(p); D-(p)
B
R
Electric field strength at point C, at
C.
r
a distance 5R/2 from axis
D.
1
(ρπR 2 )
∈0
Electric field strength at any point on the axis = 0
Column-I
A.
1
(ρπr12 )
∈0
Column-II
( )
ρR
3 ∈0
P
r1
C
a = aa^
ρR
4 ∈0
r
O
0
ρR
∈0
r
ρr ρr1
=
,E
, (where r1 = position vector of P w.r.t. C)
3ε0
3ε0
⇒ Net field, E =
E1 + E2
=
Sol.
E1
JEE (XII) Module-1 PW
⇒
ρ ρa
ρa
aˆ
=
( r − r1 ) =
3ε0
3ε0 3ε0
NOTE: Since the field at point P does not depend on the
position of point P in the cavity, we conclude that the field
inside cavity is uniform.
13. Determine and draw the graph of electric field due to
infinitely large non-conducting sheet of thickness t and
uniform volume charge density r as a function of distance
x from its symmetry place.
t
t
(b) x ≥
(a) x ≤
2
2
t

Sol. We can consider two sheets of thickness  − x  and
2


t

 + x
2

t
t
2
+x
–x
2
Where the point P lies inside the sheet.
Now, net electric field at point P:
Q1
Q2
[Q charge of left sheet; Q2 :
−
2 Aε0 2 Aε0 1
charge of right sheet.]
t

t

Aρ  + x  − ρA  − x 
2

2
 ρx
=
2 Aε0
ε0
For point which lies outside the sheet we can consider a
complete sheet of thickness t
x
σtA
ρt
dx
=
E =
2 Aε0 2ε0
Alternate: We can assume thick sheet to be
made of large number of uniformly charged
thin sheets.
Consider an elementary thin sheet of width
dx at a distance x from symmetry plane.
Charge in sheet = r Adx ( A : assumed area
of sheet)
ρAdx
Surface charge density, σ =
A
So, electric field intensity due to elementary
ρdx
:
sheet dE =
2ε0
E
r
tan = p/E0
P Electric Charges and Fields
W
t
ρdx t /2 ρdx ρx
⇒ ENet = ∫ −x t /2
− ∫x
=
2
2ε 0
2ε 0 ε 0
(b) When x >
t
ρdx ρt
⇒ ENet =
∫ t−/2t /2
=
2
2ε 0 2ε 0
14. Find force on short dipole P2 due to short dipole P1 if they
are placed at a distance r a part as shown in figure.
Sol. Force on P2 due to P1
E1=
P1
r
2kP 1
r3
P2
 dE 
F2 = ( P2 )  1 
 dr 
 d  2kP  
6kP1 P2
\ F2 = ( P2 )   3 1   or F2 = −
dr
r
r4

 
Here – sign indicates that this force will be attractive
(opposite to r)
E2 E1
P
E = E1 − E2 =
(a) When x <
15. A circular wire of radius R carries a total charge Q distributed
uniformly over its circumference. A small length of the wire
subtending angle q at the center is cut off. Find the electric
field at the center due to the remaining portion.
Q
sin ( θ )
(a)
2
4π ε 0 R 2
(b)
Q
θ
sin  
2
4π ε 0 R
2
(c)
Q
θ
sin  
4π2 ε 0 R 2
4
(d)
Q
θ
sin  
4π2 ε 0 R 2
8
2
Sol. Electric field due to an arc at its centre
is
kλ
θ
2sin   ,
R
2
Where k =
q
1
,
4πε 0
q = angle subtended by the wire at the center,
and l = Linear density of charge.
Let E be the electric field due to remaining portion.
Since intensity at the center due to the circular wire is zero.
Applying principle of superposition.
kλ
θ
2sin   nˆ + E =
0
R
2
1
Q
θ
E =
.
.2sin  
4πε0 R 2πR
2
Q
θ
sin  
= 2
2
4
R
2
π ε0
29
Exercise-1 (Topicwise)
COULOMB FORCE
1. When 1014 electrons are removed from a neutral metal
sphere, the charge on the sphere becomes
(a) 16 mC (b) –16 mC (c) 32 mC (d) –32 mC
2. Number of electrons in one coulomb of charge will be
(1e = 1.6 × 10–9 C)
(b) 6.25 × 1018
(a) 5.46 × 1029
19
(c) 1.6 × 10
(d) 9 × 1011
3. The ratio of the forces between two small spheres with
constant charge (a) in air (b) in a medium of dielectric
constant K is
(a) 1 : K
(b) K : 1
(c) 1 : K2
(d) K2 : 1
4. Four charges are arranged at the corners of a square ABCD,
as shown in the adjoining figure. The force on the charge
kept at the center O is
A
+q
B
+2q
(
+q
C
5. A total charge Q is broken in two parts Q1 and Q2 and they
are placed at a distance R from each other. The maximum
force of repulsion between them will occur, when
(b) Q2=
Q
2Q
, Q1= Q −
4
3
Q
3Q
Q
Q
(d)
=
, Q1
Q1 =
=
, Q2
4
4
2
2
6. Two charged spheres separated at a distance R exert a force
F on each other. If they are immersed in a liquid of dielectric
constant 5, then what is the new force between them?
(c)=
Q2
F
F
(b) F
(c) 5F
(d)
2
5
7. A charge q is placed at the center of the line joining two
equal charges Q. The system of the three charges will be in
equilibrium, if q is equal to
(a)
(a) −
30
Q
2
(b) −
Q
4
(c) +
Q
4
(c)
18 F
25
(d)
(
25
F
18
)
(b) 10N
(c) 104 dyne (d) 100 dyne
12. A charge particle q 1 is at position (2, – 1, 3). The
electrostatic force on another charged particle q 2 at
(0, 0, 0) is
(a) q1 q2 (2 iˆ − ˆj + 3 kˆ)
56 π ∈0
Zero
Along the diagonal AC
Along the diagonal BD
Perpendicular to side AB
Q
Q
, Q1= Q −
R
R
(b) 25F
)
(a) 50N
–2q
D
(a) Q2=
(a) 18F
11. Two charges each of 1µC are at P 2i + 3 j + kˆ m and
Q i + j − kˆ . Then the force acting on any one of them is
O
(a)
(b)
(c)
(d)
8. Two charged particles having charge 2 × 10–8 C each are
joined by an insulating string of length 1m and the system
is kept on a smooth horizontal table, what is the tension in
the string?
(a) 3.6 × 10–6N
(c) 3.4 × 10–6N
–7
(b) 4 × 10 N
(d) 3 × 10–4N
9. Point charges +4q, –q and +4q are kept on the X-axis at point
x = 0, x = a and x = 2a respectively.
(a) Only –q is in stable equilibrium
(b) All the charges are in stable equilibrium
(c) All of the charges are in unstable equilibrium
(d) None of the charges is in equilibrium
10. Charges on two spheres are +10μC and –5μC respectively.
They experience a force F. If each of them is given an
additional charge +2μC then new force between them if kept
at the same distance is
(d) +
Q
2
q1 q2
1
( − 2 iˆ + ˆj − 3 kˆ)
4π∈0 14 3
q1 q2 ˆ
(c)
( j − 2 iˆ − 3 kˆ)
56 π ∈0
q1 q2
(d)
(2 iˆ − ˆj + 3 kˆ)
56 14 π ∈0
(b)
( )
13. Two point sized metal spheres of same mass are suspended
from a common point by two light insulating strings. The length
of each string is same. The sphere are given electric charges +q
on one of them and +4q on the other. Which of the following
diagrams best shows the resulting positions of sphere?
(a)
(b)
(c)
(d)
+5q
JEE (XII) Module-1 PW
ELECTRIC FIELD AND FIELD LINES
q
are situated at the origin and
2
at the point (a, 0, 0) respectively. The point along the x-axis
where the electric field vanishes is
20. The figure shows some of the electric field lines
corresponding to an electric field. The figure suggests
14. Two point charges –q and
(a) x =
a
(c)
C
(b) x = 2 a
2
2a
2a
(d) x =
2 +1
2 −1
15. The magnitude of force, exerted by a uniform electric field
on an electron having mass me and a proton of mass mp are
represented as Fe and Fp respectively, are related as
F
m
(a) Fp = Fe
(b) e = e
Fp m p
(c) x =
B
A
Fe m p
=
Fp mc
(d)
(a)
(b)
(c)
(d)
EA > EB > EC
EA = EB = EC
EA = EC > EB
EA = EC < EB
21. Four charges are placed on corners of a square as shown in
figure having side of 5 cm. If Q is one microcoulomb, then
electric field intensity at center will be
Q
–2Q
Fe me2
=
Fp m 2p
16. ABC is an equilateral triangle. Charges +q are placed at each
corner. The electric intensity at O will be
+q
A
r
r
O
r
+q
+q
B
(a)
C
1 q
1 q
(b)
(c) Zero
4πε0 r
4πε0 r 2
(d)
1 3q
4πε0 r 2
(a)
(b)
(c)
(d)
–Q
+2Q
107
1.02 ×
N/C upwards
2.04 × 107 N/C downwards
2.04 × 107 N/C upwards
1.02 × 107 N/C downwards
22. The given figure gives electric lines of force due to two
charges q1 and q2. What are the signs of the two charges?
17. The magnitude of electric field intensity E is such that, an
electron placed in it would experience an electrical force
equal to its weight is given by
e
mg
e2
(a) mge
(b)
(c)
(d)
g
mg
e
m2
18. The distance between the two charges 25 mC and 36 mC
is 11 cm. At what point on the line joining the two, the
intensity will be zero
(a) At a distance of 5 cm from 25 mC
(b) At a distance of 5 cm from 36 mC
(c) At a distance of 10 cm from 25 mC
(d) At a distance of 11 cm from 36 mC
19. The electric field intensity required to keep a water drop of
radius 10–5 cm just suspended in air when charged with one
electron is approximately
(g = 10 Newton/kg, e = 1.6 × 10–19 coulomb)
(a) 260 volt/cm
(b) 260 Newton/coulomb
(c) 130 volt/cm
(d) 130 Newton/coulomb
P Electric Charges and Fields
W
(a)
(b)
(c)
(d)
q1 is positive but q2 is negative
q1 is negative but q2 is positive
Both are negative
Both are positive
23. Two point charges of 30 μC and 40 μC are placed 0.2 m
apart from each other. Where will be the electric field zero
on the line joining the charges from 30 μC charge?
(a)
(b)
20 3
2 +1
20 3
3+2
cm
cm
(c) 20 3 cm
5
(d) 10 cm
31
24. In which of the following cases the electric field at the center
is not zero?
30. Electric field at a point varies as r0 for
(a) An electric dipole
(b)
(a)
GAUSS’S LAW AND APPLICATION
(b) A point charge
(c) A plane infinite sheet of charge
(d) A line charge of infinite length
(c)
(d)
25. Three charges –q, +q and –q are placed at the corners of an
equilateral triangle of side ‘a’. The resultant electric force
on a charge +q placed at the centroid O of the triangle is
(a)
3q 2
4πε0 a 2
(b)
(c)
q2
2πε0 a 2
(d)
31. Electric charge is uniformly distributed along a long
straight wire of radius 1 mm. The charge per cm length of
the wire is Q coulomb. Another cylindrical surface of radius
50 cm and length 1 m symmetrically encloses the wire as
shown in the figure. The total electric flux passing through
the cylindrical surface is
5q 2
+
+
+
4πε0 a 2
3q 2
2πε0 a 2
26. Two point charges Q and –3Q are placed some distance apart.

If the electric field at the location of Q is 3E , the electric
field at the location of –3Q is



E
E
(a) 3E
(b) 2 E
(c) +
(d) −
3
3
27. A thin conducting ring of radius R is given a charge +Q. The
electric field at the center O of the ring due to the charge on
the part AKB of the ring is E. The electric field at the center
due to the charge on the part ACDB of the ring is
(a) 3 E along OK
(b) 3 E along KO
(c) E along OK
(d) E along KO
28. A charge of 4 ×10–9C is distributed uniformly over the
circumference of a conducting ring of radius 0.3 m. Calculate
the field intensity at a point on the axis of the ring at 0.4m
from its center and also at the center?
(a) 112N/C, 2N/C
(b) 118N/C, 6N/C
(c) 115.2N/C, Zero
(d) 113.2N/C, Zero
29. A circular wire of radius R carries a total charge Q distributed
uniformly over its circumference. A small length of the wire
subtending angle θ at the center is cut off. Find the electric
field at the center due to the remaining portion.
3Q
(a)
sin(θ)
2
4π ε 0 R 2
(c)
32
Q
θ
sin  
2
4π ε 0 R
4
2
Q
θ
(b)
sin  
2
2
4π ε 0 R
2
(d)
Q
θ
sin  
2
4π ε 0 R
8
2
1m
+
+
+
50cm
(a)
Q
ε0
(b)
100Q
ε0
(c)
10Q
(πε0 )
(d)
100Q
(πε0 )
32. q1, q2, q3 and q4 are point charges located at points as shown
in the figure and S is a spherical Gaussian surface of radius
R. Which of the following is true according to the Gauss’s
law?
S
q1
R
q4
q2
(a)
q3
q1 + q2 + q3
∫ ( E1 + E2 + E3 + E4 ).dA =
2ε 0





S
(b)
(q1 + q2 + q3 )
∫ ( E1 + E2 + E3 + E4 ).dA =
ε0





S
(c)
(q1 + q2 + q3 + q4 )
∫ ( E1 + E2 + E3 + E4 ).dA =
ε0





S
(d) None of these
JEE (XII) Module-1 PW
33. If the electric flux entering and leaving an enclosed surface
respectively is f1 and f2 the electric charge inside the surface
will be
(a) (f1 + f2) e0
(b) (f2 – f1) e0
(c) (f1 + f2) / e0
(d) (f2 – f1) / e0
34. Consider the charge configuration and
spherical Gaussian surface as shown in
the figure. When calculating the flux
of the electric field over the spherical
surface the electric field will be due to
(a) q2
(b) Only the positive charges
(c) All the charges
(d) +q1 and –q1
+q1
–q1
41. A point charge +q is placed at one face of a cube of side l. The
(b)
σ
ε0
(c)
σ
2ε0
(d)
Depends upon the location of the point
37. Which of the following graphs shows the variation of
electric field E due to a hollow spherical conductor of
radius R as a function of distance from the center of the
sphere
E
R
r
E
r
R
44. An electric dipole, kept at 60o with respect to a uniform
electric field of 10 7 NC –1 experience a torque of
12 × 10–20 N–m. What is the dipole moment of the dipole?
(a) 1.4 × 10–26 C–m
(b) 8 × 10–27 C–m
–26
(c) 30 × 10 C–m
(d) 1.3 × 10–27 C–m
45. An electric dipole is kept along the x-axis at the origin O. A
point P is at a distance of 20 cm from this origin such that
OP makes an angle π/3 with the x-axis. If the electric field
at point P makes an angle θ with the x-axis then the value
of θ is
(a) π/5
(b) 2π/3
r
38. A conducting sphere of radius R = 20 cm is given a charge

Q = 16 mC. What is E at center
(a) 3.6 × 106 N/C
(b) 1.8 × 106 N/C
(c) Zero
(d) 0.9 × 106 N/C
P Electric Charges and Fields
W
43. The electric field due to a dipole at a distance r on its axis
is
(a) Directly proportional to r3
(b) Inversely proportional to r3
(c) Directly proportional to r2
(d) Inversely proportional to r2
(a) 2qa along the line joining points
(x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
(b) qa along the line joining points (x = 0, y = 0, z = 0) and
(x = a, y = a, z = 0)
(d)
R
ELECTRIC DIPOLE
x
E
(c)
q
q
6ql 2
(b)
(c) Zero
(d)
2
2ε0
6l ε0
ε0
42. Two parallel line charges +λ and –λ are placed at a separation
of R in free space. The net electric field exactly mid way
between the two line charges is
2λ
5λ
1
(a) Zero
(b)
(c)
(d)
πεo R
πεo R
2πεo R
 3
 3
π
+ tan −1 
(c) tan −1 
(d)


3
 2 
 2 
46. Three point charges +q, –2q and +q are placed at points
(x = 0, y = a, z = 0), (2x = 0, y = 0, z = 0) and (x = a, y = 0,
z = 0) respectively. The magnitude and direction of the
electric dipole moment vector of this charge assembly are
(b)
R
electric flux emerging from the cube is
(a)
36. Two infinite plane parallel conducting sheets separated by a
distance d have equal and opposite uniform charge densities
s. Electric field at a point between the sheets is
(a) Zero
(a)
40. Electric flux emanating through a surface element
ds = 5 iˆ placed in an electric field E = 4iˆ + 4 ˆj + 4kˆ is
(a) 10 units
(b) 20 units
(c) 4 units
(d) 16 units
q2
35. An electric dipole is put in north-south direction in a sphere
filled with pure water. Which statement is correct
(a) Electric flux is coming towards sphere
(b) Electric flux is coming out of sphere
(c) Electric flux entering into sphere and leaving the sphere
are same
(d) Water does not permit electric flux to enter into sphere
E
39. At a point 20 cm from the center of a uniformly charged
dielectric sphere of radius 10 cm, the electric field is
100 V/m. The electric field at 3 cm from the center of the
sphere will be
(a) 150 V/m
(b) 125 V/m
(c) 120 V/m
(d) Zero
(c)
2qa along +ve x direction
(d)
2qa along +ve y direction
33
47. An electric dipole is placed at an angle of 45o with an electric
field of intensity 107 NC–1. It experiences a torque equal to
15 × 102 N–m. What will be magnitude of charge on dipole
if it is 5cm long?
(a) 1 mC
(b) 2 mC
(c) 5 mC
(d) 4 mC
interval t?
(a)
MOTION OF CHARGED PARTICLE IN
UNIFORM ELECTRIC FIELD
48. There is an electric field E in x-direction. If the work done
in moving a charge of 0.2 C through a distance of 2m along a
line making an angle of 600 with x-axis is 4J, the value of E is
(a) 2 3 N / C
(c) 4 N/C
50. A particle having charge q and mass m is projected with
velocity v= 2iˆ − 3 ˆj in a uniform electric field E = E0 ˆj.
What is the change in momentum | ∆P | during any time
(b) 5 N/C
(d) 20 N/C
(c)
qE0 t
3m
(d) Zero
51. The bob of a simple pendulum is hanging vertically down
from a fixed identical bob by means of a string of length ‘l’.
If both bobs are charged with a charge ‘q’ each, time period
of the pendulum is (ignore the radii of the bobs)
(a) 2π
49. A charged particle having mass m and charge q is released
from rest in a uniform electric field E. The kinetic energy
of the charged particle moving on a horizontal plane after
‘t’ second is
E 2 q 2t 2
2 E 2t 2
Eq 2 m
Eqm
(a)
(b)
(c)
(d)
2m
3mq
t
2t 2
(b) qE0t
13 m
(c) 2π
l
 Kq 2 
g + 2 
l m
(b) 2π
l
g
(d) 2π
l
 Kq 2 
g − 2 
l m
l
 Kq 2 
g − 2 
 lm 
Exercise-2 (Learning Plus)
1. Two small balls having equal positive charge Q (Coulomb)
on each are suspended by two insulating strings of equal
length ‘L’ metre, from a hook fixed to a stand. The whole set
up is taken in a satellite in to space where there is no gravity
(state of weight lessness). Then the angle (q) between the
two strings is
(a) 0º
(b) 90º
(c) 180º
(d) 0º < q < 180º
2. Two charges 4q and q are placed 30 cm apart. At what point
the value of electric field will be zero
(a) 10 cm away from q and between the charge
(b) 20 cm away from q and between the charge
(c) 10 cm away from q and out side the line joining the charge
(d) 10 cm away from 4q and out side the line joining them
4. The maximum electric field intensity on the axis of a
uniformly charged ring of charge q and radius R will be
1
1
q
2q
(a)
(b)
2
2
4πε
4πε0 3 3R
0 3R
(c)
→
Ey
Y
34
O
d
+
90°
P
Ex
X
+
+
+
+ 30°
(b)
L
+
x
x
O
L
+
E
E
(c)
1
3q
4πε0 2 3R 2
E
(a)
O
(d)
5. The direction (θ) of E at point P due to uniformly charged
finite rod will be
3. Two identical point charges are placed at a separation of l. P is
a point on the line joining the charges, at a distance x from
any one charge. The field at P is E. E is plotted against x for
values of x from close to zero to slightly less than l. Which
of the following best represents the resulting curve?
E
1
2q
4πε0 3 3R 2
L x
(d)
O
L x
(a)
(b)
(c)
(d)
At angle 30° from x-axis
45° from x-axis
60° from x-axis
None of these
JEE (XII) Module-1 PW
6. Two point charges a and b, whose magnitudes are same are
positioned at a certain distance from each other with a at
origin. Graph is drawn between electric field strength at
points between a and b and distance x from a. E is taken
positive if it is along the line joining from a to b. From the
graph, it can be decided that
E
x
b
a is positive, b is negative
a and b both are positive
a and b both are negative
a is negative, b is positive
+
8. A cone of radius (R) and length (L) is placed in a uniform
electrical field (E) parallel to the axis of the cone. The total
flux for the surface of the cone is given by
(a) 2pR2E
(b) pR2E
(c)
2πR 2
E
(d) zero
9. An infinite, uniformly charged sheet with surface charge
density s cuts through a spherical Gaussian surface of radius
R at a distance x from its center, as shown in the figure. The
electric flux f through the Gaussian surface is
(a)
−λ i
2πε0 R
(b)
(c)
2λ i
4πε0 R
(d) None of these
+
+
+
R
2
2
(b) 2π ( R − x )σ
ε0
π ( R − x) 2 σ
(c)
ε0
2
2
(d) π ( R − x )σ
ε0
10. Figure shows a charge Q placed at the center of open face of
a cylinder as shown in figure. A second charge q is placed at
one of the positions A, B, C and D, out of which positions
A and D are lying on a straight line parallel to open face
of cylinder. In which position(s) of this second charge,
the flux of the electric field through the cylinder remains
unchanged?
C
B
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+
+
X+
πR 2σ
ε0
λ j
2πε0 R
13. Find the force experienced by the semicircular rod charged
with a charge q, placed as shown in figure. Radius of the
wire is R and the infinite line of charge with linear charge
density l is passing through its center and perpendicular to
the plane of rod.
+ (a)
–
+
+
+
7. Electric flux through a surface of area 100 m2 lying in the
xy plane is (in V-m) if E =
iˆ + 2 ˆj + 3 kˆ
(a) 100
(b) 141.4
(c) 173.2
(d) 200
X
R
+
+
+
+
+
a
12. The charge per unit length of the four quadrant of the rings
is 2l, –2l, l and –l respectively. The electric field at the
center is
Y
+++
–2
+ 2
++
(a)
(b)
(c)
(d)
11. Two similar conducting spherical shells having charges
40 mC and –20mC are some distance apart. Now they are
touched and kept at same distance. The ratio of the initial
to the final force between them is
(a) 8 : 1
(b) 4 : 1
(c) 1 : 8
(d) 1 : 1
(a)
−λq
2π2 ε 0 R
(b)
λq
π ε0 R
(c)
λq
4π2 ε 0 R
(d)
λq
4πε0 R
2
14. In a certain region electric dipole having dipole moment
p = 2iˆ is placed at a point having coordinates (x, y) where
external electric field
=
E 3 x 2 y 2 iˆ + 3 x 3 yjˆ C m is also present.
Find magnitude of force experienced by the dipole
(a) xy 36 x 2 + 81 y 2
(b) 2 xy 36 x 2 + 81 y 2
D
Q
(a) A and D (b) B
P Electric Charges and Fields
W
(c) C
A
(c) xy 36 y 2 + 81x 2
(d) B and C
(d) 2 xy 36 y 2 + 81x 2
35
15. Two identical hemispherical bowls are placed inside an
external uniform electric field E as shown in the figure. Find
the ratio of electric flux passing through both the bowls.
20. If two balls of given masses and charges are released, which
of the following is incorrect arrangement in equilibrium?
E
(a)
(b)
+q
2m
R
+2q
+q
+q
m
(i)
(ii)
(a)
φ1
=2
φ2
(b)
φ1
=3
φ2
(c)
φ1
=4
φ2
(d)
φ1
=5
φ2
m
16. We have a spherical shell of radius R = 8 cm having
Q = –16 e charge is distributed over its surface. A point
charge q = +2e is placed at its center. The electric field at
point P at a distance r = 10 cm, is x × 10–6 NC–1. Find x.
(a) 5.0
(b) 4.0
(c) 3.0
(d) 2.0
> +++++
(i) Electric flux through the plane faces of cylinder is zero.
(ii) Electric flux through the curved surface of cylinder is
constant.
(iii) Magnitude of electric field remains constant over the
plane surface of the cylinder.
(a) (i) and (ii) are correct
(b) (i) and (iii) are correct
(c) (ii) and (iii) are correct
(d) (ii) and (iii) are correct
+2q
m
+q
2m
21. Two identical small balls each have a mass m and charge q.
When placed in a hemispherical bowl of radius R with
frictionless, nonconductive walls, the beads move, and at
equilibrium the line joining the balls is horizontal and the
distance between them is R (figure). Neglect any induced
charge on the hemispherical bowl. Then the charge on each
1
bead is: (here K =
)
4π ∈0
R
R
m
m
R
1/ 2
 mg 
(a) q = R 

K 3
1/ 2
 3 mg 
(c) q = R 

 K 
1/ 2
 mg 
(b) q =  R

 K 3
1/ 2

3 mg 
(d) q =  R

K 

22. A simple pendulum has a length l, mass of bob m. The bob
is given a charge q coulomb. The pendulum is suspended
in a uniform horizontal electric field of strength E as shown
in figure, then calculate the time period of oscillation when
the bob is slightly displace from its mean position is
E
18. The surface charge density on Earth’s surface is found to be
4.42 × 10–xcm–2. When it is noted that electric field near the
Earth’s surface is 500 V m–1, find x.
(a) 9
(b) 10
(c) 11
(d) 13
19. Select the correct statement
(a) Electric lines of forces never form a closed loop; also,
magnetic lines never form a closed loop
(b) Electric lines of forces never form a closed loop; while
magnetic lines always form closed loop or extend to
infinity
(c) Electric lines of force also exists inside a conductor
(d) Gauss’s law says that the total flux of an electric field
through an open surface is the net charge enclosed by
that surface
m
17. Choose the correct option for the given case in which a
cylindrical gaussian surface has its axis along an infinite
long linear charge, having linear charge density l.
+ + + + + + + + + + + + + + + ++ + + + + ++
(d)
+3q
+q
m
m
R
(c)
36
q1m
(a) 2π
g


 
(c) 2π 

 g − qE 
m 



 
(b) 2π 

 g + qE 
m 

(d)
2π
 qE 
g2 + 

 m 
2
JEE (XII) Module-1 PW
23. Which of the following graphs shows the correct variation of
electric field as a function of x along the axis of a uniformly
and positively charged ring of radius R and charge Q.
E(x)
26. If three infinite charged sheets of uniform surface charge
densities s, 2s and –4s are placed as shown in figure, then
find out electric field intensities at points A, B, C and D.
y
Q
3 3ε0 πr 2
(a)
r/2
A
–r/2
x
x
B
2
–4
C
D
27. A solid sphere of radius R has a volume charge density
r = r0 r2 ( Where r0 is a constant and r is the distance from
center). At a distance x from its center for x < R, the electric
field is directly proportional to
(b)
(a) 1/x2
(c)
(d)
3 3
(c)
6 3
–r/2
(b) 1/x
x3
(d) x2
28. Two mutually perpendicular infinite wires along x-axis and
y-axis carry charge densities l1 and l2. The electric line of
1
x , where P is also a point
force at P is along the line y =
3
lying on the same line then find l2/l1.
E
r/2
Y
24. Two concentric rings, one of radius R and total charge +Q
and the second of radius 2R and total charge − 8Q , lie in
x-y plane (i.e., z = 0 plane). The common center of rings
lies at origin and the common axis coincides with z-axis.
The charge is uniformly distributed on both rings. At what
distance from origin is the net electric field on z-axis zero.
2
P
O
1
29. Figure shows, in cross section, two solid spheres with
uniformly distributed charge throughout their volumes. Each
has radius R. Point P lies on a line connecting the centers of
the spheres, at radial distance R/2 from the center of sphere 1.
If the net electric field at point P is zero and Q1 is 64 mC,
what is Q2(in mC).
R
R
R
R
(b)
(c)
(d) 2 R
2 2
2
2
25. Two semicircular rings lying in same plane, of uniform linear
charge density l have radius r and 2r. They are joined using
two straight uniformly charged wires of linear charge density
l and length r as shown in figure. Find the magnitude of
electric field at common center of semi circular rings is
r
2r
P Electric Charges and Fields
W
R
P
(a)
x
1
2
30. A particle is uncharged and is thrown vertically upward
from ground level with a speed of 5 5 m/s in a region of
space having uniform electric field. As a result, it attains
a maximum height h. The particle is then given a positive
charge +q and reaches the same maximum height h when
thrown vertically upward with a speed of 13 m/s. Finally, the
particle is given a negative charge –q. Ignoring air resistance,
determine the speed (in m/s) with which the negatively
charged particle must be thrown vertically upward, so that
it attains exactly the same maximum height h.
37
31 A square loop of side ‘’ having uniform linear charge
density l is placed in xy plane as shown in the figure. There
a
is a non uniform electric field=
E
( x + ) iˆ where a and
 are constants. The resultant electric force on the loop is
having value nal. Find value of n.
y
B
C
A
D
x
32. The two ends of a rubber string of negligible mass and having
unstretched length 24 cm are fixed at the same height as
shown. A small object is attached to the string in its midpoint,
thus the depression (h) of the object in equilibrium is 5 cm.
Then the small object is charged and vertical electric field
(E1) is applied. The equilibrium depression of the object
increases to 9 cm, then the electric field is changed to E2 and
the depression of object in equilibrium increases to 16 cm.
What is the ratio of electric field in the second case to that of
in the first case (E2/E1)?
36. A sphere of radius R carries charge such that its volume
charge density is proportional to the square of the distance
from the center. What is the ratio of the magnitude of the
electric field at a distance 2R from the center to the magnitude
of the electric field at a distance of R/2 from the center
(i.e.,
Er 2=
=
R /Er R / 2 )?
(a) 1
(b) 2
(c) 4
(d) 8
37. Charge in the form of a plane sheet with density
s = 40 µC/m2 is located at z = –0.5 m. A uniform line of
charge of density l = –6 µC/m lies along the y-axis. Net flux
that crosses the surface of cube 2m on an edge, centered at
−6
the origin as shown, is _____ 10 N m 2
∈0 C
z
2m
2m
y
2m
h
x
(a) 4.25
(b) 4.20
(c) 4.30
(d) 4.35
33. A spherical insulator of radius R is charged uniformly with a
charge Q throughout its volume and contains a point charge
Q
located at its center. Which of the following graphs best
16
represent qualitatively, the variation of electric field intensity
E with distance r from the center.
E
E
(b)
(a)
R
r
R
E
39. Two identical charge q are connected by rubber cords to
the walls, as shown in figure at a distance 2a from each
other. The distance between the walls 2l, the length of
each non-deformed cord l. Determine the force constant
of the cord. Mass of charges is negligible. q = 1 µC,
a = 3 cm, l = 4 cm.
q
r
E
(c)
(d)
r
R
r
34. What is the electric field due to a line charge of linear charge
16
density l kept on y-axis from y = –3m to y =
at (4,0,0).
3
kλ
kλ
kλ
2kλ
(a)
(b)
(c)
(d)
2
4
5
2 2
35. A sphere of radius R carries charge density proportional to
the square of the distance from the center: r = Ar2, where A
is a positive constant. At a distance of R/2 from the center,
the magnitude of the electric field is
(a) A/(4pe0)
(b) AR3/(40e0)
(c) AR3/(24e0)
(d) AR3/(5e0)
38
38. Three point charges q1, q2, q3 are used to make charges
Q1 = q1 + q2, Q2 = q2 + q3, Q3 = q1 + q3. The electrostatic
forces between Q1 and q3 is FA = 2 units, between Q2 and q1
is FB = – 4 units and between Q3 and q2 is FC = – 18 units.
In all the above cases, the separation between the charges
remains the same and negative force refers to attraction.
Then the ratio of charges q1 : q2 : q3 is
(a) –1 : 2 : 3
(b) 2 : –3 : 4
(c) 4 : –3 : 1
(d) 4 : –3 : 2
2a
q
2l
(a) 250 N/m
(b)
125
N/m
3
(c)
500
N/m
3
(d)
625
N/m
3
JEE (XII) Module-1 PW
Exercise-3 (JEE Advanced Level)
4. An oil drop has a charge – 96 × 10–19 C and mass
MULTIPLE CORRECT TYPE QUESTIONS
1. Two equal negative charges –q each are fixed at the
points (0, a) and (0, – a) on the y-axis .A positive charge
Q is released from rest at the point (2a, 0) on the x-axis.
The charge Q will
1.6 × 10–15 gm. When allowed to fall, due to air resistance
force it attains a constant velocity. Then if a uniform electric
field is to be applied vertically to make the oil drop ascend
up with the same constant speed, which of the following are
correct. (g = 10 ms–2)
(a) Execute simple harmonic motion about the origin
(b) At origin velocity of particle is maximum
(Assume that the magnitude of resistance force is same in
(c) Move to infinity
both the cases)
(d) Execute oscillatory but not simple harmonic motion
(a) The electric field is directed upward
2. Two fixed charges 4Q (positive) and Q (negative) are located
at A and B, the distance AB being 3m.
+4Q
A
–Q
3m
B
(a) The point P where the resultant field due to both is zero
is on AB outside AB
(b) The point P where the resultant field due to both is zero
is on AB inside AB
(c) If a positive charge is placed at P and displaced slightly
along AB it will execute oscillations
(d) If a negative charge is placed at P and displaced slightly
along AB it will execute oscillation
3. A uniform electric field
of strength E exists in a
region. An electron (charge
–e, mass m) enters a point
A with velocity v. It moves
through the electric field
and exits at point B. Then
(c) The intensity of electric field is
1
×102 NC–1
3
(d) The intensity of electric field is
1
× 105 NC–1
6
5. An electric field converges at the origin whose magnitude
is given by the expression E = 100r N/C, where r is the
distance measured from the origin.
(a) Total charge contained in any spherical volume with its
center at origin in negative
(b) Total charge contained at any spherical volume,
irrespective of the location of its center, is negative
vV
y
(b) The electric field is directed downward
(c) Total charge contained in a spherical volume of
V
(0, 0)
radius 3 cm with its center at origin has magnitude
B(2a,d)
A(a, 0)
2
2 amv ˆ
(a) E = −
i
2
ed
x
3 × 10–13C
(d) Total charge contained in a spherical volume of
radius 3 cm with its center at origin has magnitude
3 × 10–9 C
6. An electric dipole is kept in the electric field produced by a
(b) Rate of work done by the electric field at B is
point charge.
4 m a 2 v3
d3
(c) Rate of work by the electric field at A is zero
(a) Dipole will experience a force
(d) Velocity at B is
2av ˆ
i + vjˆ
d
P Electric Charges and Fields
W
(b) Dipole will experience a torque
(c) It is possible to find a path (not closed) in the field on
which work required to move the dipole is zero
(d) Dipole can be in stable equilibrium
39
MATCH THE COLUMN TYPE QUESTIONS
7. The Column-I gives the two point charge system separated by 2a and the Column-II gives the variation of magnitude of electric
field intensity along x-axis. Match the situation in Column-I with the results in Column-II.
Column-Ι
A.
x'
B.
x'
q
+
(–a, 0)
p.
Increases as x increases in the interval 0 ≤ x < a
a
q
x
+
(a, 0)
–q
–
x
(a, 0)
q.
Decreases as x increases in the interval 0 ≤ x < a
a
r.
Zero at x = 0
s.
Decreases as x increases in the interval a < x < ∞
(0, 0)
q
+
(–a, 0)
C.
Column-II
(0, 0)
y
q + (0, +a)
x
(0, 0)
q + (0, –a)
D.
y
–q – (0, +a)
x
(0, 0)
q + (0, –a)
(a) A-(p, r, s); B-(p, s); C-(r, s); D-(q, s)
(b) A-(p, s); B-(p, r, s); C-(r); D-(q, s)
(c) A-(p, r); B-(p, s); C-(r, q); D-(q, r)
(d) A-(s); B-(p, s); C-(r, s); D-(q, s)
8. In each situation of Column-I, two electric dipoles having dipole moments p1 and p2 of same magnitude (that is, p1 = p2) are
placed on x-axis symmetrically about origin in different orientations as shown. In Column-II certain inferences are drawn for
these two dipoles. Then match the different orientations of dipoles in Column-I with the corresponding results in Column-II.
Column-Ι
A.
y
p1
p2
Column-II
p.
The torque on one dipole due to other is zero.
q.
The potential energy of one dipole in electric field of other
dipole is negative.
r.
There is at least one straight line in x-y plane (not at infinity)
which is equipotential.
x
(–a,0)
(a,0)
( p1 and p2 are perpendicular to x-axis as shown)
B.
y
p2
p1
x
(–a,0)
(a,0)
p
p
( 1 and 2 are perpendicular to x-axis as shown)
C.
y
p1
p2
x
(–a,0)
(a,0)
( p1 and p2 are parallel to x-axis as shown)
40
JEE (XII) Module-1 PW
D.
s.
y
p2
p1
Electric field at origin is zero.
x
(–a,0)
(a,0)
( p1 and p2 are parallel to x-axis as shown)
(a) A-(p, r, s); B-(p, s); C-(r, s); D-(q, s)
(c) A-(p, r); B-(p, s, q); C-(r, q, s, p); D-(q, r)
NUMERICAL TYPE QUESTIONS
9. Use Gauss law to find out the ratio of electric flux through
two concentric hollow spheres, A and B, which enclose
charges 4Q and 8Q respectively, as shown in figure.
8Q
B
4Q
A
10. Three point charges are held on the corners of an equilateral
triangle as shown in the figure.
Take Q = 2µC and L = 3 cm. What is the magnitude of
resultant force exerted on the charge +3Q in N.
(b) A-(p, r); B-(p, q, r, s); C-(p, q, r); D-(p, s)
(d) A-(s); B-(p, s, r); C-(r, s); D-(q, s, p, r)
+
+
+
P(x, y)
+
+
y
+ x
++ + ++ + ++ + ++ + ++ +
+
+
+
+
+
13. Find the magnitude of uniform electric field E in N/C
(direction shown in figure) if an electron entering with
velocity 100m/s making 30° comes out making 60°
m
(see figure), after a time numerically equal to
of electron.
e
60º
E
y
y
–2Q
30º
100m/s
x
L
L
L
Q
3Q
11. The field on either side of an infinite sheet of charge of
density of s C/m2 is E = s/2e0. Electric field in region II is
xσ
given by
then x is
ε0
2
I
–
–2
II
III
IV
14. A clock face has negative charges –q, –2q, –3q,.........,
–12q fixed at the position of the corresponding numerals
on the dial. The clock hands do not disturb the net field
due to point charges. At what time does the hour hand
point in the same direction as electric field at the center
of the dial. All the parts of the clock are of nonconducting
material.
SUBJECTIVE TYPE QUESTIONS
15. In the figure shown, a very long wire and a semicircular
ring of radius ‘R’ are placed in the same plane. The center
of the ring is at a distance ‘r’ from the wire. The wire has
uniformly distributed line charge density ‘l’ and the ring
has linear charge density ‘+l’ on one half and ‘–l’ on the
other half as shown. Find the magnitude of net torque on the
ring due to the wire.
+
12. Two infinite lines of charge with equal linear charge densities
l C/m are placed along the x and y axes, as in figure. If angle
made by resultant electric field with horizontal at point P is q
then tan q will be equal to (given y = 2x)
P Electric Charges and Fields
W
r
C
R
–
41
16. An infinitely long string uniformly charged with a linear
density λ1 and a segment of length l uniformly charged
with linear density λ2 lie in a plane at right angles to each
other and separated by a distance r0. Determine the force
with which these two interact.
17. An infinitely large non-conducting plane of uniform surface
charge density σ has circular aperture of certain radius
carved out from it. The electric field at a point which is at a
distance ‘a’ from the center of the aperture (perpendicular
σ
to the plane) is
. Find the radius of aperture
2 2 ∈0
r0
P
a
Exercise-4 (Past Year Questions)
JEE MAIN
1. A vertical electric field of magnitude 4.9 × 105 N/C just
prevents a water droplet of a mass 0.1 g from falling. The
value of charge on the droplet will be (Given g = 9.8 m/s2)
(2022)
–9
–9
(a) 1.6 × 10 C
(b) 2.0 × 10 C
(c) 3.2 × 10–9 C
(d) 0.5 × 10–9 C
2. A long cylindrical volume contains a uniformly distributed
charge of density ρ. The radius of cylindrical volume is R. A
charge particle (q) revolves around the cylinder in a circular
path. The kinetic of the particle is
(2022)
2
2
ρqR
ρqR
(a)
(b)
4 ∈0
2 ∈0
(c)
qρ
4 ∈0 R 2
2
(d) 4 ∈0 R
qρ
3. In the figure, a very large plane sheet of positive charge is
shown. P1 and P2 are two points at distance l and 2 l from
the charge distribution. If σ is the surface charge density,
then the magnitude of electric fields E1 and E2 at P1 and
P2 respectively are
(2022)
+
+
+
+
(a)
(b)
(c)
(d)
42
+
+
+
+
+
+
+
+
+
+
l
+
2l
P2
4. If a charge q is placed at the center of a closed hemi-spherical
non-conducting surface, the total flux passing through the
flat surface would be
(2022)
q
(a)
q
∈0
(b)
q
2 ∈0
(c)
q
4 ∈0
(d)
q
2π ∈0
5. Three identical charged balls each of charge 2C are
suspended from a common point P by silk threads of 2 m
each (as shown in figure). They form an equilateral triangle
of side 1 m.
The ratio of net force on a charged ball to the force between
any two charged balls will be
(2022)
p
E2
2m
2m
B3
P1
E1 = σ/∈0 , E2 = σ/2∈0
E1 = 2σ/∈0 , E2 = σ/∈0
E1 = E2 = σ/2∈0
E1 = E2 = σ/∈0
1m
B1
2m
1m
1m
B2
(a) 1 : 1
(b) 1 : 4
(c)
(d)
3:2
3 :1
JEE (XII) Module-1 PW
6. Given below are two statements:
(2022)
Statement-I: A point charge is brought in an electric field.
The value of electric field at a point near to the charge may
increase if the charge is positive.
Statement-II: An electric dipole is placed in a non-uniform
electric field. The net electric force on the dipole will not be
zero.
Choose the correct answer from the options given below:
(a) Both Statement-I and Statement-II are true
(b) Both Statement-I and Statement-II are false
(c) Statement-I is true but Statement-II is false
(d) Statement-I is false but Statement-II is True
7. The three charges q/2, q and q/2 are placed at the corners
A, B and C of a square of side ‘a’ as shown in figure. The
magnitude of electric field (E) at the corner D of the square,
is
C
A
q/2
11. The electric field in a region is given by
2
3
N
E=
E0 iˆ + E0 ˆj with E0 =
4.0 × 103 . The flux of this
5
5
C
field through a rectangular surface area 0.4 m2 parallel to
the Y – Z plane is __________ Nm2C–1
(2021)
12. A cube of side ‘a’ has point charges +Q located at each of
its vertices except at the origin where the charge is –Q. The
electric field at the center of cube is
(2021)
x
+O
+O
q
q/2
(a)
(c)
q
 1 1
+ 

4π ∈0 a 2  2 2 
4π ∈0
2
1 

1 −

2

D
(b)
q
1 

1+

2 
4π ∈0 a 
2
(d)
q
 1 1
− 

4π ∈0 a 2  2 2 
8. Two point charges A and B of magnitude +8 × 10–6 C
and –8 × 10 –6 C respectively are placed at a distance
d apart. The electric field at the middle point O between the
charges is 6.4 × 104 NC–1. The distance ‘d’ between the point
charges A and B is
(2022)
(a) 2.0 m
(b) 3.0 m
(c) 1.0 m
(d) 4.0 m
9. A positive charge particle of 100 mg is thrown in
opposite direction to a uniform electric field of strength
1 × 105 NC–1. If the charge on the particle is 40 µC and the
initial velocity is 200 ms–1, how much distance it will travel
before coming to the rest momentarily:
(2022)
(a) 1 m
(b) 5 m
(c) 10 m
(d) 0.5 m
10. Find out the surface charge density at the intersection of
point x = 3 m plane and x-axis, in the region of uniform line
charge of 8 nC/m lying along the z-axis free space. (2021)
(a) 0.424 nCm–2
(b) 47.88 C/m
(c) 0.07 nCm–2
(d) 4.0 nCm–2
P Electric Charges and Fields
W
+O
+O
–O
z
+O
a
(a)
(b)
B
+O
(c)
(d)
y
2Q
3 3π ∈0 a 2
Q
3 3π ∈0 a 2
−2Q
3 3π ∈0 a 2
−Q
3 3π ∈0 a 2
( xˆ + yˆ + zˆ )
( xˆ + yˆ + zˆ )
( xˆ + yˆ + zˆ )
( xˆ + yˆ + zˆ )
13. Two electrons each are fixed at a distance ‘2d’. A
third charge proton placed at the midpoint is displaced
slightly by a distance x(x < < d) perpendicular to the
line joining the two fixed charges. Proton will execute
simple harmonic motion having angular frequency:
(m = mass of charged particle)
(2021)
1

2
q2
(a) 
3
 2π ∈ md 
0


1
 π ∈0 md 3  2
(b) 
 2q 2 


1
 2π ∈0 md 3  2
(c) 


q2


1
 2q 2  2
(d) 
3 

 π ∈0 md 
14. A point charge +12 µC is at a distance 6 cm vertically above
the center of a square of side 12 cm as shown in figure. The
magnitude of the electric flux through the square will be
_______ × 103 Nm2 / C.
(2021)
+q
6 cm
12 cm
12 cm
43
=
E
15. The electric field in a region is given by
 3 ˆ 4 ˆ N
 E0 i + E0 j  .
5
5
c
The ratio of flux of reported field through the rectangular
surface of area 0.2 m2 (parallel to y – z plane) to that of the
surface of area 0.3 m2 (parallel to x – z plane) is a:b, where a
= __________ (round off to nearest integer) [Here iˆ, ˆj and kˆ
20. Three charged particles A, B and C with charges –4q,
2q and –2q are present on the circumference of a circle of
radius d. The charged particles A, C and center O of the circle
formed an equilateral triangle as shown in figure. Electric
field at O along x-direction is
(2020)
y
–4q
2q
A
B
150º
are unit vectors along x, y and z-axes respectively] (2021)
O
16. A charge ‘q’ is placed at one corner of a cube as shown in
figure. The flux of electrostatic field E though the shaded
area is:(2021)
d
q
q
48 ∈0
(b)
q
8 ∈0
(c)
q
24 ∈0
(d)
q
4 ∈0
a
× 10−8 C . The value of ‘a’ will be (2021)
21
18. Find the electric field at point P (as shown in figure) on the
perpendicular bisector of a uniformly charged thin wire of
length L carrying a charge Q. The distance of the point from
of the sphere is
3
L .
2
(2021)
E
a
O
P
O
(a)
(c)
Q
2 3π ∈0 L2
Q
3π ∈0 L2
(b)
3Q
4π ∈0 L2
(d)
Q
4π ∈0 L2
19. An infinite number of point charges, each carrying 1 μC
charge, are placed along the y-axis at, y = 1m, 2m.The
total force on a 1 C point charge, placed at the origin,
is x × 103 N. The value of x1 to the nearest integer, is
1
[Take
(2021)
= 9 × 109 Nm 2 /C2 ]
4πε0
44
3q
πε0 d 2
(b)
3q
4πε0 d 2
(c)
3 3q
4πε0 d 2
(d)
2 3q
πε0 d 2
(b)
x
repel each other to a distance of 0.20 m. Then charge on each
L
(a)
(a)
17. Two small spheres each of mass 10 mg are suspended from
a point by threads 0.5 m long. They are equally charged and
the center of the rod is a =
x
21. A particle of mass m and charge q is released from rest in a
uniform electric field. If there is no other force on the particle,
the dependence of its speed v on the distance x travelled by
it is correctly given by (graphs are schematic and not drawn
to scale)
(2020)
v
v
X
(a)
30º
30º
C
–2q
Z
Y
d
v
v
(c)
x
(d)
x
x
ˆ
ˆ
22. An electric dipole of moment p = (−i − 3 j + 2kˆ) ×10–29
C.m. is at the origin (0, 0, 0). The electric field due to this
dipole at r = (+iˆ + 3 ˆj + 5kˆ) (note that r . p = 0 ) is parallel
to
(a)
(c)
(
)
( +iˆ − 3 ˆj − 2kˆ )
+iˆ + 3 ˆj − 2kˆ
(b)
(
)
( −iˆ − 3 ˆj + 2kˆ )
−iˆ + 3 ˆj − 2kˆ
(2020)
(d)
23. An electric field E = 4 xiˆ − ( y 2 + 1) ˆj N/C passes through
the box shown in figure. The flux of the electric field
through surface ABCD and BCGF and marked as fI and fII
respectively. The difference between (fI – fII) is (in Nm2/C)
(2020)
Z
A(0,0,2) B(3,0,2)
D(0,2,2)
H(0,2,0)
C(3,2,2)
E
(0,0,0)
X
F(3,0,0)
G(3,2,0)
Y
JEE (XII) Module-1 PW
24. In finding the electric field using gauss law the formula
q
| E |= enc is applicable. In the formula ε0 is permittivity
ε0 | A |
of free space, A is the area of Gaussian surface and qenc is
charges enclosed by the Gaussian surface. This equation
can be used in which of the following situation? (2020)
(a) Only when the Gaussian surface in an equipotential
surface
(b) Only when E = constant on the surface
(c) Only when the Gaussian surface is an equipotential
surface and E is constant of the surface
(d) For any choice of Gaussian surface
25. A small point mass carrying some positive charge on it, is
released from the edge of a table. There is a uniform electric
field in this region in the horizontal direction. Which of the
following options then correctly describe the trajectory of
the mass? (Curves are drawn schematically and are not to
scale).
(2020)
E
x
–
+
(a)
+
x
(d) y
x
x
26. A charged particle (mass m and charge q) moves along X-axis
with velocity V0. When it passes through the origin it enters a

region having uniform electric field E = − EJˆ which extends
upto x = d. Equation of path of electron in the region x > d
is
(2020)
Y
E
O
X
V0
d
(a) y =
qEd
(x − d )
mV02
qEd  d

− x
(c) y =
2
2
mV0 

(b) y =
28. Consider the force F on a charge ‘q’ due to a uniformly
charged spherical shell of radius R carrying charge Q
distributed uniformly over it. Which one of the following
statements is true for F, if ‘q’ is placed at distance r from
the center of the shell?
(2020)
1 Qq
(a) F =
for all r
4πε0 r 2
(b)
1 Qq
> F > 0 For r < R
4πε0 r 2
1 Qq
for r > R
4πε0 r 2
1 Qq
for r < R
4πε0 R 2
29. Charges Q1 and Q2 are at points A and B of a right angle
triangle OAB (see figure). The resultant electric field at
point O is perpendicular to the hypotenuse, then Q1/Q2 is
proportional to
(2020)
A
Q1
x1
O
Q2
B
x2
x13
x22
x
(b)
(c) 1
(d) x2
3
2
x2
x1
x2
x1
30. Determine the electric dipole moment of the system of three
charges, placed on the vertices of an equilateral triangle, as
shown in the figure
(2019)
(a)
qEd 2
x
mV02
–2q
y
qEd
x
(d) y =
mV02
27. Two charged thin infinite plane sheets of uniform surface
charge density σ+ and σ– , where | σ+ |> |σ– |, intersect at
right angle. Which of the following best represents the
electric field lines for this system?
(2020)
P Electric Charges and Fields
W
+
(d)
(d) F =
x
(c) y
+
–
(c)
y
(b) y
(b)
–
(c) F =
(a) y
–
l
l
+q
l
(a)
3ql
+q
x
ˆj − iˆ
iˆ + ˆj
(b) ( ql )
(c) 2qljˆ (d) − 3qljˆ
2
2
45
31. For a uniformly charged ring of radius R, the electric field
on its axis has the largest magnitude at a distance h from
its center. Then value of h is
(2019)
R
R
(a)
(b)
5
2
(c) R
Q P
(d) R 2
A
32. Three charges +Q, q, +Q are placed respectively, at distance,
0, d/2 and d from the origin, on the x-axis. If the net force
experienced by +Q, placed at x = 0, is zero, then value of q
is
(2019)
(a) – Q/4
(b) + Q/2
(c) + Q/4
(d) – Q/2
33. Charge is distributed within a sphere of radius R with a
A −2 r /a
e
volume charge density ρ ( r ) =
, where A and a are
r2
constants. If Q is the total charge of this charge distribution
the radius R is
(2019)
Q 

(a) a log 1 −

2
aA 
π





a
1
log 
(b)

2
1− Q 
 2πaA 
(
10µC
)
and q 2 (– 25mC) are
placed on the x-axis at x = 1 m and x = 4 m respectively.


1
= 9 × 109 Nm 2 C −2  Take
4
πε
0


(c)
( 63i − 27 ˆj ) ×10
(81i − 81 ˆj ) ×10
2
2
(b)
(c) E ∝
1
D2
(d) E ∝
1
D4
(b)
(d)
( −63i + 27 ˆj ) ×10
( −81i + 81 ˆj ) ×10
(b) tan–1 (2.0)
(d) tan–1 (0.2)
38. A simple pendulum of length L is placed between the plates
of a parallel plate capacitor having electric field E, as shown
in figure. Its bob has mass m and charge q. The time period
of the pendulum is given by
(2019)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(2019)
2
2
35. Charges –q and +q located at A and B, respectively, constitute
an electric dipole. Distance AB = 2a, O is the mid point of the
dipole and OP is perpendicular to AB. A charge Q is placed
at P where OP = y and y >> 2a. the charge Q experiences an
electrostatic force F. If Q is now moved along the equatorial
 y
line to P’ such that, OP' =   the force on Q will be close
3
y

to  << 2a  (2019)
2

46
B
F
(c) 9F
(d) 27F
3
36. Four point charges –q, +q, +q and –q are placed on y-axis
at y = –2d, y = –d, y = +d and y = +2d, respectively.
The magnitude of the electric field E at a point on the x-axis
at x = D, with D >> d, will behave as
(2019)
1
(a) E ∝
D
1
(b) E ∝
D3
(a) 3F
The electric field (in V/m) at a point y = 3 m on y-axis is,
(a)
+q
(a) tan–1 (5.0)
(c) tan–1 (0.5)
1 
a

log 1 −

2
2
π
aA 

34. Two point charge q1
O
–q
37. The bob of a simple pendulum has mass 2g and a charge
of 5.0 mC. It is at rest in a uniform horizontal electric field
of intensity 2000 V/m. At equilibrium, the angle that the
pendulum makes with the vertical is (Take g = 10 m/ss)
(2019)




1
(c) a log 

1− Q 
 2πaA 
(d)
P
L
m
q
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
–
E
(a) 2π
(c) 2π
L
 qE 
g2 + 

 m 
L
qE 

g −

m 

2
(b) 2p
(d) 2π
L
qE 

g+

m 

L
g2 −
q2 E 2
m2
JEE (XII) Module-1 PW
39. An electric dipole is formed by two equal and opposite
charges q with separation d. The charges have same mass m.
It is kept in a uniform electric field E. If it is slightly rotated
from its equilibrium orientation, then its angular frequency
w is
(2019)
(a)
qE
2md
(b) 2
(c)
2qE
md
(d)
qE
md
qE
md
40. Let a total charge 2Q be distributed in a sphere of
radius R, with the charge density given by r(r) = kr, where
r is the distance from the center. Two charges A and B, of – Q
each, are placed on diametrically opposite points, at equal
distance, a, from the center. If A and B do not experience
any force, then
(2019)
3R
21/4
(c) a = 8–1/4 R
(a) a =
(b) a = R / 3
(d) a = 2–1/4 R
41. An electric dipole has a fixed dipole moment p , which
makes angle q with respect to x-axis. When subjected to
an electric field E1 = Ei, it experiences a torque T1 = −τ kˆ.
When subjected to another electric field E2 = 3E1 ˆj . It
experiences torque T2 = −T1 . The angle q is,
(2017)
(a) 60°
(b) 90°
(c) 30°
(d) 45°
42. The region between two concentric spheres of radii ‘a’ and
‘b’, respectively (see figure), has volume charge density
A
ρ = , where A is a constant and r is the distance from the
r
center. At the center of the spheres is a point charge The
value of A such that the electric field in the region between
the spheres will be constant, is
(2016)
JEE ADVANCED
43. A charge q is surrounded by a
closed surface consisting of an
inverted cone of height h and
base radius R, and a hemisphere
of radius R as shown in the figure.
R
q
The electric flux through
h
nq
the conical surface is
6 ∈0
(in SI units). The value of n is
(2022)
________.
44. A uniform electric field, E = −40 3 yNC
ˆ −1 is applied in
a region. A charged particle of mass m carrying positive
charge q is projected in this region with an initial speed
of 2 10 × 106 ms −1. This particle is aimed to hit a target
T, which is 5 m away from its entry point into the field as
q
shown schematically in the figure. Take
= 1010 Ckg −1 .
m
Then
(2020)
E
u
T
5m
(a) The particle will hit T if projected at an angle 45º from
the horizontal
(b) The particle will hit T if projected either at an angle 30º
or 60º from the horizontal
(c) Time taken by the particle to hit T could be
well as
a
Q
Q
2πa 2
Q
(b)
2π(b 2 − a 2 )
(a)
b
5 ms
2
(d) Time taken by the particle to hit T
5
ms
3
45. A circular disc of radius R carries surface charge density
r

σ( r ) =
σ0 1 −  , where s 0 is a constant and r is the
 R
distance from the center of the disc. Electric flux through
a large spherical surface that encloses the charged disc
completely is f0. Electric flux through another spherical
(c)
2Q
π(a 2 − b 2 )
surface of radius
(d)
2Q
πa 2
the ratio
P Electric Charges and Fields
W
5
ms as
6
R
and concentric with the disc is f. Then
4
φ0
is .....
φ
(2020)
47
46. Two identical non-conducting solid spheres of same mass
and charge are suspended in air from a common point by
two non-conducting, massless strings of same length. At
equilibrium, the angle between the strings is a. The spheres
are now immersed in a dielectric liquid of density 800 kg m3
and dielectric constant 21. If the angle between the strings
remains the same after the immersion, then
(2020)
(a) electric force between the spheres remains unchanged
(b) electric force between the spheres reduces
(c) mass density of the spheres is 840 kg m–3
(d) the tension in the strings holding the spheres remains
unchanged
47. A charged shell of radius R carries a total charge Q. Given
f as the flux of electric field through a closed cylindrical
surface of height h, radius r and with its center same as
that of the shell. Here, center of the cylinder is a point on
the axis of the cylinder which is equidistant from its top
and bottom surfaces. Which of the following option(s)
is/are correct? [∈ 0 is the permittivity of free space]
(2019)
(a) If h > 2R and r > R then Φ =
(c) The electric flux through the shell is 2 R λ ε0
(d) The electric field is normal to the surface of the shell at
all points
49. The electric field E is measured at a point 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.
Column-I contains different relations between E and d.
Column-II describe different electric charge distributions,
along with their locations. Match the functions in Column-I
with the related charge distributions in Column-II. (2018)
A.
B.
Column-I
E is independent
of d
Column-II
p. A point charge Q at the
origin
E∝ 1
q. A small dipole with point
charges Q at (0,0, l)
and –Q
(0, 0, –l).
d
Q
∈0
3R
(b) If h < 8 R and r =
then Φ = 0
5
5
(c) If h > 2R and r = 4 R then Φ = Q
5 ∈0
5
(d) If h > 2R and r =
(a) The electric flux through the shell is 3 R λ ε0
(b) The z-component of the electric field is zero at all the
points on the surface of the shell
Q
3R
then Φ =
5 ∈0
5
Take 2l << d
C.
D.
48. An infinitely long thin non-conducting wire is parallel to the
z-axis and carries a uniform line charge density l. It pierces a
thin non-conducting spherical shell of radius R in such a way
that the arc PQ subtends an angle 120º at the center O of the
spherical shell, as shown in the figure. The permittivity of
free space is e0. Which of the following statements is (are)
true?
(2018)
z
P
E∝
1
d2
r.
E∝
1
d3
s.
t.
R
120º
Q
48
O
(a)
(b)
(c)
(d)
An infinite line charge
coincident with the
x-axis, with uniform
linear charge density l.
Two infinite wires
carrying uniform linear
Charge density parallel
to the x-axis. The one
along (y = 0, z = l) has a
charge density +l and the
one along (y = 0, z = – l)
has a charge density – l.
Take 2l << d
Infinite plane charge
coincident with the
xy-plane with uniform
surface charge density
A-(t); B-(r, s); C-(p); D-(q)
A-(t); B-(r); C-(p, s); D-(q)
A-(t); B-(r); C-(p, q); D-(s)
A-(s), B-(q, r); C-(p); D-(t)
JEE (XII) Module-1 PW
ANSWER KEY
CONCEPT APPLICATION
1. (b)
2. (d)
3. (d)
11. (b)
12. (c)
13. (c)
4. (b,c)
5. (a)
6. (c)
14. (b)
15. (a)
16. (d)
7. (c)
8. (a)
9. (d)
10. (a,c)
EXERCISE-1 (TOPICWISE)
1. (a)
2. (b)
3. (b)
4. (c)
5. (d)
6. (a)
7. (b)
8. (a)
9. (c)
10. (c)
11. (d)
12. (b)
13. (a)
14. (c)
15. (a)
16. (c)
17. (b)
18. (a)
19. (b)
20. (c)
21. (a)
22. (c)
23. (b)
24. (b)
25. (d)
26. (c)
27. (c)
28. (c)
29. (b)
30. (c)
31. (b)
32. (b)
33. (b)
34. (c)
35. (c)
36. (b)
37. (a)
38. (c)
39. (c)
40. (b)
41. (d)
42. (b)
43. (b)
44. (a)
45. (d)
46. (a)
47. (b)
48. (d)
49. (c)
50. (b)
10. (a)
51. (c)
EXERCISE-2 (LEARNING PLUS)
1. (c)
2. (a)
3. (d)
4. (c)
5. (a)
6. (a)
7. (c)
8. (d)
9. (d)
11. (a)
12. (a)
13. (b)
14. (d)
15. (a)
16. (d)
17. (a)
18. (a)
19. (b)
20. (a,b)
21. (a)
22. (d)
23. (b)
24. (d)
25.
32. (a)
33. (a)
34. (b)
35. (b)
36. (b)
7. (a)
8. (b)
26.
−σ ˆ −3σ ˆ
−σ7 ˆ σ ˆ
j , EB =
j , EC =
j , ED =
j
2ε0
2ε0
2ε0
2ε0
27. (c)
28. [3]
37. [0148] 38. (d)
29. [72]
30. [9]
1 λ
4πεo r
31. [10]
39. (d)
EXERCISE-3 (JEE ADVANCED LEVEL)
1. (d)
11. [2.5]
2. (a,d)
12. [0.5]
3. (a,b,c,d) 4. (b,c)
13. [100]
14. [9:30]
5. (a,b,c)
15. τ =
6. (a,c)
9. [0.33]
10. [0.208]
2
λ R R+r
λ1λ 2 

n 
16. F
n 1 +  17. R = a
=
πε0
2πε 0  r0 
 r 
EXERCISE-4 (PAST YEAR QUESTIONS)
JEE Main
1. (b)
2. (a)
3. (c)
4. (b)
5. (d)
6. (a)
7. (a)
8. (b)
9. (d)
10. (a)
11. [640]
12. (c)
13. (a)
14. [226]
15. [1]
16. (c)
17. [20]
18. (a)
19. [12]
20. (a)
21. (b)
22. (a)
23. [48]
24. (c)
25. (b)
26. (c)
27. (d)
28. (c)
29. (c)
30. (d)
31. (b)
32. (a)
33. (b)
34. (a)
35. (d)
36. (d)
37. (c)
38. (a)
39. (c)
40. (c)
41. (a)
42. (a)
45. [6.40]
46. (a,c)
47. (a,b,d) 48. (a,b)
JEE Advanced
43. [3]
44. (b,c)
P Electric Charges and Fields
W
49. (b)
49
CHAPTER
Electrostatic Potential
and Capacitance
2
INTRODUCTION
Potential energy of a system of particles is defined only in
conservative fields. As electric field is also conservative, we
define potential energy in it.
The electrostatic interaction energy of this system can be given
as work done by external agent in bringing q2 from infinity to the
given separation r from q1.
=
Wext
∞
Electrostatic potential energy is defined in two ways.
(i) Interaction energy of charged particles of a system.
(ii) Self energy of a charged object (will be discussed later)
ELECTROSTATIC INTERACTION ENERGY
Electrostatic interaction energy of a system of charged particles
is defined as the external work required to assemble the particles
from infinity to a given configuration.
When some charged particles are at infinite separation, their
potential energy is taken as zero as there is no interaction between
them.
U∞ = 0
When these charges are brought close to a given configuration,
external work is required if the force between these particles
is repulsive and energy is supplied to the system hence final
potential energy of system will be positive. If the force between
the particles is attractive work will be done by the system and final
potential energy of system will be negative.
Let us take some illustrations to understand this concept in detail.
→ →
. dx
∫ Fext=
r
 1 1  kq q
=
−kq1q2  −  =1 2
r
∞ r 
Wext =
U=
kq1q2
= U r
kq1q2
r
[Interaction energy]
[P.E. of two charges kept at distance r]
Potential Energy for a System of
Charged Particles
When more than two charged particles are there in a system, the
interaction energy can be given by sum of interaction energy of
all the pairs of particles.
U =k
∑
All pairs
qi q j
rij
Note: Number of pairs corresponding to n charges =
For example
n(n − 1)
2
q1
Figure shows two +ve charges q1 and q2 separated by a distance r.
q1
kq1q2
 1
dx kq1q2  − 
∫ x 2=
 x ∞
∞
If a system of three particles having charges q1, q2 and q3 is given
as shown in figure.
Interaction Energy of a System of
Two Charged Particles
x
r
r
dx
q2
a=0
Felec Fext
Felec = Fext =
From work energy theorem
Wext + Welec = DK (change in K.E.) = 0
kq1q2
r2
Wext = –Welec = –(–DU) = DU = U – U∞, [ Welec = –DU]
r13
r12
q2
q2
r23
q3
The total interaction energy of this system can be given as
U=
Kq1q2 Kq1q3 Kq2 q3
+
+
r12
r13
r23
Train Your Brain
Example 1: Calculate the total interaction energy of the
system shown in the figure
a (+)
q
q
(–)
(–)
a
(–)
(–)
–q
–q
(+)
Sol. Square (4 charges)
 1 − 1 + 1 − 1   −1 − 1   −2kq 2
U system kq 2 
=
+ =
a
  a 2  
2a

Example 2: Calculate the total interaction energy of the
system shown in the figure
q
q
q
q
q
q
q
a
Sol. Cube (8) Charges
Variety
distance
a
a
q
Number of pairs
side
Face diagonal
2 a 12
So, U system =
−kq 2 kq 2 kq 2 −kq 2
+
−
=
r
r
2r
2r
Example 5: Find out speed of particles when separation
between them is r.
Released
+q
–q
3 a m
2m
2r
Sol. By Mechanical Energy conservation:
Ui + Ki = Uf + Kf
0−
kq1q2 1 2 1
kq q
= mv1 + 2mv22 − 1 2 2r
2
2
r
By Momentum conservation
4
mv1 = 2mv2 v
⇒ v2 = 1 2
Solving (i) & (ii), we get,
v2 = q
12kq 2 12kq 2 4kq 2
+
+
a
2a
3a
Q
a
2Q
a
–q
...(ii)
k
k
& v1 = 2q
6mr
6mr
Concept Application
Example 3: If the electric potential energy of the given
(shown in figure) is positive then prove that 2Q > 3q
a
...(i)
(as E.F is internal force/action-reaction pair)
1 a 12
Main diagonal
−kq 2 −kq 2 kq 2 −3kq 2
+
+
=
r
r
2r
2r
Sol. U1 =
U2 =
4×3
=6
2
Number of pairs =
Example 4: Two negative charge, each of magnitude q are
2r distance apart. A positive charge q is lying at the middle
them. The potential energy of the system is U1. If the two
nearest charges are mutually interchanged and the potential
U
energy becomes U2, then 1 will be:
U2
–q
–q q
q
–q
–q
r
r
r
r
1. 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
Sol. U(system) = Sum of potential energies of all possible
pairs
=
k (Q)(2Q) k 2Qq kQq kQ
−
−
=
(2Q − 3q )
a
a
a
a
Given that U is positive
⇒U>0
⇒
kQ
(2Q − 3q ) > 0 ⇒ 2Q > 3q ⇒ 2Q > 3q
a
P Electrostatic Potential and Capacitance
W
+q
(a)
−q
1+ 2
(c) –2q
+q
a
(b)
−2q
2+ 2
(d) +q
51
2. Four charges q each are placed at four corners of a
square of side a. Find the potential energy of one of
the charges.
a
B
A
a
C
q
a
q
D
1  q2

(a) K  2 +

2 a

1  q2

(b) K 1 +

2 a

1  q2
(c) K  2 +

2 a

1  q2

(d) K  2 2 +

2 a

ELECTRIC POTENTIAL
Electric potential is a scalar property of every point in the region
of electric field. At a point in electric field, electric potential is
defined as the interaction energy of a unit positive charge.
If at a point in electric field a test charge q0 has potential energy U,
then electric potential at that point can be given as
U
V=
joule/coulomb
q0
“Work done in bringing a unit positive charge from infinity to the
given point against the electric forces.”
Properties:
(i) Potential is a scalar quantity, its value may be positive,
negative or zero.
joule
(ii) S.I. Unit of potential is volt =
and its dimensional
coulomb
formula is [M1L2T–3I–1].
(iii) Electric potential at a point is also equal to the negative
of the work done by the electric field in taking the point
charge from reference point (i.e. infinity) to that point.
(iv) Potential decreases in the direction of electric field.
Potential Difference Between
Two Point in an Electric Field
Note:
™ If VB < VA, then positive charges have a tendency to move
towards B (low potential point), it implies that electric forces
carry the charge from high potential to low potential points.
™ Electric potential decreases in the direction of electric fields.
As electric field is conservative,
work done and hence potential
difference between two points is
path independent and depends
only on the position of initial and
final points
B
W1
W2
A
W3
W1 = W2 = W3
Electric Potential due to a Point Charge
Electric potential at P at a distance r in the surrounding of a point
U
charge q. Vp =
q
O
P
r
q
qo
Where U is the P.E. of charges q and q0, which can be given as
Kqq0
U=
r
U Kq
Thus potential at point P is V=
=
P
q0
r
Potential Due to System of Charges
Consider a system of charges, q1, q2,..... qn with position vectors
r1, r2, ...., rn relative to some origin.
q1
q5
r
r1P 5P
q4
r4P
r3P
P
r2P
q3
q2
Potential at a point due to a system of charge is the
sum of potentials due to individual charges
The potential V1 at P due to the charge q1 is V1 =
r1P is the distance a between q1 and P.
1 q1
where
4πε0 r1P
Similarly, the potential V2 at P due to q2 and V3 due to q3 are given
Potential difference between two points in electric field can be
1 q2
1 q3
by V2 =
=
, V3
defined as work done in displacing a unit positive charge from
4πε0 r2 P
4πε 0 r3 P
one point to another.
B
A
Where r2P and r3P are the distance of P from charges q2 and q3
VA
VB
respectively and so on for the potential due to other charges. By
If a unit +ve charge is displaced from point A to point B as shown,
the superposition principle, the potential V at P due to the total
B
charge configuration is the algebraic sum of the potentials due to
work required can be given as VB – VA = − ∫ E ⋅ d r
the individual charges.
A
V = V1 + V2 + ......... + Vn
If a charge q is shifted from point A to point B, work done against
electric forces can be given as W = q(VB – VA)
q 
1  q1 q2
If in a situation work done by electric forces is asked, as we use
=
+
+ .... + n 

4πε 0  r1P r2 P
rnP 
W = q(VA – VB)
52
JEE (XII) Module-1 PW
Train Your Brain
Example 6: A charge Q is placed at the center of a circle of
a radius ‘r’. What will be the work done in taking a charge
q from A to diametrically opposite point B?
1 Qq
Sol. Potential energy of q at A = UA =
4π ∈0 r
PE of q at B
(d) +10 Volt
q  1 1 1 1 1

 + + + + + .... 
4π ∈0  20 21 22 23 23

x
dx
+ + + + + +
Q
+ + + +
L
V = ∫ dV =
r+L
∫
r
Electric Potential due to a Charged Ring
Case I : At its center
To find potential at the center C of the ring, we first find potential
dV at center due to an elemental charge dq on ring which is given
Kdq
as dV =
R
+ + +
+
+
+
+
R
P Electrostatic Potential and Capacitance
W
Total potential at C is
V
Q
0
0
V = ∫ dV =
∫
+ +
+
+
(d) 4.1 × 105 V
+
+
(c) 2.1 × 105 V
+
q2 = +4.0µC 20 cm q3 = –4.0µC
(b) 3.1 × 105 V
(a) 1.1 × 105 V
C
+ +
20 cm
+ + + + +
+
Concept Application
20 cm
P
KQ
KQ
KQ  r + L 
dx =
[n x]rr + L =
n 

Lx
L
L
 r 
dq
3. What is the electric potential at the center of the
triangle shown in figure?
q1 = +4.0µC
r
For this we consider an element of width dx at a distance x from
the point P. Charge on this element is
Q
dq = dx
L
The potential dV due to this element at point P can be given by
using the result of a point charge as
KQ
Kdq
dV =
=
dx
Lx
x
Net electric potential at point P can be given as
+
2q
q
 1 
⋅
= =
 1 − 1/ 2  4π ∈0 2π ∈0
Figure shows a charged rod of length L, uniformly charged with a
charge Q. Due to this we will find electric potential at a point P at
a distance r from one end of the rod shown in the figure.
+
++
B
+
A
q 1 1 1 1

 + + + + .... 
4π ∈  1 2 4 8

q
=
4π ∈0
(c) 20 Volt
+
1 Qq
4π ∈0 r
Q
The net electrostatic potential at origin (x = 0) due to
the infinite array is,
=
(b) –20 Volt
Electric Potential due to a Charged Rod
r
∴ Work done = UB – UA = 0
Example 7: When a 2µC charge is carried from point A
to point B, the amount of work done by the electric field
is 50µJ. What is the potential difference between them and
which is at a higher potential?
Sol. Wd = q(VA – VB)
⇒ 50 × 10–6 = 2 × 10–6(VA – VB) ⇒ VA – VB = 25 volt
 VA – VB > 0
∴ VA > VB
Example 8: Infinitely large number of point charges each
equal to q are placed at positions x = 1, 2, 4, 8, ...... Calculate
the electrostatic potential at the origin.
Sol. Electrostatic potential at a point distant r from a charge
1 q
q is given by
4π ∈0 r
=
V
(a) –10Volt
+
+ +
= UB =
4. A charge of 10 C is moved in an electric field of
a fixed distribution from point A to another point
B slowly. The work done by external agent in
doing so is 100 J. What is the potential difference
VA – VB?
Kdq KQ
=
R
R
As all dq’s of the ring are situated at same distance R from the
ring center C, simply the potential due to all is added as being a
scalar quantity.
KQ
EC (Electric potential at ring center) =
.
R
53
Note: Even if charge Q is non-uniformly distributed on ring, the
electric potential at C will remain same.
Case II: At a Point on Axis of Ring
Electric potential at a point P on the axis of ring as shown, we can
directly state the result as here also all points of ring are at same
x 2 + R 2 from the point P.
Example 9: Find min velocity v0 such that particle crosses
the ring.
+Q
+ +
+ + + + + +
+
+ +
R
+
+ + + +
+ +
+ +
distance r =
Train Your Brain
r
+ R
v0
P
x
+
+
KQ
=
r
R2 + x2
2R
Applying energy conservation
kqQ
1 2 kQq
mv0 +
0+
=
R
2
2R
+
x
Electric Potential due to a Uniformly Charged Disc
Figure shows a uniformly charged disc of radius R with surface
charge density s.
R
x
P
⇒ v0
=
electric potential at a point on its axis at a distance 2 2 R
from its center.
Sol.
VP =
0
x2 + y 2
σ
[ x 2 + R 2 – x]
2 ∈0
Note: If x < < R then Vp =
54
ydy
σR
2 ∈0
=
R
σ  2
x + y2 
2∈0 
 0
+ + + +
+ +
+ +
+
+
dq = s. 2py dy
Due to this elemental ring, the electric potential at point P can be
given as
Kdq
K .σ.2πy dy
dV =
=
2
2
x +y
x2 + y 2
Net electric potential at point P due to whole disc can be given as
R
+ + + + + +
+
+ +
+ +
To find electric potential at point P we consider an elemental ring
of radius y and width dy, charge on this elemental ring is
σ
2kQq 
1 
1 −

mR 
2
Example 10: A ring of radius R is having two charges q
and 2q distributed uniformly on its two half parts. Find the
dy
∫ 2 ∈0 .
kQ
Sol. Potential at P =
KQ
R
R
+
KQ
Vp
y
P
qin
R
Potential at P can be given as; =
VP
Graphs of Vp vs x
V = ∫ dV =
+
r
P
x=2 2
+
Distance of P from periphery of ring is
R 2 + (2 2 R) 2 =
3R
Electric potential = Potential due to upper half +
Potential due to lower half
Kq 2 Kq
+
3R 3R
3Kq Kq
=
=
3R
R
VP =
JEE (XII) Module-1 PW
(iii) Inside the surface
Concept Application
5. A ring of radius R carries a charge +q. A test charge
–q0 is released on its axis at a distance 3R from its
center. How much kinetic energy will be acquired by
the test charge when it reaches the center of the ring?
1 qq0
4πε 0 R
(a)
(b)
1 qq0
4πε0 2 R
1 qq0
1 qq0
(d)
4πε 0 3R
4πε0 3R
6. Figure show three circular arcs, each of radius R and
total charge as indicated. The net electric potential at
the center of curvature is
dV 
q

−
so V =
∵ E =

4πε0 R
dr 
(Same as that on the surface)
Graph of V versus r
V
KQ/R
(c)
+Q
–2Q
45°
30°
R
+3Q
(a)
Q
2πε0 R
(b)
Q
4πε0 R
(c)
2Q
πε0 R
(d)
Q
πε0 R
Electric Potential due to a Conducting Sphere or a
Uniformly Charge Shell
(i) Outside the sphere
According to definition of electric potential, at point P
r
r
q
dr
V = − ∫ E . dr = − ∫
r2
4
πε
0
∞
∞
q 

∵ Eout = 4πε r 2 
0


r
r
q
1
q 1 
q
V=
dr =
−
=
4πε0 ∞∫ r 2
4πε0  r  ∞ 4πε0 r
(ii) On the surface
R
O
r
P
dV
= 0 or V = constant
dr
Inside the surface E = 0,
r
R
Electric Potential due to Solid Non-conducting Sphere
(i) Out the sphere
Same as conducting sphere.
(ii) Inside the sphere
r
V=
− ∫ E ⋅ dr
∞
r
r
R

 R  kq 
 kqr  
−  ∫ E1dr + ∫ E2 dr  =
−  ∫  2 dr + ∫  3  dr 
⇒V=
∞
R

∞  r 
R R 

  1  R kq  r 2  r 
V=
−  kq  −  + 3   
R  2 R 
  r
 1
r2
R2 
kq
−kq  − + 3 − 3  ⇒ V = 3 [3R 2 − r 2 ]
⇒V=
 R 2R 2R 
2R
Graph of V versus r
V
3KQ/2R
KQ/R
r
R
Note: At the center r = 0, V=
centre
3 kq 3
= VS .
2 R 2
Potential due to a group of concentric shells
To calculate the potential due to a group concentric shells, we
can use superposition principle. For any point, we can simply add
the potentials due to individuals shells. Consider a pair of two
uniformly charged concentric shells having radii ‘a’ and ‘b’ and
carrying charges q1 and q2 respectively let a < b.
q2
q1
O
(r > R)
R
q
V=
− ∫ E ⋅ dr =
−∫
dr
4
r2
πε
0
∞
∞
R
R
q 1
V=
−
∫ 2
πε
4
0 ∞ r
a b
q 

∵ Eout = 4πε r 2 
0


R
q 1 
q

⇒V =
 dr =
4πε 0  r  ∞
4πε 0 R

P Electrostatic Potential and Capacitance
W
We will calculate potential at three different points A, B and C.
A lies inside the r inner shell B lies in the space between the two
shells and C lies outside the outer shell.
55
Case I: Potential at point C
OC = r where r > b
As the point C, lies outside the outer shell potential at C due to
q2
outer shell is V2 =
4πε 0 r
Similarly potential at C due to inner shell is V1 =
q2
q1
4πε 0 r
q1
a
O
r
Example 11: A total charge Q is given to (Two concentric
shells) so that their surface charge densities are equal.
Deduce an expression for potential at their common center.
B
b A
a
C
b
Sol. Let surface charge density of each shell be s then
Charge on A + charge on B = Q
q1 + q2
4πε0 r
Case II: Potential at point B
Let OB = r, where a < r < b.
So, potential at C is VC =
⇒ s4pa2 + s4pb2 = Q or s =
The point B due to outer shell is V2 =
q2
.
4πε0 b
V0 =
q2
Putting expression for from s equation (i)
q1
V0 =
(a + b)Q
4π ∈0 (a 2 + b 2 )
Example 12: Find VA, VB, VC, and if VA = VC then what is
the required condition?
O
B
s
–s
s
a
q1
q
+ 2
4πε0 r 4πε 0 b
c
So, potential at A, due to outer shell is V2 =
= V=
VSurface
in
q2
4πε 0 b
Similarly, potential at A, due to inner shell is V1 =
q1
O r
a
A
b
So, potential at A =
is VA
56
q1
q
+ 2 .
4πε0 a 4πε0 b
B
A
C
b
Sol. For conducting sphere or shell
Case III: Potential at point A
Let OA = r, where r < a.
The point A lies inside the outer shell.
q2
...(i)
σ
σ
σ
a + b = (a + b).
∈0
∈0
∈0
b
So, potential at B =
is VB
Q
4π(a 2 + b 2 )
Now potential at common center
= Potential due to A + potential due to B
As the point B, lies inside the outer shell,
a
Train Your Brain
VOut =
q1
4πε0 a
σ
R
ε0
σ R2
ε0 r
where R = radius of sphere or shell and r = distance from
from the center
VA =
σa σb σc σ
(a − b + c)
−
+ =
∈0 ∈0 ∈0 ∈0
VB=
σa 2 ab 2 σc σ  a 2 − b 2 + c 2 
−
+ =


c
∈0 b 2 ∈0 ∈0 ∈0 

⇒ Now if VA = VC
a 2 − b2 − c2
c
2
2
⇒ c(a – b) = a – b ⇒ c = a + b
⇒ (a – b + c) =
JEE (XII) Module-1 PW
Concept Application
7. In a hollow spherical shell potential (V) changes with
respect to distance (r) from center
(b) V
(a) V
r
(c) V
r
(d) V
r
r
8. In an arrangement of two concentric conducting
shells, with center at origin and radii a and b
(a < b), charges Q1 and Q2 are given to inner and
outer shell, then the potential V at a distance r from
the origin is (r < a)
(a) =
V
kQ1 kQ2
+
a
b
(b) =
V
kQ1 kQ2
+
r
b
 Q + Q2 
(c) V = k  1

 r 
kQ1 kQ2
(c) =
V
+
a
r
RELATION BETWEEN ELECTRIC FIELD
INTENSITY AND ELECTRIC POTENTIAL
For Uniform Electric Field
B
→
⇒ dV = –Exdx – Eydy – Ezdz
⇒
∂V
∂V
∂V
−
−
−
Ex =
; Ey =
; Ez =
∂x
∂y
∂z
Potential difference between two points A and B in Non Uniform
Electric Field
E = E iˆ + E ˆj + E kˆ
x
y
z
∂V
∂V
∂V
E x –=
=
, E y –=
, Ez –
∂x
∂y
∂z
 ∂
∂
∂ 
= – iˆ V + ˆj
V + kˆ V 
∂y
∂z 
 ∂x
 ∂ ˆ ∂
∂ 
= – iˆ
+j
+ kˆ  V
x
y
z
∂
∂
∂

E = –∇ V = – grad V
Note: Potential is scalar quantity but the gradient of potential is a
vector quantity.
Potential difference between two points A and B

E
Note:
™ Area under E-x curve gives negative of change in potential
(–DV ).
™ Negative of slope of V-x curve gives the electric field at that
point.
™ Negative sign in the expression dV =− E ⋅ dr signifies that
as one moves in the direction of electric field, potential
decreases.
The negative sign in the expression dV =− E ⋅ dr signifies that as
one moves in the direction of electric field, potential decreases.
In cartesian form, E = Ex iˆ + E y ˆj + Ez kˆ
dr = dx iˆ + dy ˆj + dzkˆ
→
VB – VA = – E . AB
Train Your Brain
A
Non Uniform Electric Field
If electric potential and electric field depends only on one
coordinate, say r:
dV
E=–
rˆ [Note E is radial field]
dr
where r̂ is a unit vector along increasing r.
Example 13: The potential function of an electrostatic field
is given by V = 2x2. Determine the electric field strength at
the point (2m, 0, 3m).
dV
Sol. Ex =
−
⇒ Ex =
−4 x
dx
dr is along the increasing direction of r.
Electric field at x = 2m
⇒ Ex = –8NC–1 ⇒ E = –8 iˆ NC −1
Example 14: V = x2 + y, find E .
The above result can also be expressed in differential form as
Sol.
rB
– ∫ E.dr
dV
=
–
E
.
dr
⇒
V
–
V
=
B
A
∫
∫
rA

−dWE 
dV =− E ⋅ dr as dV =
q0 

r The potential of a point V = – E.dr
∫
∂V
∂V
∂V
= 2 x,
= 1 and
=0
∂x
∂y
∂z
 ∂V ˆ ∂V ˆ ∂V 
E = –  iˆ
+j
+k
 = –(2 xiˆ + ˆj )
∂y
∂z 
 ∂x
Electric field is nonuniform.
∞
P Electrostatic Potential and Capacitance
W
57
Example 15: Potential in the x-y plane is given as
V = 5(x2 + xy) volts. Find the electric field at the point
(1, –2).
∂V
= – (10 x + 5y) = –10 + 10 = 0,
∂x
∂V
Ey = −
= – 5x = –5 V/m
∂y
∴ E = (− 5 ˆj )V / m
EQUIPOTENTIAL SURFACES
As shown in figure if a charge is shifted from a point A to B on a
surface. M which is perpendicular to the horizontal uniform field
Sol. Ex = −
Example 16: Graph shows variation of potential V/s
distance. Calculate electric field at r = 1m, 1.5m, 2.5m, 5m.
V
(volts)
A E=0 B
2 E=–ve
E=+ve
1
C
4 5 6
3
0 1 2
r (m)
−dv
Sol. E = =
− slope =
− tan θ
dr
Distance
r=1
r = 1.5
r = 2.5
r=5
The work done in shifting
will obviously, be zero as
electric force is normal to
the direction of displaceE ment.
M
B
Equipotential
lines
A
As NO WORK is done in moving from A to B:
™
A and B are at same potential
™
All the points of surface M are at same potential
™
Surface M as equipotential surface.
Following figures show equipotential surfaces in the surrounding
of point charge and a line charged wire.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
E (V/m)
–1
–1
0
1
q
Concept Application
9. The potential at the point x (measured in µm) due
to some charges situated on the x-axis is given
by V(x) = 20(x2 – 4) volt. The electric field E at
x = 4µm
(a) (10/9) volt/µm and in the +ve x direction
(b) (5/3) volt/µm and in the –ve x direction
(c) (5/3) volt/µm and in the +ve x direction
(d) (10/9) volt/ µm and in the –ve x direction
10. The variation of potential with distance R from a fixed
point is as shown below. The electric field at R = 5 m is
Point charge
Line charge
Spherical equipotential surfaces Cylindrical equipotential surfaces
Every surface in electric field on which at every point direction

of E is normal to the surface can be regarded as equipotential
surface.
Figure shows two equipotential
surfaces in a uniform electric
field E.
d
Potential in volts
B
5
4
A
E
3
2
M1
1
0
1
2
3
4
5
6
Distance R in metres
(a) 2.5 volt/m
(b) –2.5 volt/m
(c) 2/5 volt/m
(d) –2/5 volt/m
58
M2
Potential difference between two points A and B = Potential
difference between the two equipotential surfaces on which the
points lie, given as
VA – VB = Ed
JEE (XII) Module-1 PW
Electric Potential Difference due to Infinite Wire
Figure shows a line charge with linear charge density l C/m.
+
+
+
+
+
+
+
+
+
+
+
+
r2
Y
P
E
x
r1
Potential difference between
two points X and Y which lie
on equipotential surfaces M1
& M2 = Potential difference
between surfaces M1 & M2.
We consider a point P at
a distance x from wire as
shown.
Electric Field at point P
M2
M1
X
EP =
Cm
– Vy
= Vx=
r2
r1
r1
Edx ∫
∫=
2K λ
dx
x
r 
Vx – Vy = 2 K λ n  2 
 r1 
Train Your Brain
Example 17: In a uniform electric field E = 10 N / C as
shown in figure, find: (i) VA – VB (ii) VB – VC.
A
2m
2m
B
400
3
200 ˆ
j N/C
E = 200iˆ −
3
cos 30º = 200
30º
200
3
=
400
3
sin 30º
400
3
Concept Application
11. A charge of 5C experiences a force of 5000 N when it
is kept in a uniform electric field. What is the potential
difference between two points separated by a distance
of 1 cm along the electric field line.
(a) 10 V
(b) 250 V
(c) 1000 V
(d) 2500 V
12. Equipotential surfaces are shown in figure. Then the
electric field strength will be
10V 20V 30V
30°
20
10
0cm
30
x
(a) 100 Vm–1 along X-axis
(b) 100 Vm–1 along Y-axis
(c) 200 Vm–1 at an angle 120° with X-axis
(c) 50 Vm–1 at an angle 120° with X-axis
C
2m
3
⇒ 20 = E × 0.1 ×
2
400
⇒E=
3
2K λ
x
Now the potential difference between surface M1 and M2 can be
given as
r2
Sol. VA – VB = E × 0.1 × cos 30°
Sol.
(i) VB > VA, so VA – VB will be negative
If d denotes effective displacement between two
points along the field, then
dAB = 2cos60° = 1m
∴ VA – VB = –EdAB = (–10)(1) = –10V
(ii) VB > VC, so VB – VC will be positive
Now dBC = 2.0 m
∴ VB – VC = 10(2) = 20V
Example 18: Write down the Electric field in vector form?
13. The potential difference between point A and B in the
given uniform electric field is
a
B
E
b
E
A
(a) Ea
(b) E (a 2 + b 2 )
(c) Eb
(d)
( Eb /
2)
ELECTRIC POTENTIAL DUE TO A DIPOLE
40
At an Axial Point
20
A
60º
10cm
0
B
Potential at P due to dipole (with p = 2aq)
(–a, 0)
–q
P Electrostatic Potential and Capacitance
W
O
(a, 0)
P
q
r
59
Vnet
=
kq
kq
−
(r – a) (r + a)
2akq
⇒ Vnet =
Vnet =
⇒V=
+q
(r 2 − a 2 )
kp
(As P = 2aq) and (r >> 2a)
r2
kp . r
q
x
x
q
O
a
a
Potential Due to Dipole at a General Point
P
r2
r
q
os
A
–
–q
ac
a
θ p
O
a
θ2
U1
θ1
∫ dU =
∫ pE sin θd θ ⇒ U
=
r1
We also know that the net torque on a dipole in electric field can
be given as
=
τ pE sin θ
t = 0 on dipole in electric field is zero in two situations when θ =
0° and θ = 180°.
Stable Equilibrium
B
kP cos θ
=
(a << r)
r2
kp cos θ
Potential at P due to dipole =
r2
Potential Energy of a Dipole in an External Field
Potential energy can be associated with the orientation of an
electric dipole in an electric field. Charge in potential energy is
related to the work done by electric field as
dU = −dWE = −τ ⋅ d θ
dU pE sin θ d θ
⇒ =
60
− U1 = − pE[cos θ2 − cos θ1 ]
We have discussed that when a dipole is in an electric field E, the
potential energy of dipole can be given as
U=
− pE cos θ
+
+q
kq (r + a cos θ − r + a cos θ)
(r 2 – a 2 cos 2 θ)
⇒ dU = −(− pE sin θkˆ) ⋅ (d θ kˆ)
2
Let θ = 90° is taken as reference position or zero of potential
energy.
∴ θ = 90° ⇒ U1 = 0
⇒ U2 – 0 = –pE[cosθ2 – cos 90°] ⇒ U = –pEcos θ
U =− p ⋅ E
qE
+
kq
kq
Vnet
=
−
(
r
–
a
cos
θ
)
(
r
+
a
cos θ) –q
Stable and Unstable Equilibrium of Dipole in
Electric Field
r
–q
⇒
U2
q
–q
r3
At an equatorial point, electric potential due to dipole is always
zero because potential due to +ve charge is cancelled by –ve
charge.
P
+q
q
At a Point on Perpendicular Bisector
q
+
–
qE
–
Figure (a)
When θ = 0 as shown in figure (a). when t = 0.
™ If we slightly tilt the dipole from its equilibrium position in
anticlockwise direction as shown by dotted position.
™ The dipole experiences a clockwise torque which tends to
rotate the dipole back to its equilibrium position.
™ This shows that at θ = 0, dipole is in stable equilibrium. We
can also find the potential energy of dipole at, it can be given
as U = –pE (minimum)
 At θ = 0, potential energy of dipole in electric field is minimum
⇒ Position of stable equilibrium.
JEE (XII) Module-1 PW
Unstable Equilibrium
When q = 180º, net torque on dipole is zero and potential energy
of dipole in this state is given as
U = pE (maximum)
–
n
=
qE
+
From this equilibrium position if dipole is slightly displaced
in anticlockwise direction.
Torque on dipole also acts in anticlockwise direction away
from equilibrium position.
Thus here dipole is in unstable equilibrium.
Note:
™ θ = 0°, Umin = –pE, t = 0. Stable Equilibrium Position
™
θ = 180°, Umax = +pE, t = 0. Unstable Equilibrium Position
™
θ = 90°, U = 0, t = pE.
(0,a) +q
q
+q O
(0, –a) –q
Example 19: Prove that the frequency of oscillation of
an electric dipole of moment p and rotational inertia I is
1  pE 

 for small amplitude about its equilibrium
2π  I 
position in a uniform electric field strength E.
Sol. Let an electric dipole (with charges q and –q 2a distance)
be placed in a uniform electric field of strength E.
+q
F
2a
P
q
E
–q
Restoring torque on dipole τ = − pE sin θ = − pE θ
(as q is small)
Here –ve sign shows the restoring tendency of torque,
 t = Ia
τ PE
θ
 Angular acceleration = α= =
I
I
For SHM a = w2q;
P Electrostatic Potential and Capacitance
W
dipole moment
p = q(2a)
Potential at point P due to charge q (placed at origin) is
kq
r
Potential due to remaining charge system (dipole) can be
written as
V0 =
=
Vd
Train Your Brain
F
y
Sol.
Figure (b)
Thus at q = 180º, dipole is in unstable equilibrium. This can also
be shown by figure (b).
™
ω
1  pE 
=


2π 2π  I 
Example 20: Three charges –q, +q and +q are situated in
X-Y plane at points (0, –a), (0, 0) and (0, a) respectively.
Find the potential at a distant point r(r >> a) in a direction
making an angle q from the Y-axis:
–
+
™
pE
I
Thus, frequency of oscillation of dipole is
qE
™
Comparing we get ω =
kp
kq ⋅ 2a cos θ
cos
=
θ
2
r
r2
Hence, VP = V0 + Vd
=
kq  2a cos θ 
1+

r 
r
Example 21: An electric dipole in a uniform electric field E
is turned from q = 0 position to q = 60° position. Find work
done by the field.
Sol. Work done by external agent (W)
= pE(cosq1 – cosq2)
(where p is the dipole moment of the dipole)
=
W pE (cos θ − cos
=
60°)
1
pE
2
1
− pE
Work done by the field = −W =
2
Example 22: A dipole of dipole moment P lies in a uniform
electric field E such that dipole direction is along field. If
dipole is rotated through 180º such that dipole direction
becomes opposite to the field direction. Find the work done
by the electrostatic field.
Sol. U i = − P.E = − PE cos θ = − PE
Uf = – P.E. cos (180°) = PE
work done by the field = – ΔU
= Ui – Uf = –2 PE
61
™
Concept Application
14. Electric potential at an equatorial point of a small
dipole with dipole moment P (r, distance from the
dipole) is
P
(a) Zero
(b)
4πε0 r 2
(c)
P
4πε0 r 3
(d)
2P
4πε0 r 3
15. Four points a, b, c and d are set at equal distance. The
electrostatic potential Va, Vb, Vc and Vd would satisfy
the following relation
a
+q
d
b
–q
c
(a) Va > Vb > Vc > Vd
(c) Va > Vc = Vb = Vd
(b) Va > Vb = Vd > Vc
(c) Vb > Vd > Va > Vc
CONDUCTORS
™
™
Inside a conductor, electrostatic field is zero: In the
previous chapter, we have already discussed that “when there
is no electric current inside or on the surface of a conductor,
the electric field inside the conductor is zero”.
At the surface of a charged conductor, electrostatic field
must be normal to the surface at every point: If the field is
not normal to the surface, it will have a non zero component
along the surface. Hence the free charge on the surface will
move due to electrostatic force on it. But free charge on the
surface in electrostatics remains at rest. So the electrostatic
field at the surface of a charged conductor must be normal to
the surface.
–
–
+
–
+
–
++
+ –––
++ ––
+ ––
+
+ –– E= 0 + –
–
+ –
+ ––
–– +++ –
+
–
+
+
Metallic sphere in electric field
™
62
The interior of a conductor can have no excess charge in
the static situation.
™
Electrostatic potential is constant throughout the
volume of the conductor and has the same value
(as inside) on its surface : Since the field E = 0 inside the
conductor and its surface has no non zero component along
dV
dV
the surface, so the potential gradient
= 0 because −
dr
dr
is equal to the component of the field along dr. If we take dr
dV
along the surface or inside the surface, then −
0 ⇒V
=
dr
is constant
σ
Electric field at the surface of a charged conductor: E = nˆ
ε0
where s is the surface charge density which may be positive
or negative and n̂ is the unit vector normal to the surface in
the outward direction.
Electric Field Due To A Conductor Near Surface
Suppose we have a conductor and at any ‘A’, local surface charge
density = σ. We have to find electric field just outside the conductor
surface.
+ +
+
+ A
+
E=?
+
+
+
+
+
+
+
+
+
+
+ +
For this, let’s consider a small cylindrical Gaussian surface, which
is partly inside and partly outside the conductor surface, as shown
in figure. It has a small cross section area ds and negligible height.
+
(2)
(3)
Ein = 0
ds
E
(1)
Cylindrical
gaussian
surface
Applying Gauss’s theorem for this surface:
φnet =
qin σdS
=
∈0
∈0
Flux through surface (1) f1 = EdS
(because E is normal to the surface of conductor)
™ Flux through surface (2) f2 = EdS
(E is normal to curved Gaussian surface)
™ Flux through surface (3) f2 = 0
(as E inside the conductor = 0)
σdS
σ
⇒ E=
So, EdS=
ε0
ε0
™
JEE (XII) Module-1 PW
Electric field just outside the surface of conductor:
σ
E =
(direction will be normal to the surface)
ε0
σ
In vector form: E = nˆ (Here, n = unit vector normal to the
ε0
conductor surface)
Electrostatic Pressure At The Surface Of The
Conductor
Suppose a conductor is given some charge. Due to repulsion, all
the charges will reach the surface of the conductor. But the charges
will still repel each other. So an outward force will be felt by each
charge due to others. Due to this force, there will be some pressure
at the surface, which is called electrostatic pressure.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
σ
⇒ Es + Er =
ε0
Now, lets see the electric field just inside the metal surface. Here,
electric field due to the remaining charges (Er) will be in the same
direction (normally outward), but the electric field due to the small
element will be in opposite direction (normally inward)
So net electric field just inside the metal surface = Er – Es and we
know that electric field inside a conductor = 0
ds
Es
+
+
So Net electric field just outside the surface = Es + Er and we have
σ
proved that electric field just outside the conductor surface =
ε0
+
+
To find the electrostatic pressure, lets take a small surface element
having Area ‘ds’.
Force on this element due to the remaining charges:
dF = Er(dq)
ds
Er
So, Er − Es =0 ⇒ Er =Es
from eqn. (ii) and eqn. (iii), we can say that :
2 Er =
σ
σ
⇒ Er =
2ε0
ε0
Now, we can easily find the pressure from eqn.(i)
=
P
( Er )( σ=)
σ
σ2
( σ=)
2ε0
2ε0
So, electrostatic pressure at the surface of the conductor P =
σ2
2ε0
where, σ = local surface charge density.
 electric field at   charge of 



dF =  that place due to   the small 
 remainig charges   element 



Let electric field at that point due to the remaining charges = Er
and charge of the small element = dp = σds
⇒ dF=
( Er )( dq )= ( Er )( σds )
So, pressure on this small element
dF ( Er )( σds )
=
⇒ P = ( Er )( σ )
ds
ds
Now to find pressure, we have to find Er (electric field at that
position due to the remaining charges)
Suppose, electric field due to the small element near the surface = Es
Electric field due to the remaining part near the surface Er
At a point just outside the surface, electric field due to the small
element (Es) will be normally out wards, and electric field due to
the remaining part (Er) will also be normally out wards.
Es E
r
P=
ds
P Electrostatic Potential and Capacitance
W
ELECTROSTATIC SHIELDING
Electrostatic shielding is a method of shielding any sensitive
instrument/building from the external field produced by external
charges. The instrument is enclosed inside a conductor with a
cavity (Fraday’s cage). The charge on the outer surface of the
conductor redistributes to cancel the external electric field inside
the conductor and cavity.
(i) There is no net charge anywhere on the surface of the cavity.
i.e., σ = 0
(ii) The potential inside (or on the surface of) the cavity
= The potential inside the material of the conductor
= The potential on the surface of the conductor
= V0 (constant)
V =V0
++
– –
– V =V 0(constant)++ E
E –
+
–
+
–
V =(v constant)
+
–
E
+
–
–
+ +
––
+
–
+
E
–
E
++
conducting body
E
0
63
(iii) The electric field in (or on the surface of) the cavity
= The electric field inside material of the conductor
= zero
But the electric field on the surface of a charged conductor is
not zero.
[Figure shows an uncharged conductor in an external
electric field E. The free electrons in the conductor distribute
themselves on the surfaces as shown in the figure so that the
net electric field inside the conductor becomes zero and the
net field at the surface is perpendicular to the surface.]
Train Your Brain
Example 23: Three parallel plates each of area A are kept
as shown in figure and charges Q, – 2Q and 3Q are given
to them as shown
Q
+
+ + + +
+ +
2
0
+ + +
+ +
+
r2
r1
+
E2 =
+ + +
Corona Discharge
Except for spherical surfaces, the charge is not distributed
uniformly on the surface of a conductor.
E1 =
1
0
Figure shows a charged conductor with nonuniform
distribution of charge. σ1 and σ2 are the surface charge
densities at the portions where radii of curvature are
r1 and r2 then
σ1 r2
=
.
σ2 r1
At the sharp points or edges, the surface charge density (σ) is very

σ
high and hence the electric field  E =  becomes very strong.
ε0 

The air around such sharp points may become ionised, producing
the corona discharge in which the charge jumps from th conductor
to the air because of the dielectric breakdown of the air.
Note: (i) The corona discharge from the pointed end of the
conductor at the top of a building protect the building
from lightning strikes.
(ii) It is electrostatic shielding that protects a person from
lightning strikes if he is in a car.
Figure shows an uncharged conducting plate placed in a region
having electric field E. Due to electric force on the free electrons
in the conductor, they redistribute themselves till the electric field
produced due to the redistribution neutralizes the original field.
Hence the net field inside the conductor becomes zero.
–q
+q
+
+
+
+
+
+
+
+
+
+
+
+
+
+
–
–
–
E
–
–
–
E
–
–
– Enet=0
–
–
–
– metal
+q
–q
+ –
–
+ –
+
–
+
+ –
–
+
–
+
+ –
+ –
+ –
–
+
+ –
–
+ –
Figure: A metallic slab between two charged plate. The field E
due to the charged plate is directed towards right and the field E
due to the induced charge in the slab is directed towards left, and
hence the net field inside the slab becomes zero.
64
a
3Q
–2Q
c
b
e
d
Q
P
f
R
d
2d
(i) Find the resulting charge distribution on all the six
surfaces. (Neglect the edge effect).
(ii) Draw the graph of Electric Field as the function of
x. (Take x = 0 at P)
(iii) Draw the graph of Electric Potential as the function
of x. (Take V = 0 at P)
Sol. (i)
qa
qb
qc
qd
qe
Q
P
qf
R
As the total charge on each plate remains the same.
⇒ qa + qb = Q ⇒ qc + qd = –2Q ⇒ qe + qf = 3Q
⇒ qa + qb + qc + qd + qe + qf = 2Q
As P lies inside the conductor, electric field at point
P is zero.
At P, charge qa will given an electric field towards right.
All other charge qb, qc ..... etc. will give the electric field
towards left. So,
1
0
( qa − qb − qc − qd − qe − qf ) =
Ep = 0 ⇒
ZA ∈0
⇒
1
( 2q a − 2Q ) =0 ⇒ q a =Q ⇒ qb = 0
ZA ∈0
By considering a Gaussian at P and Q, we can show that
qb = –qc ⇒ qc = 0 ⇒ qd = –2Q
Similarly qc = –qd = 2Q ⇒ qf = Q
(ii) As E =
q
2A ∈0
E
Q
A0
–Q
–2Q A0
A0
d
3d
x
JEE (XII) Module-1 PW
Shorting of Conductors
A low resistance connection between two conductors supplying
electrical power to a circuit.
Q2,V
Q1,V
(iii) As ∆V =−E . r
V
4Qd
0A
O
d
a
x
3d
b
Electrostatic equilibrium
(V = common potential)
Concept Application
16. Two infinite sheets of uniform charge density +s and
–s are parallel to each other as shown in the figure.
Electric field at the
+
–
+
+s +
–s
–
Let Q1 and Q2 are the charges at electrostatic equilibrium
At this state, the potential of the spheres are same,
⇒
Q1
Q2
Q
a
=
⇒ 1 =
4πε 0 a 4πε 0 b
Q2 b
Also, Q = Q1 + Q2
(Total charge on the two sphere)
Qa
Qb
=
, Q2
( a + b)
( a + b)
=
⇒ Q1
–
Earthing of Conductor
+
–
+
–
The process of transferring of charge from a charged object to the
earth is called earthing.
An earthed conductor is always at zero potential due to flow of
charge either from the earth to the conductor or from the conductor
to the earth
(a) Points to the left or to the right of the sheets is
zero
(b) Midpoint between the sheets is zero
(c) Midpoint of the sheets is σ/ε0 and is directed
towards right
(d) Midpoint of the sheets is 2σ/ε0 and is directed
towards right
Note:
™ Earth is assumed to be a conducting sphere having radius
R = 6400 km. Its potential V = 0.
™ Earthing is to capture high electric sparks and moving them
to ground to minimize its effect.
17. Consider a non-spherical conductor shown in the figure which is given a certain amount of positive charge.
The charge distributes itself on the surface such that
the charge densities are σ1, σ2 and σ3 at the region 1,
2 and 3 respectively. Then
A conducting shell with charge q.
Figure (a)
3
1
2
(c) σ3 > σ1 > σ2
(b) σ2 > σ3 > σ1
(d) σ2 > σ1 > σ3
B
(a) σA < σB < σC < σD
(c) σA > σC > σB > σD
D
Q
(b) σA = σB = σC > σD
(d) σA = σB = σC = σD
P Electrostatic Potential and Capacitance
W
+q
Figure (b)
– –
–
– –
q –
+
+ + ++
–
–
–
–
–q
–
–
–
–
–
–
–
– –
– –
–
++
C
–
–
–q –
–
–
++++
P
–
– –
– –
The conducting shell with charge
q has been surrounded by another
larger conducting shell which is
uncharged.
™ Charges –q and +q are induced
on the its inner and outer surfaces
but net charge in the outer shell is
still zero.
The free charge on the outer surface
goes to the earth but the inner charge
remains bounded to the charge on
the inner shell so that the potential of
the outer shell connected to the earth
becomes zero.
™
–
++
A
q
++++
18. A conductor PQ as shown in figure is charged. σA, σB,
σC and σD are surface charge densities of points A, B,
C and D respectively.
–
–
–
–
–
– –
–
– –
++ + + +
(a) σ1 > σ2 > σ3
3
–
Figure (c)
(Outer shell of figure
(b) is earthed)
65
1
++ +
+++++
+++
++
++
Two concentric conducting shells.
Some charge q1 is given to the outer
shell. No charge is developed on the
inner shell.
++++++++
R +
++
+
++
+++++
++
R2
q1
+q
–q
q
+ ++++
++
+++
++
Figure (d)
™
+++++
+++
+ q
++ q
R + 1
++ +
+++++
++
+++++
+++
++
++
++
2
1
–– –– –
–
–
–
–
–
–
– R2
–
–
–
–
–
–– – –
–
™
+ ++++
++
+++
++
–
–
–
–
–
Q
–
–
– + +
+
–
+
+
+
+
Ep
– –
+ +
r
A positive point charge Q has been
placed at a distance r from the center
of the sphere. The point charge Q
exerts force on the electrons in the
sphere and hence the free electrons
redistribute themselves so that the
left half is negatively charged and the
right half is positively charged. The
charge distribution is non-uniform.
+
R
Ei
+
+
(b)
Figure (f)
At the center of the sphere, the
field due to the point charge Q is
1 Q
EP =
, toward right.
4πε0 r 2
Since the field inside the conductor is
zero, the field due the induced charge
on the sphere is equal in magnitude of
EP but opposite in direction.
i.e., Ei =
Qi
R
Q
4πε0 r 2
+q+Q
r
–q
q
(iii) Resultant field, due to q (which is inside the cavity) and
induced charge on S1, at any point outside S1 (like B, C)
is zero. Resultant field due to q + Q on S2 and any other
change outside S2, at any point inside of surface 32
(like A, B) is zero
S2
.B
S1
.q .A
–q
.C
q+Q
(iv) Resultant field in a charge free cavity in a closed conductor
is zero. There can be charges outside the conductor and on
the surface also. Then also, this result is true. No charge
will be induced on the inner most surface of the conductor.
. The potential at
any point of the sphere = potential at
its center.
The sphere is earthed, so
V = 0 hence
1  Q Qi
+
4πε0  r R
Q

0 or, Qi = – QR/r
=

(c)
Figure (g)
(Outer shell of figure
(f) is earthed)
Charge Distribution in Conductor With Cavity
(i) If a charge q is kept in the cavity then –q will be induced
on the inner surface and +q will be induced on the outer
surface of the conductor (it can be proved using Gauss
theorem)
66
(ii) If a charge q is kept inside the cavity of a conductor and
conductor is given a charge Q then –q charge will be induced
on inner surface and total charge on the outer surface will
be q + Q. (it can be proved using Gauss theorem)
R2
1  q2 q1 
0 or q2 = −q1
 + =
R1
4πε0  R2 R1 
Figure (e)
–
The inner shell is earthed
and hence some charge q2
is developed on it so that its
potential becomes zero.
On the surface of the inner
shell, the net potential is
No
charge
Train Your Brain
Example 24: Two concentric spheres of radii R1 and R2
(R1 < R2) have charges q1 and q2 respectively distributed
uniformly over their surfaces.
(i) Find the potential difference between the two
surfaces.
(ii) Find the charges on all the surfaces of the shells.
(iii) What will happen if the spheres are connected by
a conducting wire?
JEE (XII) Module-1 PW
Sol.
(iii)When the shells are connected, they attain equal
potential by redistributing their charges. Let x and y be
the final charges on them.
x + y = q1 + q2
q2
q1
R1
x
y
x
y
+
=
+
4πε 0 R1 4πε 0 R 2 4πε 0 R 2 4πε 0 R 2
2
R
O
(i) Potential of inner sphere:
V1 = potential due to q1 + potential due to q2
=
V1
q1
+
q2
q1
+
q2
4πε 0 R1 4πε 0 R2
Potential of outer sphere:
V2 = potential due to q1 + potential due to q2
=
V2
4πε 0 R2
Solve to get x = 0 and y = q1 + q2
Hence all the charges from inner shell flows to the outer
shell.
Example 25: A charge q kept in front of a neutral metallic
sphere. Find the electric field at the center of sphere due
to the induced charges on the surface of the sphere shown.
d
4πε 0 R2
q  R − R1 
Potential Difference: V1 − V2 =1  2

4πε 0  R1 R2 
Note: That potential difference does not depend on the
charge of the outer sphere.
Sol. The net electric field inside the conductor is zero. This
is produced by induced charges and the external charge.
+ E
q
+
+
+q1 + q
+
– – – 2+
–
–
+ –
q1
q +
+
–
+– 1
–
No
– +
–
Charge
+
+ – +
–
+
+
– +
+ ––
–
– – –
+
+
+
qencl
=
0 =
0
( as E 0 is a conductor ) ⇒ qencl =
ε0
Hence there is no charge on the inner most surface.
So the charge q, lies entricly on the outer surface of
the inner shell.
2. By passing another Gaussian surface through the
q
outer shell. ∫ E . d A = encl
ε0
qencl
=
0 =
( as E 0 is a conductor ) ⇒ qencl =
0
ε0
As there is charge q1 inside this Gaussian surface,
there must be a charge –q1 also on the inner surface
of another shell.
Hence charge on outer surface of outer shell is
q1 + q2.
P Electrostatic Potential and Capacitance
W
q
R
(ii)
By passing a Gaussian surface through the inner shell. We
see that by Gauss’s Law.
qencl
1. ∫ E . d A =
ε0
(as V1 = V2)
EiN
–
–
–
q
⇒ EInduced + Eq = 0
⇒ EInduced = −
q
4πε 0 d 2
Example 26: Two concentric hollow metal spheres have
radii R1 and R2. The outer sphere is given a positive charge
Q and the inner is earthed. What is the charge on the inner
sphere? (R2 > R1)
Q
R1
R2
Sol. Let us suppose the charge on the inner sphere be x, then
1 x Q
+
the potential of the inner sphere=
will be
4πε0  R1 R2 
This potential has to be made equal to zero because the
inner sphere is grounded
So,
1 x Q
+
=
0
4πε 0  R1 R2 
∴
x = −Q
R1
R2
67
Concept Application
19. Figure shows two conducting thin concentric shells of
radii r and 3r. The outer shell carries charge q. Inner
shell is neutral. Find the charge that will flow from inner
shell to earth after the switch S is closed.
q
q
q
(c) −
(d) +
3
2
2
20. Figure shows a system of three concentric metal shells
A, B and C with radii a, 2a and 3a respectively. Shell
B is earthed and shell C is given a charge Q. Now if
shell C is connected to shell, A then the final charge
on the shell B, is equal to
(a)
−
q
3
(b) +
3a
2a
(a) –4Q/3
(b) –8Q
C
(c) –5Q/3
(d) –3Q/7
21. Two equally charged spheres of radii a and b are
connected together. What will be the ratio of electric
field intensity on their surfaces?
(a) a/b
(b) a2/b2
(c) b/a
(d) b2/a2
CONCEPT OF CAPACITANCE
When a conductor is charged then its potential rises. The increase
in potential is directly proportional to the charge given to the
conductor.
Q ∝ V or Q = CV
The constant C is known as the capacity of the conductor.
SI Unit :- coulomb/volt or farad (F)
1 farad = 9 ×
1011
statfarad (CGS unit).
Capacitor or Condenser
(a) A capacitor is a device that stores electric energy or a
capacitor is a pair of two conductors of any shape, which
are close to each other and have equal and opposite
charge.
68
++
+
+ ++
++
+ ++
++
–
–
–
–
–
–
–
–
(c) A capacitor get’s charged when a battery is connected
across the plates. Once capacitor get’s fully charged, flow
of charge carriers stop in the circuit and in this condition
potential difference across the plates of capacitor is same
as the potential difference across the terminals of battery.
(d) net charge on a capacitor is always zero, but when we
speak of the charge Q on a capacitor, we are referring
to the magnitude of the charge on each plate.
Parallel Plate Capacitor
The arrangement consist of two thin conducting plates, each of
area A and separated by a d distance. When charge q is given to
first plate, a charge –q is induced on the inner face of other plate
and positive on the outer face of plate. As this face is connected
to earth, a net negative charge is left on this plate. Thus, the
arrangement is equivalent to two thin sheets of charge.
a
B
(b) The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive plate divided by
the magnitude of the potential difference V between the
Q
plates i.e., C =
V
The electric field between the plates is
E=
σ
, where s is the charge density.
ε0
V
;
d
V is the potential difference between the plates.
As E =
+Q
Area = A
+
+
+
+
+
+
+
+
+
–Q
–
–
–
– σ
– ε0
–
–
–
=
=
V Ed
=
V
E=
d
V
–
C
=
C=
σd
ε0
Qd
Q
=
as σ
Aε0
A
Q
Q
=
V  qd 
 Aε 
 0
ε0 A
d
Figure (a)
Energy Stored in A Capacitor
During the charging of a capacitor, work has to be done to add
charge to the capacitor against its potential. This work is stored in
the capacitor as electric energy.
JEE (XII) Module-1 PW
Suppose during the charing of capacitor its potential at any instant
q
is given by V =
small amount of work done in adding a charge
C
q
dq is given by dW = dq
C
Q
q
The work done in giving a charge Q to the condenser is W = ∫ dq
C
0
Q
 q2 
∴ W = 
 2C  0
∴
W=
∴ =
U
Q2
2C
1
Q2 1
2
=
CV=
QV
2C 2
2
Energy Density in Parallel Plate Capacitor
1
CV 2
U
A
2
where C =
∴ Energy density=
u =
ε0 ,V =
Ed
Ad
Ad
d
u=
1  ε0 A   E 2 d 2


2  d   Ad
ε0 A (8.85 × 10 –12 C 2 / N ⋅ m 2 )(0.0280m 2 )
=
C =
0.550 × 10 –3 m
d
We obtain the capacitance of the parallel plate capacitor
C = 4.51 × 10–10 F.
Since Q = CV. we have Q = (4.51 × 10–10 F)
(20.1 V) = 9.06 × 10–9 C.
Example 29: If the distance between the plates of a capacitor
is made half and the area of plates is doubled, then the
capacitance will become:
(a) Twice
(b) Four times
(c) Half
(d) One-Fourth
1
and C ∝ A
d
A
⇒ C∝
d
Sol. (b) C ∝
The volume of a parallel plate capacitor is Ad
∴
Sol. Using the formula
 1
ε0 E 2
=
 2
C1 A1 d 2
A1 d1 / 2 1
.
⇒
=
=
=
C2 A2 d1 2 A1 d1
4
⇒ C2 = 4C1
Force Between the Plates of Parallel Plate Capacitor
σ
As the field due to charge on the plates on the other is E =
,
2ε0
the force exerted by one plate on the other
2
 σ  σ 
=
F=
qE =
(σA) 

A

 2ε0   2ε0 
Concept Application
22. A thin metal plate P is inserted between the plates of
a parallel plate capacitor of capacity C in such a way
that its edges touch the two plates figure. The capacity
now becomes
2
F σ
.
So force per unit area   =
 A  2ε 0
P
Train Your Brain
Example 27: A capacitor of 0.75 µF is charged to a voltage
of 16 V. What is the magnitude of the charge on each plate
of the capacitor?
Sol.Using q = CV, we get
q = CV = (0.75 × 10–6F) = 1.2 × 10–6 C
Example 28: A parallel-plate capacitor is constructed with
plates of area 0.0280 m2 and separation 0.550 mm. Find
the magnitude of the charge on each plate of this capacitor
when the potential difference between the plates is 20.1 V.
+Q +
+
+
+
+
+
–Q –
–
–
–
P Electrostatic Potential and Capacitance
W
–
–
(b) 2C
(c) Zero
(d) ∞
23. The capacitance of a capacitor is C when the distance
between the plates is halved, the capacitance will
becomes :
(a) Half
(b) Twice
(c) One-fourth
(d) Four-times
Case I: In series
d
–
C
2
Expression For Equivalent Capacitance
+
E
(a)
Consider three capacitors connected in series. Consider the shown
arrangement of three capacitors in series. Let a V potential is
applied to circuit and +Q be the charge, which will come on all
the capacitors. Let V1, V2, V3 be the individual potentials of the
capacitors and their respective capacitance is C1, C2, C3.
69
C2
C1
V1
C3
∴ They can be replaced by a single capacitor of
capacitance given by,
V3
V2
C = 10µF + 20µF = 30µF
V
Thus, the circuit can be shown like this. Now, the
capacitors 15µF and 30µF are arranged in a row (series
combination).
where C is the effective capacitance.
∴ The equivalent capacitances is given by C such that,
1 1 1
=
+
or C = 10µF.
C 15 30
A
C
B
Q Q Q Q
= +
+
C C1 C2 C3
1
1
1
1
= +
+
C C1 C2 C3
∴
∴
Case II: In parallel
15µF
Consider three capacitors connected in parallel. A voltage V is
applied across the combination Q1, Q2, Q3 be the charge on each
capacitor and C1, C2, C3 be their respective capacitance.
A
Q1 = C1V, Q2 = C2V, Q3 = C3V
As Q = CV
Q1
C1V
Q2
C2V
B
C3
(a)
Sol. Figure (a) has been simplified as in figure (b) which is
a parallel combination of C1, C2 and C3.
C1
V
C2
A
which C is the net capacitance of the combination,
∴ Q = Q 1 + Q2 + Q3
B
C3
Ceq
A
(c)
B
(b)
∴ CV = C1V + C2V + C3V
Where Ceq = C1 + C2 + C3
∴ C = C1 + C2 + C3
If C1 = C2 = C3 = C,
Train Your Brain
Example 30: In the arrangement shown. Find the equivalent
capacitance between A and B.
10µF
A
B
15µF
20µF
Sol. First, mark different junctions (nodes) in the circuit. A
node is a point where more than two wires are connected.
As we can clearly see that, capacitors 10µF and 20µF
are connected between same points C and B (Parallel
combination)
10µF
70
C2
C1
Q3 C3V
A
30µF
Example 31: Find the equivalent capacitance of the
combination between A and B in the figure.
15µF C 20µF
B
then Ceq = 3C
Concept Application
24. Four capacitor each capacity 4µF are connected as
shown in figure VP – VQ = 15 volts. The energy stored
in the system is :
4µF
4µF
4µF
P
Q
4µF
(a) 2400 erg
(b) 1800 erg
(c) 3600 erg
(d) 5400 erg
JEE (XII) Module-1 PW
25. The resultant capacitance between A and B in the
following figure is equal to
3mF
3mF
Equivalent Capacitance of Infinite Grid of Identical
Capacitors
The equivalent capacitance between points A and B.
3mF
A
2mF
2mF
3mF
A
B
B
3mF
3mF
3mF
(a) 1 µF
(b) 3 µF
(c) 2 µF
(d) 1.5 µF
26. In the circuit shown in the figure, the potential
Q
4
Now let Q charge comes out from point B, then charges get
collected equally from four branches.
difference across the 4.5 mF capacitor is
3mF
3mF
3mF
A
2mF
2mF
Capacitors of capacitance C0 each
By superposition method, Let charge Q enters at A and in gets
distributed equally in four branches.
3mF
∴
q AB =
∴
q AB =
Q
4
\ By superposition, QnuVAB) =
B
3mF
3mF
8
volts
3
(c) 6 volts
(a)
3mF
⇒
Q Q Q
and Q = Ceq × V
+ =
4 4 2
Ceq × V
= C0 × V ⇒ Ceq = 2C0
2
(b) 4 volts
(d) 8 volts
Equivalent Capacitance of Wheatstone Brdige
When five capacitors are connected in Wheatstone bridge
C
C
arrangement as shown, such that 1 = 3 , the bridge is balanced
C2 C4
and C5 becomes ineffective. Nor charge is stored on C5. Therefore
C1, C2 and C3, C4 are in series. The two series combinations are
in parallel between A and C. Hence equivalent capacitance can be
calculated.
B
Train Your Brain
Example 32: Find the equivalent capacitance of the
combination between A and B.
C
C
C
C/2
C/2
Sol. The circuit is a balanced Wheatstone bridge
C
C/2
C
C2
C1
C5
A
C/2
C/2
Here C1 = C2 = C
C
C=
C=
3
4
C4
C3
D
P Electrostatic Potential and Capacitance
W
∴
C/4
C
2
C1 C3
=
C2 C4
Hence Ceq =
C C 3C
+ =
2 4
4
71
Example 33: Find the equivalent capacitance of the
combination between A and B in the figure.
2µF
A
Sol. (d) Given circuit can be drawn as
Here C, D are equipotential points
∴ 24 µF capacitor holds no charge as
4µF B
5µF
3µF
6µF
Sol. The simplified form of the given combination has been
shown in the figure.
4µF
D
4µF
2µF
D
2µF
BA
A
3µF
3µF
6µF
C
B
C
6µF
C1 2µF 1
C3 3µF 1
=
Here=
⇒=
=
C2 4µF 2
C4 6µF 2
Thus,
C1 C3
=
C2 C4
Hence
=
Ceq
∆V = 0
12 × 6
9 × 18
∴ Ceq =
+
= 10 µF
12 + 6
9 + 18
Concept Application
CC
C1C2
+ 3 4
C1 + C2 C3 + C4
10
 2 × 4 3× 6 
4

= 
+
µF
 µF =  + 2  µF =
3
 2+ 4 3+ 6 
3

27. The effective capacitance between points X and Y
6µF
6µF
X
Example 34: In the connection shown in the adjoining
figure. The equivalent capacity between A and B will be
6µF
12µF
A
9µF
B
24µF
(c) 18µF
(d) 24µF
9µF
3µF
B
2µF
12µF
24µF
18µF
72
(b) 12µF
A
10
 2 × 4 3× 6 
4

= 
+
µF
 µF =  + 2  µF =
3
 2+ 4 3+ 6 
3

Example 35: In the connection shown in the adjoining
figure. The equivalent capacity between A and B will be
6µF
8µF
12µF
20µF
10µF
(a) 6µF
1µF
CC
C1C2
Sol. Hence
=
Ceq
+ 3 4
C1 + C2 C3 + C4
(a)
(b)
(c)
(d)
8µF
(b) 12µF
(d) 10µF
A
6µF
28. The capacitance between the points A and B of the
combination is
12µF
(a) 8µF
(c) 20µF
Y
B
(a) 9 µF
4
(c)
32
µF
12
5µF
6µF
(b)
87
µF
12
(d)
92
µF
12
DIELECTRIC
Dielectrics are insulating (non-conducting) materials which
transmits electric effect without conducting.
Dielectrics are of Two Types
(a) Polar dielectrics: A polar molecule has permanent electric
dipole moment ( p ) in the absence of electric field also.
But a polar dielectric has net dipole moment zero in the
absence of electric field because polar molecules are
randomly oriented as shown in figure.
JEE (XII) Module-1 PW
–
+
+
–
–
–
+
+
+
+
–
–
The electric field E0 between the plates due to the charges +Q and
–Q on the capacitor is
–
E=
0
+
+
–
In the presence of electric field polar molecules tends to
line up in the direction of electric field, and the substance
has finite dipole moment e.g. water, alcohol, CO2, NH3,
HCl etc. are made of polar atoms/molecules.
(b) Non polar dielectric: In non-polar molecules, Each
molecule has zero dipole moment in its normal state.
When electric field is applied, molecules becomes induced
electric-dipoles e.g. N2, O2, Benzene, Methane etc. are
made of non-polar atoms/molecules.
In general, any non-conducting material can be called as
a dielectric but broadly non-conducting material having
non-polar molecules referred to as dielectric.
(c) Polarization of a dielectric slab: It is the process of
inducing equal and opposite charges on the two faces of the
dielectric on the application of electric field.
+
+
+
+
+
–+
–+
–
–
–
–
–
–+
–+
F
–+
–+
–+
–+
–+
–+
E
Electric field between the plate with air
= K
E ' Electric field beween the plates with medium
(iii) K is also known as relative permittivity (εr) of the
material
(d) Dielectric breakdown and dielectric strength: If a very
high electric field is created in a dielectric. The dielectric
then behaves like a conductor. This phenomenon is known
as dielectric breakdown.
The maximum value of electric field (or potential gradient)
that a dielectric material can tolerate without it’s electric
breakdown is called it’s dielectric strength.
kV
.
mm
V
but
m
Effect of Dielectric on Capacitance
Figure shows a parallel plate capacitor with plate area A and
separation d between the plates.
A dielectric slab of dielectric constant K is inserted between the
plates filling the entire space between the plates. The plates of the
capacitor are given charge +Q and –Q and hence induced charges
–QP and +QP appear on the surfaces of the slab.
P Electrostatic Potential and Capacitance
W
EP
+QP
–QP
+
+
+
+
+
EP
K
+
The net field E in the dielectric is
E
–Q
–
–
–
–
–
–
E
Q
=
K ε0 AK The potential difference between the plate is
Qd
V = E ⋅d =
ε 0 AK
E=
0
or C = CK0
(i) Electric field between the plates in the presence of
dielectric medium is E′ = E – Ei where E = Main field,
E′ = Induced field.
(ii) Dielectric constant of dielectric medium is defined as
practical unit is
+Q
(directed towards right)
K ε0 A
Q
Q
=
=
V  Qd 
d
 ε AK 
 0

(Capacitance with dielectric)
So, the capacitance is C
=
E
S.I. unit of dielectric strength of a material is
σ
Q
=
ε0 ε0 A
...(i)
...(ii)
ε0 A
is the capacitance without the dielectric. Thus,
d
capacitance is increased to K times when the space between the
plates is filled with a dielectric constant K. The product e0K in
equation (i) is called are permittivity of the medium and is denoted
ε
by e. So, ε = ε0 K or K =
ε0
where C0 =
Thus, we see that =
K
ε A
E0 C
ε
and C0 = 0
= =
d
E C0 ε0
(capacitance of parallel plate capacitor)
Effect of Dielectric Slab (Inserted Along the Length
of Plates)
When a dielectric slab is placed, between the plates of capacitor its
polarisation take place. Thus a charge –Qi appears on its left face
and +Qi appears on its right face, as shown in diagram.
+Q
–Qi
–Qi
–Q
σ
+
–
E0 =
+
–
–
+
ε
0
–
+ +
–
+ E –
E
–
E
0
0 – EP
–
+ +
–
+ 0 –
+
–
– EP = σi
+
–
+
+
–
ε0
–
+
–
+
+
–
– σ =Qi
+
–
+
i
t
A
d
This causes a decrease in electric field between the plates.
73
E0
K
Here K is dielectric constant (or er that is relative permittivity)
The net field inside a dielectric is E, such that E =
(vi) If a dielectric slab is partially filled between the plates
⇒ C'=
Not field is also given by E = E0 – EP (where EP is electric field
due to polarisation)
E0
= E0 − EP
K
E
E=
E0 − 0
P
K
σi σ
σ
=
−
ε0 ε0 K ε0
∴
⇒
or,
A
1

Qi Q 1 − 
⇒ =
 K
Potential difference between the plates can be calculated as,
ε0 A
t 

d −t + 
K


+
+
+
+
+
L
K
–
–
–
–
–
E
d
(vii) If a number of dielectric slabs are inserted between the
plate as shown.
t 

=
V E0 (d − t ) + =
Et E0  d − t + 
K

∴ =
V
t 
Q 
t 
σ
d −t + =

d −t + 
ε0 
K  Aε0 
K
As C =
Q
V
⇒ C=
ε0 a
d −t +
t1
t
K
Note: When t = d, C =
C'=
K ε0 A
d
Capacity of Various Capacitor Parallel plate
capacitor
E
(i) Electric field between the plates: =
t2 t3
d
ε0 A
 t

t
t
d − (t1 + t2 + t3 + ....) +  1 + 2 + 3 + ..... 
 K1 K 2 K3

(viii) When a metallic slab is inserted between the plates
It consists of two parallel metallic plates (may be circular,
rectangular, square) separated by a small distance. If A = Effective
overlapping area of each plate.
C'=
ε0 A
(d − t )
t
σ
Q
=
ε0 Aε0
(ii) Potential difference between the plates: V=E×d=
σd
ε0
ε A
(iii) Capacitance: C = 0
d
(iv) If a dielectric medium of dielectric constant K is filled
completely between the plates then capacitance increases
by K times i.e. C ' = K ε0 A
d
⇒ C′ = KC
(v) The capacitance of parallel plate capacitor depends on
1
A(C ∝ A) and d  C ∝  . It does not depend on the
d


charge on the plates or the potential difference between
the plates.
74
K1 K2 K3
A
A
K=
–
–
–
–
–
d
If metallic slab fills the complete space between the plates
(i.e. t = d) or both plates are joined through a metallic
wire then capacitance becomes infinite.
(ix) Force between the plates of a parallel plate capacitor.
=
|F|
σ2 A
Q2
CV 2
= =
2ε 0 2ε 0 A
2d
(x) Energy density between the plates of a parallel plate
capacitor
Energy 1
=
ε0 E 2
Energy density =
Volume 2
JEE (XII) Module-1 PW
Variation of different variable (Q, C, V, E and U)
of parallel plate capacitor when dielectric is
introduced
Battery is removed
Battery remains connected
ε0 L
dC ε L
 L + ( K − 1) x  ⇒ = 0 ( K − 1)
d
dx
d
C
As, =
⇒
−Q 2 d ( K − 1)
dU −Q 2 ε0 L
=
×
−
=
K
1
(
)
dx 2C 2
d
2ε0 L[ L + ( K − 1) x]2
−dU
,
dx
As, F =
A
A
K
Q 2 d ( K − 1)
∴F=
2ε0 L[ L + ( K − 1) x]2
K
d
d
Figure (a)
figure (b)
Capacitance
Charge
Potential
Electric field
Energy
C′ = KC
Q′ = Q
V′ = V
E′ = E/K
U′ = U/K
C′ = KC
Q′ = KQ
V′ = V/K
E′ = E
U′ = KU
Force on a Dielectric Slab Placed Between the
Plates of a Parallel Plate Capacitor
Case I: Isolated capacitor
Consider a charged isolated capacitor. The capacitance is C = ε0 A
d
Here A = L × L (L is side of its plate)
When a dielectric slab is kept in it as shown, the capacitance
becomes
(+)
x
d
k

ε0
KLx + L ( L − x )} ∵
=
C
{
d

( K , A1 + K 2 A2 ) ε0 
d


ε0 L
C
or, =
{L + ( K − 1) x}
d
If slab moves inside (without change in kinetic energy), x
increases, which increases the capacitance, causing a decrease in

Q2 
electrostatic energy  as U =
 . Thus, the slab experiences a
2C 

−dU
force F =
, which pulls the slab inside.
dx
As, U =
2
C
Again, =
ε0 L
dC ε L
= 0 ( K − 1)
 L + ( K − 1) x  and
dx
d
d
Now, Q = CV ⇒ dQ =
dC.V
(+)
x
V
d
k
(–)
2
Area = L
Work done by battery = dW = dQ × V = dCV2
(–) Area = L × L
⇒=
C
Case II: A constant potential difference is maintained across the
capacitor
In this case, a constant potential difference is being maintained.
When the slab moves inside by a small distance dx, capacitance
increases and thus battery supplies extra charge to the capacitor
thereby increasing its potential energy.
During the displacement, an external force (external agent) of
equal magnitude F should be applied in opposite direction, so that
the dielectric moves with no acceleration.
Note: Slab moves inside the capacitor without change in kinetic
energy.
2
Q
dU −Q dC
⇒
=
×
2C
dx 2C 2 dx
P Electrostatic Potential and Capacitance
W
Area = L2
Work done by external force dW2 = –Fdx
\ Change in potential energy dU = Work done by battery + Work
done by external force = dCV2 – Fdx
[Battery and external force (external agent) both are “external” to
the system of capacitor]
dC 2 dU
dU dC 2
V −F =
V −
or =
or F
dx
dx
dx
dx
As U =
dU 1 dC 2
1
CV 2 ⇒
=
⋅V
dx 2 dx
2
⇒=
C
ε0 L
dC ε L
L + ( K − 1) x} ⇒ = 0 ( K − 1)
{
d
dx
d
⇒=
F
dC 2 1 dC 2 1 dC 2
V −
V=
V
dx
2 dx
2 dx
1 ε L ( K − 1) 2
∴ F =0
V
d
2
The force is independent of x.
75
Key Note
In case (1), the force depends x as F ∝
1
l + ( k − 1) x 
2
whereas in case (2) force is independent of x.
Sol. (a) The arrangement shown in fig. is equivalent to
two capacitors joined in series. Let their capacitance
be C1 and C2 respectively. Then
C1 = ε0
K1 A
d /2
and C2 = ε0
Now
Train Your Brain
Example 36: If 50µF be the capacity of a capacitor in air and
110 µF in oil then the dielectric constant of oil will be:
(a) 0.45
(b) 0.55
(c) 1.10
(d) 2.20
Sol. (d) C ∝ K
C
K
⇒ 1 = 1
C2 K 2
⇒
50 × 10−6
110 × 10−6
=
1
K2
[For air K1 = 1]
⇒ K2 = 2.20
Example 37:
(a) Two dielectric slabs of dielectric constant K1 and K2
have been filled in between the plates of a capacitor
as shown in fig. What will be the capacitance in
each case.
d
d
1
1
1
= +
=
+
C C1 C2 2ε0 K1 A 2ε0 K 2 A
d  K 2 + K1 
=


2ε0 A  K1 K 2 
C=
2ε0 A  K 2 K1 


d  K 2 + K1 
(b) Let A be the area of each plate of the capacitor
and d be the distance between the two plates. If
the capacitance be C1 and C2 respectively, then
K1 A / 2
K A/ 2
and C2 = ε0 2
d
d
Let C be the equivalent capacitance, then
C1 = ε0
C = C1 + C2 = ε 0
K1 A
K A
+ ε0 2
2d
2d
ε0 A
( K1 + K 2 )
2d
{∴ Two condensers are in parallel}
Example 38: An air-cored capacitor of plate area A and
separation d has a capacity C. Two dielectric slabs are
inserted between its plates in two different manners as
shown. Calculate the capacitance in the figure
=
Cequ
(+)
(b) A capacitor if filled with two dielectric of same
dimensions but of dielectric constant 2 and
3 respectively.
K2 A
d /2
K1
K2
(t1)
(t2)
(–)
d
Sol. Let the charge on the plates are Q and –Q
76
Electric field in the free space is E=
0
σ
Q
=
ε0 Aε0
∴ Electric field in first slab is E
=
1
E0
Q
=
.
K1 Aε0 K1
JEE (XII) Module-1 PW
The potential difference between the plates is
E = E0(d – t1 – t2) + E1t1 + E2t2.
Concept Application
t
t 

=
V E0  d − t1 − t2 + 1 + 2 
K1 K 2 

E0
E0 

=
E1 =
, E2
 As,
K1
K 2 

∴
V
=
∴
C=
Q
Aε0
29. A capacitor is filled with dielectrics as shown in the
diagram. Which of the options is correct?
k1
k2
t1
t2 

 d − t1 − t2 + K + K 

1
2 
E
ε0 A
d − t1 − t2 +
t1
t
+ 2
K1 K 2
Example 39: Calculate the capacitance of the parallel plate
capacitance shown in the following figure, between points
X and Y.
X
30. Separation between the plates of a parallel plates
capacitor is d and the area of each plate is A. When a
slab of material of dielectric constant k and thickness
t(t < d) is introduced between the plates, its capacitance
becomes.
C1
C2
Y
X
X
C3
Y
C′
Y
Sol. The capacitor can be broken apart into the
following circuit.
(i) Series (C2 and C3)
A
A
C2 =
ε 0 K 2 , C3 =
ε 0 k3
d
d
1
1
1
+
⇒ =
′
C C2 C3
(ii) Parallel (C1 || C′)
A
C1 = ε0 k1
2d
X
X
C1
C
C′
Y
Y
⇒ Equivalent capacitance between X and Y
CC
is C = C1 + C′ = C1 + 2 3
C2 + C3
 k1k2 + 2k2 k3 + k3 k1  ε0 A
So the capacitance is 
 d .
2(k2 + k3 )


P Electrostatic Potential and Capacitance
W
(a) Electric field inside dielectric k1 is equal to that
of k2
(b) Surface change density on the plates is uniform
(c) Potential difference across k1 is equal to the
potential difference across k2
(d) Electric field inside the free space is nonuniform
(a)
(c)
ε0 A
 1
d + t 1 − 
 k
(b)
ε0 A
 1
d − t 1 − 
 k
(d)
0A
 1
d + t 1 + 
 k
ε0 A
 1
d − t 1 + 
 k
31. In a parallel plates condenser, the radius of each
circular plates is 12 cm and the distance between the
plates is 5 mm. There is a glass slab of 3 mm thick and
of radius 12 cm with dielectric constant 6 between its
plates. The capacity of the condenser will be
(a) 144 × 10–9 F
(b) 40 pF
(c) 160 pF
(d) 1.44 µF
Redistribution of Charge When two Charged
Capacitor are connected to each other
Case I: Two capacitor C1 and C2 charged to potentials, V1 and
V2, are connected as shown, so that their like terminals are
connected to each other.
+
+
+
+
C1V1
–
–
–
–
+–
+–
+–
+–
C2V2
K is
Closed
+
+
+
+
V
+
+
+
+
C1V
–
–
–
–
–
–
–
–
C2V
77
Consider the part of circuit inside dotted loop. This part is isolated
from other part. Therefore, total change of this part remains
constant. Hence, C1V1 + C2V2 = C1V + C2V
or
V=
C1V1 + C2V2
C1 + C2
During this redistribution of charge, there is a loss of energy in the
form of heat and electromagnetic radiations.
Loss of energy ∆U = Ui – Uf
⇒ ∆U=
1
1
1
2
C1V12 + C2V2=
(C1 + C2 )V 2
2
2
2
1  C1C2
2  C1 + C2

2
 (V1 − V2 ) 
C1V1 + C2V2 

as V = C + C 

1
2

+
+
+
+
–
–
–
K
–
C2V2
C1
+–
+–
+–
+–
V
+–
+–
+–
+–
C2
K is
Closed
Again following the conservation of charge,
C1V1 – C2V2 = C1V + C2V or, V =
C1V1 − C2V2
C1 + C2
As calculated earlier, loss of energy can be calculated equal to
=
∆U
1 C1C2
(V1 + V2 ) 2
2 C1 + C2
Example 40: Two capacitors of capacitance C1 = 2µF and C2
= 8µF are connected in series and the resulting combination
is connected across 300 volts. Calculate the charge, potential
difference and energy stored in the capacitor separately.
Sol. If C is the equivalent capacitance,
1
1
1 1 1 5
=
+
= + = ∴
then
C C1 C2 2 8 8
8
C= = 1.6µF
5
Charge q = CV = 1.6 × 10–6 × 300 = 4.8 × 10–4 C
Potential across C=
V=
1
1
q
= 240 volt
C1
Potential across C=
V=
2
2
q
= 60 volt
C2
1
C1=
V12 5.76 × 10 –2 joule
2
1
C2
C2=
V21 1.44 × 10 –2 joule
Energy stored in =
2
78
(ii) Initial and final charge
(iii) Initial and final potential difference
(v) Initial and final energy stored in the capacitor
Sol. (i) Initial capacitance = C
Final capacitance = KC (after inserting dielectric)
Cf
=k
⇒
Ci
(ii) Charge on isolated capacitor remains same, therefore
Qf
=1
Qi
(iii) As, Q = CV
Et V1 1
⇒ = =
Ei Vi K
E f Vf
V
1
= =
(iv) E = ⇒
d
Ei Vi K
(v) U =
Train Your Brain
Energy stored in =
C1
(i) Initial and final capacitance
(iv) Initial and final electric field between the plates
Case II: The two capacitors are connected in a manner, so that
their unlike terminals are connected together.
C1V1
+–
+–
+–
+–
Example 41: An air cored parallel plate capacitor having a
capacitance C is charged by connecting it to a battery. Now,
it is disconnected from the battery and a dielectric slab of
dielectric constant K is inserted between its plates so as to
completely fill the space between the plates. Compare
Uf
C
Q2
1
⇒
= i =
2C
Ui C f K
Example 42: A parallel plate capacitor of capacitance C
has been charged, so that the potential difference between
its plates is V. Now, the plates of this capacitance are
connected to another uncharged capacitor of capacitance
2C. Find the common potential acquired by the system and
loss of energy.
Sol. Here, C1 = C, C2 = 2C,
V1 = V and V2 = 0
⇒ Common potential = V =
=
C1V1 + C2V2
C1 + C2
CV + 0 V
=
.
3C
3
And=
∆U
1 C × 2C
1
× (V =
− 0) 2
CV 2
2 3C
3
Example 43: Two parallel plate capacitors A and B having
capacitance of 1µF and 5µF are charged separately to the
same potential of 100 Volt. Now the positive plate of A is
connected to the negative plate of B and the negative plate
of A to the positive plate of B. Find the final charges on each
capacitor.
JEE (XII) Module-1 PW
Sol. As for a capacitor q = CV. So initially the charge on
each capacitor
q1 = C1V1 = (1 × 10–8) × 100 = 100µC
and q2 = C2V2 = (5 × 10–6) × 100 = 500µC
Now, when two capacitors are joined to each other such that
positive plate of one is connected with the negative of the
other, by conservation of charge
q = q1 + q2 = |q1| – |q2| = (500 – 100)µC = 400µC.
So common potential
34. In the given circuit, all the capacitors are initially
uncharged. After closing the switch S1 for a long time
suddenly S2 is also closed and kept closed for a long
time. Total heat produced after closing S2 will be
C
S1
E
C
S2
(q1 + q2 )
400 × 10 –6
200
=
=
V =
V
(C1 + C2 ) (1 + 5) × 10 –6
3
and hence after sharing charge on each capacitor
(a) 4 CE2
(c) 2 CE2
200 1000
=
µC
3
3
Concept Application
32. In the figure initial status of capacitor and their
connection is shown. Which of the following is
incorrect about this circuit?
10V
D
B C
A
3µF
2µF
C
D
(a) Final charge on each capacitor will be zero
(b) Final total electrical energy of the capacitors will
be zero
(c) Total charge flown from A to D is 30µC
(d) Total charge flown from A to D is – 30µC
33. The circuit was in the shown state for a long time.
Now if the switch S is closed then the net charge that
flows through the switch S, will be:
50V
2µF
4µ F
S
2µF
(a)
4µ F
400
100
µC (b) 100 µC (c)
µC (d) 50µC
3
3
P Electrostatic Potential and Capacitance
W
(b) 1/2 CE2
(d) 0
35. Two capacitors of 2 µF and 3 µF are charged to
150 volt and 120 volts respectively. The plates of
a capacitor are connected as shown in the figure. A
discharged capacitor of capacity 1.5 µF falls to the
free ends of the wire. Then after the system comes in
steady state
1.5µF
3µF
120V
150V
2µF
A
(a) Charge on the 1.5 µF capacitor will become
180 µC
B
A
C
(b) Charge on the 2 µF capacitor will become
120 µC
(c) +ve charge flows through A from left to
right
(d) +ve charge flows through A from right to
left
CAPACITY OF AN ISOLATED
SPHERICAL CONDUCTOR
When charge Q is given to a spherical conductor of radius R, then
potential at the surface of sphere is
+ + +Q
Q
1 Q
V=
⇒ = 4πε0 R
+
+
4πε0 R
V
R
+
1
O
C = 4πε0 R =
.
R
+
+
9 × 109
+
+
+
15V
E
E
200 200
= µC
3
3
(5 × 10 –6 ) ×
q2′ =
C2V =
C
+
q1′ =
C1V =×
(1 10 –6 ) ×
E
79
Key Note
b
If earth is assumed to be a conducting sphere having radius
R = 6400 km. It’s theoretical capacitance C = 711 µF. But for
all particle purpose capacitance of earth is taken infinity and
its potential V = 0.
Q–O
l
Spherical Capacitor
It consists of two concentric conducting spheres of radii a and b
(a < b). Inner sphere is given charge + Q, while outer sphere is
earthed
1. Potential difference: Between the spheres is
V
=
Q
Q
−
4πε0 a 4πε0 b
a
+Q
 ab 
2. Capacitance: C = 4πε0 

b−a
3. If outer sphere is given a charge + Q while inner sphere is
earthed
Induced charge on the inner sphere
a
Q ' = − .Q and capacitance of
b
+Q
b
b2
the system C '= 4πε0 .
b−a
Capacitance is equivalent to the sum of capacitance of
spherical capacitor and spherical conductor
b2
ab
= 4πε0
+ 4πε0 b
b−a
b−a
Cylindrical Capacitor
It consists of two co-axial cylinders of radii a and b (a < b),
inner cylinder is given charge +Q while outer cylinder is earthed.
Common length of the cylinders is l then
80
Q1
C1
V1
U1
r2
r1
O
O
Q1 = C1V1
Q2 = C2V2
Q2
C2
V2
U2
If these two spheres are connected through a conducting wire,
then alteration of charge, potential and energy takes place.
a
2πε0 b
log e  
a
When two charged conductors are joined together through a
conducting wire, charge begins to flow from one conductor to
another from higher potential to lower potential.
Suppose there are two spherical conductors of radii r1 and r2,
having charge Q1 and Q2, potential V1 and V2, energies U1 and U2
and capacitance C1 and C2 respectively.
b
C=
Redistribution of Charges And Loss of Energy
This flow of charge stops when they attain the same potential. Due
to flow of charge, loss of energy also takes place in the form of
heat through the connecting wire.
–Q
i.e. 4πε0 .
Grouping of capacitors
Q'1
C1
V
U'1
r2
r1
O
O
Q1 = C1V1
Q'2
C2
V
U'2
Q2 = C2V2
(a) New charge: According to the conservation of charge
Q1 + Q2 = Q'1 + Q'2 = Q (say), also
Q '1 C1 r1
= =
Q '2 C2 r2
 r 
 r 
⇒ Q '2 = Q  2  and similarly Q '1 = Q  1 
r
+
r
 1 2
 r1 + r2 
(b) Common potential: Common potential
Total charge Q1 + Q2 Q '1 + Q '2 C1V1 + C2V2
=
= =
=
V
C1 + C2
C1 + C2
Total capacity C1 + C2
(c) Energy loss: The loss of energy due to redistribution of
charge is given by
∆U = Ui – Uf =
C1C2
(V1 − V2 ) 2
2(C1 + C2 )
JEE (XII) Module-1 PW
(iii) Two sphere comes into parallel configuration.
Train Your Brain
4πε0 b
C1
q =
∴
=
×Q
×Q
C1 + C2
4πε0 a + 4πε0 b
Example 44: Can a sphere of radius 1 cm, placed in air be
given a charge of 1 coulomb?
Sol. Potential of sphere of 1 cm radius is
b
×Q
a+b
⇒=
q
1
q
= 9 × 109 × −2 = 9 × 1011 volt
10
r
This potential is so high that the surrounding air gets
ionized, thereby charge leaks to the medium.
[air gets ionized at 3 × 106 volt].
Hence the answer is No.
Example 45: Two conductors having capacities 2µF and
5µF and potentials 2 volts and 10 volt respectively. The
ratio of their charges after connecting there by a wire will be
(a) 2 : 5
(b) 5 : 2
(c) 1 : 5
(d) 5 : 1
V =k
Concept Application
36. Two concentric conducting spheres of radii R and
2R are carrying changes Q and –2Q respectively. If
the change on inner sphere is doubled, the potential
between the two spheres will
– –
–
+
–
– +
+Q
–
– +
+
–
– +
+ –
+
–
– – –
–2Q
q
2 × 10−6 2
=
Sol. =
(a) 1
q2 5 × 10−6 5
[Charge distributes in the ratio of capacities]
Example 46: Consider situation shown in figure in which
two isolated conducting spheres are connected. Initially
both keys K1 and K2 are open. At t = 0, K1 is closed and
after a long time charge on sphere of radius a is Q. Now K4
is opened and K2 is closed, then charge an sphere of radius
(a) Becomes two times
(b) Become four times
(c) Be halved
(d) Remain same

1 
b is q.  K =

4πε0 

K1
37. Two conducting spheres, each given a charge q are kept
for apart as shown. The amount of charge that crosses
the switch S, when it is a closed, is (the connection
between the spheres is conducting)
a
q
q
K2
R
b
Find
(i) Terminal voltage of battery.
(ii) Energy loss when key K, is closed for the first time.
(iii) Value of q
Q
KQ
Sol. (i) Vsphere
= V=
=
T
a
4πε0 a
(ii) U sphere =
q
3
(b)
2q
3
(c)
3q
4
(d) Zero
Equivalent Capacitance of Conducting Plates
Charge on each capacitor remains same and equals to the main
charge supplied by the battery but potential difference distributes
i.e. V = V1 + V2 + V3
Q2
8πε0 a
Series Grouping
Work done by battery = VT × Q =
∴ Loss of energy =
(a)
2R
2
Q
4πε0 a
Q2
Q2
Q2
−
=
4πε0 a 8πε0 a 8πε0 a
P Electrostatic Potential and Capacitance
W
™
If n identical capacitors each having capacitances C are
connected in series with supply voltage V then Equivalent
C
capacitance Ceq =
and potential difference across each
n
capacitor V ' =
V
n
81
If n identical plates are arranged as shown below, they
constitute (n – 1) capacitors in series. If each capacitor has
ε A
ε0 A
capacitance 0 then Ceq =
d
(n − 1)d
The faces 1 and 2 form a capacitor out of which, 1 is
connected to V volts and 2 is grounded. Similarly, faces 3
and 4 from a capacitor
+ – + – + – + –
–
+
+– +– +– +–
+ – + – + – + –
+– +– +– +–
In this situation except two extreme plates each plate is
common to adjacent capacitor.
™ Here, effective capacitance Ceq is even less than the least of
the individual capacitance.
ε A
So, net charge on plate A is  0  V
 d 
Net charge on plate B is equal to sum of charges on faces 2
 2ε A 
and 3, which is equal to −  0  V
 d 
™
Parallel Grouping
If n identical plates are arranged such that even numbered
plates are connected together and odd numbered plates are
connected together, then (n – 1) capacitors will be formed and
they will be in parallel grouping.
™
4
1
3
6
B
7
5
Equivalent capacitance C' = (n – 1)C
ε0 A
d
This type of combination is used when high capacity is
required.
If Cp is the effective capacity when n identical capacitors are
connected in parallel and Cs is their effective capacity when
Cp
connected in series, then
= n2
Cs
where C = capacitance of a capacitor =
™
ε A
Net charge on plate C is  0  V
 d 
Example 48: Four plates of the same area of cross-section
are joined as shown in the figure. The distance between each
plates is d. The equivalent capacity across AB will be
A
2
™
ε A
So, we have =
=  0 V
Q CV
 d 
Train Your Brain
Example 47: Find the charge on each plate, given area of
each plate is A and separation between two consecutive
plates is d.
(a)
2ε0 A
d
(b)
3ε0 A
d
(c)
3ε0 A
2d
(d)
ε0 A
d
Sol. (b) The arrangement shown in the figure is equivalent
to three capacitance in parallel hence resultant capacitance
3ε A
= 0 .
d
Concept Application
38. Find the equivalent capacitance between point a and
bs in the following arrangement of the plates.
A
d
a
b
V
Sol. Let the plates are A, B and C. This is an arrangement
of two capacitors. First, we identity how the capacitors are
connected.
A
1 2
V
B
3 4
C
+Q –Q
1
⇒
82
V
4 3
⇒
V
+Q
–Q
4 ε0 A
2 ε0 A
3 ε0 A
1 ε0 A
(b)
(c)
(d)
3 d
3 d
2 d
2 d
39. A relatively simple arrangement of combination of
plates happens to be when the alternate plates are
interconnected as follows:
1
a
2
b
3
(a)
4
(a) 3C
(b) 2C
(c) 4C
(d) 6C
JEE (XII) Module-1 PW
Short Notes
For a Conducting/non Conducting Spherical shell
Relation between E & V
−∂V
E = −grad V = − ∇V , E =
;
∂r
∂V ∂V ∂V ˆ
−
E=
i−
j−
k , V = ∫ − E.dr
∂x
∂y
∂z
1 q
1 q
=
=
:E
,V
For r ≥ R
2
4π ∈0 r
4π ∈0 r
For r < R : E = 0, V =
Electric Potential Energy of two Charges
U=
+
+ C
At a point which is at a distance r from dipole midpoint and
making angle q with dipole axis.
Potential V =
q
R
+
+
+
For a Charged Circular Ring
1 q1q2
4π ∈0 r
Electric Dipole
™
1 q
4π ∈0 R
+
1 p cos θ
4π ∈0 r 2
q
+
x
R
P
+
1
qx
q
=
EP =
,Vp
2
2 3/ 2
Equipotential Surface and Equipotential Region
4π ∈0 ( x + R )
4π ∈0 ( x 2 + R 2 )1/ 2
In an electric field the locus of points of equal potential is called
R
an equipotential surface. An equipotential surface and the electric
Electric field will be maximum at x = ±
2
field line meet at right angles. The region where E = 0, Potential
of the whole region must remain constant as no work is done in
For a charged Long Conducting Cylinder
displacement of charge in it. It is called as equipotential region
like in conducting bodies.
q
™ For r ≥ R E =
π∈ r
Gauss’s Law
Electric flux: φ = ∫ E.d s
Expression for Gauss's Law:
™
For r < R : E = 0
∫ E.d s = ∑ en
∈0
 
q
+
+
+
+
+
+
+
+
(Applicable only on closed surface)
Net flux emerging out of a closed surface is
Potential due to Various Bodies
qen
ε0
+
For a Conducting Sphere 1 q
1 q
=
=
:E
,V
For r ≥ R
2
4π ∈0 r
4π ∈0 r
1 q
For r < R : E = 0, V =
4π ∈0 R
q
R
+
+
+
+
q
+ + R
For a non-conducting Sphere
+
+ +
+ +
+ +
1 q
1 q
+ + +
For r ≥ R : E =
,V=
2
4π ∈0 r
4π ∈0 r
+
2
2
1 qr
1 q (3R − r )
=
=
,V
For
r<R:E
4π ∈0 R 3
4π ∈0
2R3
Vcenter
= V=
max
3 kq
= 1.5Vsurface
2 R
P Electrostatic Potential and Capacitance
W
q
+
+
+
R +
+
+
+
+
σ
∈0
™
Electric field intensity at a point near a charged conductor E =
™
Mechanical pressure on a charged conductor P =
™
For non-conducting infinite sheet of surface charge density
s, E =
™
™
σ2
2 ∈0
σ
2 ∈0
For conducting infinite sheet of surface charge density s,
σ
E=
∈0
∈
Energy density in electric field U = 0 E 2
2
83
Name/Type
Point charge
Formula for Potential
Note
™
kq
r
™
Graph
q is source charge.
V
r is the distance of the point from the point
charge.
r
Ring (uniform/
nonuniform charge
distribution)
R
At the axis
Uniformly charged
hollow conducting/
nonconducting/solid
conducting sphere
Uniformly charged
solid nonconducting
sphere
™
kQ
kQ
r
For r ≤ R, V =
kQ
R
For r ≥ R, V =
For r ≤ R,
=
(
Not defined
™
™
™
)
™
™
)
™
™
™
™
Infinite nonconducting
thin sheet
Not defined
Infinite charged
conducting thin sheet
Not defined
84
R is radius of sphere
V
r is the distance from center of sphere to the
kQ/R
point
Q is total charge = s4pR2.
R
™
3
ρ
3R 2 − r 2
6ε 0
(
V
r
™
kQ
r
kQ 3R 2 − r 2
2R
Q is source charge.
x is the distance of the point on the axis from
center of ring
R2 + x2
For r ≥ R, V=
V=
Infinite line charge
™
At center: kQ
™
™
™
™
R is radius of sphere
r is distance from center to the point
3
Vcenter = Vsurface .
2
4
Q is total charge =ρ πR3 .
3
Inside the sphere, potential varies parabolically
Outside the sphere potential varies
hyperbolically.
V
3kQ/2R
kQ/R
R
Absolute potential is not defined.
Potential difference between two points is
given by formula:
VB – VA =
−2k λln ( rB /rA )
Absolute potential is not defined.
Potential difference between two points is
given by formula
σ
VB − VA =
−
( rB − rA )
2ε0
Absolute potential is not defined.
Potential difference between two points is
given by formula
σ
VB − VA =
− ( rB − rA )
ε0
JEE (XII) Module-1 PW
Solved Examples
1. Two particles, each of charge q, and masses m1 and m2
are released from rest, from points A and B, r distance
apart. Calculate the final speed of each particle. (Neglect
gravitational interaction of the particles)
Sol. Both particles move under mutual electrostatic repulsion.
Let v1 and v2 are final speeds (when the separation becomes
large)
As there is no external force,
A
F
B F
r
m 1, q
m 2, q
⇒ ∆p = 0 [law of conservation of linear momentum]
⇒ – m1v1 + m2v2 = 0
∴ m1v2 = m2v2 = p (say)
Also, ∆K + ∆U = 0
[conservation of mechanical energy]
1
q2 
1
 
⇒  m1v12 + m2 v22 − 0  + 0 −
0
=
2
2
  4πε0 r 
⇒
Also, by conservation of mechanical energy
1 2 1
1 Q ( −e )
=
mv0
mv′2 +
2
2
4πε0 r0
⇒ v '2 = v02 +
2Qe
4πε0 ( mrc )
2
v b

v0 b 
2Qe
or,  0  =
v02 +
∵ v′ =

4πε0 ( mrc ) 
rc 
 rc 
⇒ v02 b 2 = v02 ro2 +
⇒ v02 rc2 +
2Qθ
rc − v02 b 2 =
0
4πε0 m
2
 2Qe 
2Qe
4 2
+ 
 + 4v0 b
πε
4πε0 m
4
m
0


⇒ rc =
2v02
−
3. An uncharged conductor of inner radius R1 and outer radius
R2 contains a point charge q at the center as shown in figure
p2
p2
q2
+
=
2m1 2m2 4πε0 r
S2
1/2
p2
⇒=
2Qe
rc
4πε0 m
 2m m
2m1m2
q2
q2 
p  1 2 ⋅
⋅
or=

m1 + m2 4πε0 r
 m1 + m2 4πε0 r 
S1 q R1
o
R2
A
B
1/2
p
q2 
1  2m1m2
⇒
v
=
=
⋅


1 m
m1  m1 + m2 4πε0 r 
1
1/2
1  2m1m2
q2 
⋅


m2  m1 + m2 4πε0 r 
2. An electron (mass m, charge –e) is moving towards a heavy
nucleus of charge Q, fixed at its place. When the electron is
far from the nucleus, its speed is v0 and the impact parameter
is b. What is the closest distance of the electron from the
nucleus?
=
v2
C
(i) Find E and V at points A, B and C
(ii) If a point charge Q is kept outside the sphere at a
distance ‘r’ (>> R2) from center, then find out resultant
force on charge Q and charge q.
Sol. (i) At point A:
+q
Sol. The path of the electron is shown. rq is distance of closest
approach, v′ is the speed at this instant.
Clearly, v′ is perpendicular to radial distance
The angular momentum of electron about the nucleus
remains conserved, as torque about nucleus is zero.
⇒ mv0b = mv′rc
–q
q
e
b
Electric field at ‘A’ due to –q of S1 and + q of S2 is zero
individually because they are uniformly distributed. Electric
field is only due to q at center.
v0
Q
v
rc
P Electrostatic Potential and Capacitance
W
EA =
kq
kq k (−q ) kq
+
; VA = +
OA
OA
R1
R2
2
85
At point B : VB =
Kq K ( −q ) Kq Kq
+
+
=
, EB = 0
OB
OB
R2
R2
Kq Kq At point=
C : VC
OC
=
, EC
OC
OC 3
(ii) Force on point charge Q :
(Note: Here, force on ‘Q’ will be only due to ‘q’ of S2 (see
result (iii))
KqQ
F0 = 2 r̂ (r distance of ‘Q’ from center ‘O’)
r
Force on point charge q:
Fq = 0 (Electrostatic shielding)
Sol. Due to spherical symmetry, the field at any point will be
radially (if it is nonzero). So, to calculate the field, we take
a concentric sphere of radius r as a Gaussian surface in each
case.
+2q
+2q
–2q
E=
E=
1 – 2q
.
4 0 d 2
O
E= 0
E= 0
a
b c
1 – 2q
.
4 0 c 2
1 – 2q
E=
.
4 0 b 2
1 –2q
.
4 0 r 2
(b) (i) The charge on the inner surface of the small
shell is zero so that the charge inside a spherical
Gaussian surface passing through the material of this
shell becomes zero.
(ii) Since there is no charge inside the material of the
conductor, total excess charge of this shell remains on
its outer surface. Hence, this is –2q.
(iii) The charge on the inner surface of the larger shell
will be +2q so that the total charge inside a Gaussian
surface passing through its material becomes zero.
(iv) Since the total charge on the larger shell is 4q, so the
charge on the outer surface will be 4q – 2q = 2q.
5. If electric potential at the corner of a cube having uniform
volume charge density is V0 then find the electric potential
at the center of this cube.
Q
a
Where, α is a positive constant and Q is total charge of cube.
∴ Q = r × a3
P
2
1 2q
.
4 0 r 2
r
d
E=
Sol. Let potential at center, VC = α
a bc
r r
d
P1
(v) For r > d, the charge contained in the Gaussian surface
through the point P5 is –2q + 4q = 2q
E=
4. A small conducting spherical shell with inner radius a and
outer radius b is concentric with a larger conducting spherical
shell with inner radius c and outer radius d. The inner shell
has total charge –2q and the outer shell has charge +4q. (a)
Calculate the electric field in terms of q and the distance r
from the common center of the two shells for (i) r < a, (ii)
a < r < b, (iii) b < r < c, (iv) c < r < d, (v) r > d. Show your
result in a graph of E as a function of r. (b) What is the total
charge on the (i) inner surface of the small shell (ii) Outer
surface of the small shell (iii) inner surface of the large shell
(iv) Outer surface of the large shell?
P4 P3
(iv) For c < r < d, the point P4 lies inside the conductor so
the field is zero.
r
P5
 ρ× a 3 
2
⇒ VC = α 
 = αρa
 a 
P
(a) (i) The charge inside the Gaussian surface through P1 is
zero. So,
q
= 0 or E= 0
ε0
(ii) For a < r < b, the point P2 lies inside the conductor, so
E = 0.
(iii) For b < r < c, the charge inside the Gaussian surface
through P3 is –2q.
φ= E ⋅ 4πr 2 =
1 −2q
−2q
=
⋅
So, E ⋅ 4πr 2 =
or,
E
4πε0 r 2
ε0
Negative sign indicates that the field is radially inward.
86
a
Now, suppose point P is center of cube of side length 2a.
then, VP = 8 × potential at corner due to one cube of side
length a
⇒ VP = 8V0 = αr × (2a)2
⇒ αρa2 = 2V0 = VC
∴ Potential at center of given cube is 2V0.
JEE (XII) Module-1 PW
6. A conducting spherical thin shell of radius R is given a charge
4Q. A point charge –Q is placed at a distance of R/2 from
the center O. Find the electric potential at point P which is
at a distance 2R from the center along the line joining center
and point charge as shown in the figure.
+4Q
 1

K1 ×  ρ× × a 2 × H 
 3

Sol. V0 =
H
 1

K 2 ×  ρ× × a 2 × H 
 3

E0 =
H2
–Q
O
P
R/2
2R
Sol.
++ +
+ ++
+
+ + 3Q
+ +
+Q
+ +
+
+ +
+
++
+
+
+
–Q
+
+ +
+
+
+ +
+
+
+
+
+ + +
+ +
a
Potential and field due to upper half part
 1  a 2  H  
K1 × ρ× ×   ×   
 3  2   2   V0
=
V1 =
4
H
 
Charge distribution on shell
 2
Potential at point P is only due to charges on outer surface
 1  a 2  H  
of shell because potential due to point charge inside the shell
K 2 × ρ× ×   ×   
and due to induced charge on inner surface of shell is zero
 3  2   2   E0
E1 =
=
at all points outside the shell. And, on outer surface, +3Q
2
 H2 
charge is uniformly distributed.
 4 


3Q
∴ VP =
∴ Magnitude of electric field due to remaining portion,
4πε0 × ( 2 R )
E
E2 = E0 − E1 = E0 − 0
7. A right pyramid of square base and height H has uniform
2
charge distributed everywhere within its volume. Modulus
E0
of electric field and potential at the apex P of pyramid are
⇒ E2 =
2
E0 and V0 respectively. A symmetrical portion of height
And, potential due to remaining portion,
H
h =   from apex is removed (shown in figure).
 2
V
V2 = V0 − V1 = V0 − 0
Truncated pyramid
4
P
h
H
3V
⇒ V2 =0
4
8. Find charge on each spherical shell after joining the inner
most shell and outer most shell by a conducting wire. Also
find charges on each surface.
5Q
3
Truncate pyramid
Find
(a) Magnitude of electric field at P due to remaining
portion.
(b) Potential at P due to remaining portion.
P Electrostatic Potential and Capacitance
W
2
–2Q
Q
1
R
2R
3R
87
Sol. Let the charge on the innermost sphere be x.
Finally potential of shell 1 = Potential of shell 3
∴
4 joules. Calculate the distance of the farthest point D which
the negative charge will reach before returning towards C.
kx k ( −2Q ) k ( 6Q − x ) kx k ( −2Q ) k ( 6Q − x )
+
+
=+
+
R
2R
3R
3R
3R
3R
+q
A
6Q – x
3
D x
–2Q
x
2
1
+Q
2
∴ Charge on innermost shell =
Q
and Charge on outermost
2
11Q
shell = 2
Charge on middle shell = – 2Q
∴ Final charge distribution is as shown in figure.
4Q
+3Q/2
–3Q/2
–Q/2
Q/2
9. Two point charges q and –2q are placed at a distance 6a
apart. Find the locus of the point in the plane of charges
where the field potential is zero.
Sol. Let us take the charges on X-axis:
q at A(0, 0) and –2q at B (6a, 0)
P(x, y)
A (0, 0) B(6a, 0)
X
q
(–2q)
Potential at a point P(x, y) is
2
−2q
( x − 6a ) + y 2
2
C
B
Sol. The kinetic energy is lost and converted to electrostatic
potential energy of the system as the negative charge goes
from C to D and comes to rest at D instaneously. The total
energy (TE) is conserved.
TE at D = TE at C
0 + (–q)VD = 4 + (–q) VC
4 = q(VC – VD)
2q 2
4
=
4πε 0
1
1 
 −

9 + x 2 
 5
(
)(
2
4=
2 5 × 10−5 9 × 109
45
4 =9 −
⇒x=
9 + x2
Sol.
A



9 + x 
1
2
72 = 8.48 m
B
+q
+q
O
O
E
EA
EB
45° 45°
ED EC
D –q F –q C
=
V0
q+q−q−q
∑q
=
= 0,
4πε0 a / 2
4πε0 a / 2
(
2
10. Two fixed equal positive charges, each of magnitude
5 × 10–5 C are located at points A and B, separated by a
distance of 6 m. An equal and opposite charge moves
towards them along the line COD, the perpendicular
bisector of line AB. The moving charge, when it reaches the
point C at a distance of 4 m from O, has a kinetic energy of

)  15 −
11. Four charges +q, +q –q, –q are placed respectively at the
corners A, B, C and D of a square of side a arranged in
the given order. Calculate the electric potential and the
intensity at O, the center of the square. If E and F are midpoint of the sides BC and CD, what will be the work done
in carrying charge e from E to F?

2q

E0 2 
=
 4πε0 a / 2
q
4q
⇒ 2
=2
2
x
y
+
x
−
6
( a) + y2
⇒ the locus is (x – 6a)2 = 4x2 + 3y2
88
O 4m
+q
3R
q
V
=
+
2
2
+
4
πε
x
y
4πε 0
0
V=0
–q
3m
R
2R
⇒ x =
3m
=
)


cos 45°
2 

4 2q
4π ∈0 a 2
WEF = e(VF – VE)
=
where
VE = 0, VF
W= eVF − 0=
4πε 0
(
2q
5a/2
)
+
−2q
4πε 0 a / 2

2qe  2
− 2

4πε 0 a  5

JEE (XII) Module-1 PW
12. A charge q is distributed uniformly over the volume of a
Charge on ball
Qx 2
=
V =
ball of radius R. Find:
4πε 0 x
4πε0 R 3
(a) The energy stored in the ball
The work done to add a charged layer of thickness dx
(b) The energy in the surrounding space
from infinity is
(c) Hence or otherwise, find the total energy of the system
Q x2
=
dW V=
dq
Sol. To calculate the energy stored in an electric field,
( ρ dV )
4πε0 R 3
2
we can use energy per unit volume = 1/2 ke0E


Assuming k = 1,
Qx 2  Q 
2
dW
=
 4πx dx
1
2
3 
4
⇒
U
=
ε
E
dV
4πε0 R  πR 3 
0


2
3

Let us recall the expression for the field of a uniformly
Total work done= W= ∫ dW
distributed charge in a non-conducting sphere.
 Qr
 4πε R 3
0
E=
 Q

2
 4πε 0 r
for r ≤ R
W
=
for r > R
(a) Energy inside the ball
1
2
U=
1
∫ 2 ε0 E dV
Consider an element of radius r and thickness dr
2
R
1  Qr 
U1 =
4πr 2 dr
ε0 
3 
∫
2
4
πε
R
0


0
R
Q2
U1 =
r 4 dr
6 ∫
8πε 0 R 0
Q2
U1 =
40πε0 R
(b) Energy outside the ball
1
U=
ε E 2 dV
2
∫
2 0
2
∞
1  Q 
U2 =
4πr 2 dr
ε0 
2 
∫
2
4
r
πε
0


R
Q2
U2 =
8πε0
∞
R
3Q 2
3Q 2
4
x
dx
=
∫0 4πε0 R6
20πε0 R
Hence total energy of the system =
Paragraph for Questions 13 to 15
Two solid hemispheres of charge +Q and –Q and radius R, are
combined to make a complete sphere.
13. Find out the maximum value of electric field strength
at a distance x from center of the sphere. It is given that
x >>> R
(a)
3QR
8πε0 x 3
(b)
QR
8πε0 x 3
(c)
3QR
5πε0 x 3
(d)
QR
5πε0 x 3
Sol. (a)
It is a dipole. Find out dipole moment assuming point charges
on COM.
Solid hemisphere has its center of mass at 3R/8
So combined hemispheres can be considered as a dipole.
dr
∫R r 2
Q2
U2 =
8πε 0 R
(c) Total energy = U1 + U2
Q2
Q2
⇒ U1 +=
U2
+
40πε 0 R 8πε 0 R
3Q 2
⇒ U1 + U 2 =
20πε 0 R
3R/4
–Q
+Q
 3R  3QR
=
P Q=


4
 4 
Maximum electric field
Emax =
2kP
x3
2k  3

QR 
3 
x 4

Alternatively: Total energy = work done to assemble
the parts of the ball starting from infinity
The potential at the surface of the ball at the instant
when it has a radius x is
=
P Electrostatic Potential and Capacitance
W
3Q 2
20πε0 R
3QR
8πε0 x 3
89
14. If we investigate potential at a distance x from center of this
sphere, it is given that x >>> R.
Choose the correct statement (s):
(a) Maximum potential at a distance x from center is
3QR
16πε 0 x 2
15. Two spherical conducting shells are given charges Q1 and
Q2. These two spheres are concentric and kept eccentrically
in a larger spherical shell of charge Q3 as shown in figure.
Find out the charge density at the outer surface of conductor
in the given arrangement. Given that Q1 = Q2 = Q3 = Q.
(b) Minimum potential at a distance x from center is
3QR
−
16πε0 x 2
(c) Maximum potential at a distance x from center is
QR
8πε 0 x 2
(d) Minimum potential at a distance x from center is
−
QR
8πε0 x 2
Sol. (a,b)
Electric potential will be maximum and minimum at axial
point.
Q
r1 1
r2
r
r3
(a) Charge density will be non-uniform on outer surface
(b) Charge density is uniform and equal to
3Q
4πr32
(c) Charge density is uniform and equal to
Q
4πr32
(d) Charge density is non-uniform and equal to
some point of outer surface
At point a, for potential
a
Q3
Q2
Q
at
12πr32
Sol. (b)
Charge density will be uniform.
P
KP cos θ
V=
x2
V=
P cos θ
4πε 0 x 2
Maximum potential
=
Minimum potential =
P
3QR
=
2
4πε0 x
16πε 0 x 2
−P
4π ε 0 x 2
= −
3QR
16πε0 x 2
3QR
3QR
−
2 < Potential at distance x <
16πε0 x
16πε 0 x 2
90
Q1 will be completely on outer surface of inner sphere
uniformly, hence –Q1 will be uniformly on inner surface of
second sphere, (Q1 + Q2) will be charge on outer surface of
second sphere uniformly. Therefore, (–Q1 – Q2) on inner
surface of outer sphere but non-uniformly, as second sphere
is kept eccentrically. Finally (Q1 + Q2 + Q3) charge will be
uniformly distributed on outer surface as inner charges will
not show any effect.
Thus, charge density on outer surface, σ =
As Q1 = Q2 = Q3 = Q
Q1 + Q2 + Q3
4πr32
Hence,
σ =
3Q
4πr32
JEE (XII) Module-1 PW
Exercise-1 (Topicwise)
ELECTROSTATIC POTENTIAL AND
POTENTIAL ENERGY
1. A hollow metal sphere of radius 5 cm is charged so that the
potential on its surface is 10 V. The potential at the center
of the sphere is
(a) 0 V
(b) 10 V
(c) Same as at point 5 cm away from the surface
(d) Same as at point 25 cm away from the surface
2. If a unit positive charge is taken from one point to another
over an equipotential surface, then
(a) Work is done on the charge
(b) Work is done by the charge
(c) Work done is constant
(d) No work is done
6. Two point charges 100 mC and 5 mC are placed at points
A and B respectively with AB = 40 cm. The work done
by external force in displacing the charge 5 mC from
π
B to C, where BC = 30 cm, angle ABC = and
2
1
9
2
2
= 9 × 10 Nm /C
4πε0
(a) 9 J
(c)
(b)
9
J
25
81
J
20
(d) −
9
J
4
7. Three charges Q, (+q) and (+q) are placed at the vertices
of an equilateral triangle of side l as shown in the figure. If
the net electrostatic energy of the system is zero, then Q is
equal to
Q
10
× 10−9 C are placed at each of the four
3
corners of a square of side 8 cm. The potential at the
intersection of the diagonals is
3. Charges of +
l
l
(a) 150 2 volt
(b) 1500 2 volt
(c) 900 2 volt
(d) 900 volt
4. Four equal charges Q are placed at the four corners of a
square of each side ‘a’. Work done in removing a charge
–Q from its center to infinity is
(a) 0
(b)
2Q 2
4πε0 a
(c)
2Q 2
πε0 a
(d)
Q2
2πε0 a
(b) Electric field is zero but potential is not zero
(c) Electric field is not zero but potential is zero
(d) Neither electric field nor potential is zero
P Electrostatic Potential and Capacitance
W
+q
 q
(a)  − 
 2
(b) (–q)
(c) (+q)
(d) Zero
8. A hollow conducting sphere is placed in an electric field
produced by a point charge placed at P as shown in figure.
Let VA, VB, VC be the potentials at points A, B and C on
sphere Then
A
C
5. Two charge +q and –q are situated at a certain distance. At
the point exactly midway between them
(a) Electric field and potential both are zero
l
+q
P
B
(a) VC > VB
(b) VB > VC
(c) VA > VB
(d) VA = VC
91
RELATION BETWEEN FIELD AND POTENTIAL
9. The electric potential V at any point O (x, y, z all in metres)
in space is given by V = 4x2 volt. The electric field at the
point (1m, 0, 2m) in volt/metre is
(a) 8 along negative X-axis
(b) 8 along positive X-axis
(c) 16 along negative X-axis
(d) 16 along positive Z-axis
10. A uniform electric field having a magnitude E0 and direction
along the positive X-axis exists. If the potential V is zero at
x = 0, then its value at X = + x will be
(a) Vx = +xE0
(b) Vx = –xE0
2
(c) Vx = +x E0
(d) Vx = –x2E0
11. Two plates are 2 cm apart, a potential difference of 10 volt
is applied between them, the electric field between the
plates is
(a) 20 N/C
(b) 500 N/C
(c) 5 N/C
(d) 250 N/C
12. There is an electric field E in X-direction. If the work
done on moving a charge 0.2 C through a distance of
2 m along a line making an angle 60° with the X-axis is 4 J,
what is the value of E
16. The value of electric potential at any point due to any
electric dipole is
p×r
p×r
(a) k · 2
(b) k · 3
r
r
p ·r
p ·r
(c) k · 2
(d) k · 3
r
r
17. Two charges +3.2 × 10–19 C and –3.2 × 10–9 C kept 2.4 Å
apart forms a dipole. If it is kept in uniform electric field
of intensity 4 × 105 volt/m then what will be its electrical
energy in stable equilibrium
(a) +3 × 10–23 J
(b) –3 × 10–23 J
–23
(c) –6 × 10 J
(d) –2 × 10–23 J
CONDUCTORS AND SELF-ENERGY AND
ELECTROSTATIC POTENTIAL
18. The self energy of a conducting shell of radius R and
charged Q is
(a) KQ2/R (b) KQ2/2R (c) 2KQ2/R (d) 3KQ2/5R
19. An uncharged sphere of metal is placed in between two
charged plates as shown. The lines of force look like
+
+
+
+
+
+
+
–
–
–
–
–
–
+
+
+
–
A
+
+
+
+
(a) 100 Vm–1 along X-axis
–
–
–
–
–
–
(b) 100 Vm–1 along Y-axis
+
+
+
+
+
(a)
(b) 4 N/C
3N/C
(c) 5 N/C
(d) None of these
13. Equipotential surfaces are shown in figure. Then the electric
field strength will be
Y
(cm)
10V
30V
30º
O
10
(c) 200
(d) 50
20V
Vm–1
Vm–1
X
20
30
(cm)
–
B
+ +
at an angle 120° with X-axis
at an angle 120° with X-axis
ELECTRIC DIPOLE
14. An electric dipole when placed in a uniform electric field E
will have minimum potential energy, if the positive direction
of dipole moment makes the following angle with E
(a) p
(b) p/2
(c) Zero
(d) 3p/2
15. An electric dipole of moment p is placed normal to the
lines of force of electric intensity E , then the work done in
deflecting it through an angle of 180° is
(a) pE
(c) –2pE
92
(b) +2pE
(d) Zero
C
+
(a) A
+
+
(b) B
+
D
+
+
(c) C
+
(d) D
JEE (XII) Module-1 PW
20. Conduction electrons are almost uniformly distributed
within a conducting plate. When placed in an electrostatic

field E , the electric field within the plate
(a) Is zero
(b) Depends upon E

(c) Depends upon E
(d) Depends upon the atomic number of the conducting
element
21. A ball having a charge –100e is placed at the center of a
hollow spherical shell which has a net charge of –20e. What
is the charge on the shell’s outer surface?
(a) + 80e
(b) –20e
(c) –100e
(d) –120e
22. A solid conducting sphere of radius a has a net positive
charge 2Q. A conducting spherical shell of inner radius b
and outer radius c is concentric with the solid sphere and
has a net charge –Q. The surface charge density on the inner
and out surface of th spherical shell will be
Y
X
(a)
2Q
ε0 A
(b)
Q
towards left
2 Aε0
(c)
Q
towards right
2 Aε0
(d)
Q
towards right
Aε0
25. Three concentric conducting spherical shells carry charges
as follows +4Q on the inner shell, –2Q on the middle shell
and – 5Q on the outer shell. The charge on the inner surface
of the outer shell is
(a) 0
(b) 4Q
(c) –Q
(d) –2Q
26. A charge Q is uniformly distributed over a large plastic
plate. The electric field at a point P close to the center and
just above the surface of the plate is 10 V/m. If the plastic
plate is replaced by a copper plate of the same geometrical
dimensions and carrying the same uniform charge Q, the
electric field at the point P will become
(a) Zero
(b) 5 V/m
(c) 10 V/m
(d) 20 V/m
27. A and B are two conducting concentric spherical shells. A is
given a charge Q while B is uncharged. If now B is earthed
as shown in Figure. Then
(a) − 2Q , Q
2
2
4πc
B
(b) − Q , Q
4πb 2 4πc 2
++ ++
(c) 0, Q
2
A
+ ++
4πc
(d) None of these
23. If charges are distributed in the sphere, where charge density
r

is ρ = ρ0 1 +  where ‘r’ is the distance from the center
 R
of the sphere then
R
r
(a) Sphere is conducting
(b) Sphere is non conducting
(c) Information is incomplete
(d) None of these
P Electrostatic Potential and Capacitance
W
(a)
(b)
(c)
(d)
+
++ +
++ ++
++
4πb
24. Two conducting plates X and Y, each having large surface
area ‘A’ as shown in figure (on one side) are placed parallel
to each other. The plate X is given a charge Q whereas the
other is neutral. The electric field at a point in between the
plates is given by
The charge appearing on inner surface of B is –Q
The field inside and outside A is zero
The field between A and B is not zero
The charge appearing on outer surface of B is zero
CAPACITANCE, PARALLEL PLATE
CAPACITOR, SPHERICAL CAPACITOR
28. The charge stored in a capacitor is 20mC and the potential
difference across the plates is 500 V. Its capacity is
(a) 0.04 µF
(b) 10–2 µF
(c) 2 × 10–6 µF
(d) 250 µF
93
29. What is the value of capacitance, if the thin metallic plate
is introduced between two parallel plates of area A and
separated at distance d?
(a)
0 A
d t
(b)
ε0 A
d
(c)
2ε0 A
d
(d)
3ε0 A
2d
30. The capacitance of a parallel plate capacitor is 60 μF. If the
distance between the plates is tripled and area doubled then
new capacitance will be
(a) 10 μF
(b) 40 μF
(c) 80 μF
(d) 100 μF
31. A 500 μF capacitor is charged at a rate of 125 μC/sec. The
potential difference across the capacitor will be 10 V after
an interval of
(a) 80 sec
(b) 50 sec
(c) 20 sec
(d) 40 sec
32. Between the plates of parallel plate condenser there is 1mm
thick medium sheet of dielectric constant 4. It is charged at
100 volt. The electric field in volt/ meter between the plates
of capacitor is
(a) 100000
(b) 100
(c) 25000
(d) 400000
33. The radius of the circular plates of a parallel plate capacitor
is R. Air is dielectric medium between the plates. If the
capacitance of the capacitor is equal to the capacitance of a
sphere of radius R, then the distance between the plates is
(a) R/4
(b) R/2
(c) R
(d) 2R
34. Sixty-four spherical drops each of radius 2 cm and carrying
5C charge combine to form a bigger drop. Its capacity is
36. Two spherical conductors A and B of radii a and b
(b > a) are placed concentrically in air. B is given a charge +Q
and A is earthed. The equivalent capacitance of the system is
+ + B
+
A
+
+
+
a
+
+
+ b
+
+
+ + +
+Q
 ab 
(a) 4πε0 

b − a 
(b) 4πε 0 [ a + b ]
 b2 
(d) 4πε 0 

b−a 
(c) 4πε 0 b
COMBINATION OF CAPACITORS
37. When two capacitors are joined in series the resultant
capacity is 2.4 µF and when the same two are joined in
parallel the resultant capacity is 10 µF. Their individual
capacities are
(a) 7µF, 3µF
(b) 1µF, 9µF
(c) 6µF, 4µF
(d) 8µF, 2µF
38. The equivalent capacitance between P and Q is
10µF
10µF
10µF
10µF
P
10µF
5µF
5µF
5µF
Q
(a) 10 µF
(b) 20 µF
(c) 5 µF
(d) 15 µF
39. The equivalent capacity between the points X and Y in the
circuit with C = 1µF
(a) 8 × 10−11 F
9
(b) 90 × 10–11 F
(c) 1.1 × 10–11 F
(d) 9 × 1011 F
35. The spheres shown in the figure are connected by a
conductor. The capacitance of the system is
ab 
(a) 4πε0 

b
 −a 
(c) 4πε 0 b
94
(b) 4πε0 a
 a2 
(d) 4πε0 

b−a
(a) 2 µF
(b) 3 µF
(c) 1 µF
(d) 0.5 µF
40. The resultant capacity between the points P and Q of the
given figure is
16
µF
3
(a) 4 µF
(b)
(c) 1.6 µF
(d) 1 µF
JEE (XII) Module-1 PW
41. Charge ‘Q’ taken from the battery of 12V in the circuit is
(a) 72 µC
(b) 36 µC
(c) 156 µC
(d) 20 µC
42. The equivalent capacitance of the network given below is
1 µF. The value of ‘C’ is
(a) 3 µF
(b) 1.5 µF
(c) 2.5 µF
(d) 1 µF
43. Four capacitors 4 μF, 6 μF, 5 μF, 3 μF are connected in series
to a source of 150V. The potential difference in volt, across
the 5 μF capacitor will be
(a) 30 volt
(b) 40 volt
(c) 50 volt
(d) 60 volt
47. The following arrangement consists of five identical metal
plates parallel to each other. Area of each plate is A and
separation between the successive plates is d. The equivalent
capacitance between P and Q is
5ε 0 A
(b) 7 ε0 A
d
3 d
ε0 A
5
ε
A
4
0
(c)
(d)
3
d
3 d
48. If each plate has area A and separation between successive
plates is d from equivalent capacitance between A and B is
(a)
Aε 0
3Aε 0
(b) Aε0
(c) 4Aε0
(d)
d
d
d
2d
49. A capacitor is made of a flat plate of area A and a second
plate having a stair like structure as shown in diagram. The
capacitance of the arrangement is
(a)
44. 4 plates of the same area of cross-section are joined. The
distance between plates is d. The effective capacitance of
combination is
(a) 3ε0 A
d
(b)
ε0 A
d
4ε0 A
(d) None of these
(c)
d
45. Three capacitors of capacitances 2μF, 5μF, 8μF are connected
first in parallel combination and then in series combination,
The ratio of their equivalent capacitances will be
(a) 15 : 1
(b) 20 : 1
(c) 13 : 1
(d) 12 : 1
DIELECTRIC AND ENERGY STORED IN
CAPACITORS
50. A parallel plate capacitor is made of two dielectric blocks
in series. One of the blocks has thickness d1 and dielectric
constant K1 and the other has thickness d2 and dielectric
constant K2 as shown in figure. This arrangement can be
thought as a dielectric slab of thickness d (= d1 + d2) and
effective dielectric constant K. The K is
46. For circuit the equivalent capacitance between A and B is
(each capacitor is of 3μF)
(a) 1 μF
(b) 1.5 μF
(c) 6 μF
(d) 8 μF
P Electrostatic Potential and Capacitance
W
11 ε0 A
(b) 18 ε 0 A (c)
(d) ε 0 A
18
d
3d
11 d
(a) ε0 A
d
d1
K1
d2
K2
(a)
K1d1 + K2 d2
(b)
K1d1 + K2 d2
(c)
K1K 2 ( d1 + d 2 )
( K1d 2 + K 2 d1 )
(d)
2K1 K2
d1 + d2
K1 + K2
K1 + K2
95
51. A condenser is charged to a p.d. of 120 volt. Its energy
is 1 × 10–5 joule. If the battery is there and the space
between plates is filled up with a dielectric medium
(er = 5), its new energy is
(a) 10–5 J
(b) 2 × 10–5 J
(c) 3 × 10–5 J
(d) 5 × 10–5 J
57. A parallel plate capacitor with air as the dielectric has
capacitance C. A slab of dielectric constant K and having
the same thickness as the separation between the plates
1
is introduced. So as to fill
of the capacitor. The new
3
capacitance will be
52. A 2µF condenser is charged to 500V and then the plates
are joined through a resistance. The heat produced in the
resistance in joule is
(a) 50 × 10–2 Joule
(a)
(b) 25 × 10–2 Joule
(c) 0.25 × 10–2 Joule
53. The plates of a parallel plate capacitor have an area of 90 cm2
each and are separated by 2 mm. The capacitor is charged
by connecting it to a 400 V supply. Then the density of the
energy stored in the capacitor is
(a) 0.113 Jm–3
(b) 0.226 Jm–3
(c) 0.152 Jm–3
(d) 0.176 Jm–3
54. In a capacitor of capacitance 20 μF the distance between
the plates is 5 mm. If a dielectric slab of width 2 mm and
dielectric constant 3 is inserted between the plates, then the
new capacitor will be
(a) 22.3μF
(b) 25 μF
(c) 19 μF
(d) 27.2 μF
55. Two parallel plates of area A are separated by two different
dielectric. The net capacitance is
(a)
ε0 A
d
(b)
7 ε0 A
20 d
(c)
20 ε0 A
7 d
(d)
2ε0 A
d
56. A capacitance of an air capacitor is 25 μF the separation
between the parallel plates is 8 mm. An aluminum plate
of thickness 2mm thickness is introduced symmetrically
between the plates. The new capacitance will be
96
C
C
K 2
3
(c) 2C
C
(d)
2K 3 3
58. A parallel plate capacitor is filled with dielectrics. What is
its capacitance?
(b)
(d) 0.5 × 10–2 Joule
(ε0 = 8.8 × 10–12 F/m)
( K + 1)
(a) 10 μF
(b) 33.3 μF
(c) 44 μF
(d) 50 μF
(a)
d/2
K1
d/2
K1
2 K1 K 2
K1 + K 2
(c) 2 o A K1 K 2 d K1 K 2 (b)
K1 K 2
K1 + K 2
(d) 2o A 2 K1 K 2 d 2 K1 K 2 SHARING OF CHARGE, CAPACITOR CIRCUITS
59. Two spheres of radii 1 cm and 2 cm have been charged
with 1.5 × 10–8 and 0.3 × 10–7 coulomb of positive charge
respectively. When they are connected with a wire, then
charge:
(a) Will flow from the first to the second
(b) Will flow from the second to the first
(c) Will not flow at all
(d) May flow either from first to second, or from the second
to first, depending upon the length of the connecting
wire
60. A capacitor of 30 µF charged to 100 V is connected in parallel
to capacitor of 20 µF charged to 50 volt. The common
potential is
(a) 75 V
(b) 150 V
(c) 50 V
(d) 80 V
61. Two spherical conductors each of capacity C are charged to
potential 25 V and 10 V. These are connected by a wire then
the loss in energy will be (C = 5 F)
(a) 125 J
(b) 425 J
(c) 700 J
(d) 281.25 J
JEE (XII) Module-1 PW
62. A capacitor of capacity C1 is charged upto V volt and then
connected to an uncharged capacitor of capacity C2. Then
final potential difference across each will be
(a) C1V1 + C2V2
C1 + C2
(c) C1 + 2C2
(b) C1
C1 C2
V
C C2 (d) 1
V
C1C2 63. Few capacitors are connected to a battery of 12V. What will
be charge on each capacitor (in μC)?
(a)
(b)
(c)
(d)
20, 14
12, 20
20, 20
12, 12
64. A light bulb, a capacitor and a battery are connected together
as shown in figure with switch initially open when the switch
S is closed which one of the following is true
(a) The bulb will light up for an instant when the capacitor
start charging
(b) The bulb will light up when the capacitor is fully charged
(c) The bulb will not light up at all
(d) The bulb will light up and go off at regular intervals
65. In the circuit the potential difference across the 4.5μF
capacitor is
(a) 8/3 volt
(c) 6 volt
66. The correct combination is
(symbols have their usual meanings)
(b) 4 volt
(d) 8 volt
P Electrostatic Potential and Capacitance
W
(a)
(b)
(c)
(d)
Q1 = Q2 = Q3 and V1 = V2 = V3 = V
Q1 = Q2 + Q3 and V = V1 + V2 + V3
Q1 = Q2 + Q3 and V = V1 + V2
Q2 = Q3 and V2 = V3
67. Two identical capacitors are connected in parallel across a
potential difference V. After they are fully charged the battery
is removed and the positive of first capacitor is connected
to negative of second and negative of first is connected to
positive of other. The loss in energy will be
(b) CV2
(a) 1 CV 2
2
CV 2
(c)
(d) Zero
4
68. Three plates A, B, C each of area 50 cm2 have separation
3 mm between A and B and 3mm between B and C. The
energy stored when the plates are fully charged is
(a) 1.6 × 10–19 J
(c) 5 × 10–9 J
(b) 2.1 × 10–9 J
(d) 7 × 10–9 J
69. A 2 μF capacitor is charged to a potential difference of 30
V. The battery is removed and it is connected to a series
combination of two uncharged capacitors as shown in figure.
The final potential difference across C1 will be
(a) 10 V
(c) 30 V
(b) 20 V
(d) 5 V
97
Exercise-2 (Learning Plus)
1. Three equal charges are placed at the three corners of an
isosceles triangle as shown in the figure. The statement
which is true for electric potential V and the field intensity
E at the center of the triangle is
q
6. Two identical particles of mass m carry a charge Q
each. Initially one is at rest on a smooth horizontal
plane and the other is projected along the plane directly
towards first particle from a large distance with speed v.
The closest distance of approach will be
(a)
1 Q2
4πε0 mv
(b)
1 4Q 2
4πε0 mv 2
(c)
1 2Q 2
4πε0 mv 2
(d)
1 3Q 2
4πε0 mv 2
O
q
q
(a) V = 0, E = 0
(c) V ≠ 0 , E = 0
(b) V = 0, E ≠ 0
(d) V ≠ 0, E ≠ 0
2. A particle A has charge +q and particle B has charge
+4q with each of them having the same mass m. When
allowed to fall from rest through same electrical potential
difference, the ratio of their speed vA : vB will be
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
3. A charged particle ‘q’ is shot from a large distance with
speed v towards a fixed charged particle. It approaches Q
upto a closest distance r and then returns. If q were given a
speed ‘2v’, the closest distance of approach would be
q
Q
v
(a) r
(b) 2r
r
(c)
(d) r
2
4
4. Figure represents a square carrying charges +q, +q, –q, –q
at its four corners as shown. Then the potential will be zero
at points
P
+q
+q
A
–q
(a) A, B, C, P and Q
(c) A, P, C and Q
B
Q
C
–q
(b) A, B and C
(d) P, B and Q
5. A particle of mass 2 g and charge 1mC is held at rest on a
frictionless horizontal surface at a distance of 1m from a
fixed charge of 1 mC. If the particle is released it will be
repelled. The speed of the particle when it is at distance of
10m from the fixed charge is
98
(a) 100 m/s
(b) 90 m/s
(c) 60 m/s
(d) 45 m/s
7. Figure shows equi-potential surfaces for a two charges
system. At which of the labelled point will an electron have
the highest potential energy?
B –q
+q
D
A
(a) Point A
(c) Point C
C
(b) Point B
(d) Point D
8. In a regular polygon of n sides, each corner is at a distance
r from the center. Identical charges are placed at (n – 1)
corners. At the center, the intensity is E and the potential is
V. The ratio V/E has magnitude.
(a) r n
(b) r(n – 1)
(c) (n –1)/r
(d) r(n –1)/n
9. Two identical thin rings, each of radius R meter are coaxially
placed at distance R meter apart. If Q1 and Q2 coulomb are
respectively the charges uniformly spread on the two rings,
the work done in moving a charge q from the center of one
ring to that of the other is
(a) Zero
(b) q (Q1 − Q2 )( 2 − 1) / ( 2.4πε0 R)
(c) q 2 (Q1 + Q2 ) / 4πε0 R
(d) q (Q1 − Q2 )( 2 + 1) / ( 2.4πε0 R)
JEE (XII) Module-1 PW
10. If the electric potential of the inner metal sphere is
10 volt & that of the outer shell is 5 volt, then the potential
at the center will be
15. The electric potential V as a function of distance x
(in metre) is given by
V = (5x2 + 10x – 9) volt.
The value of electric field at x = 1 m would be
(a) – 20 volt/m
a
(b) 6 volt/m
b
(c) 11 volt/m
(a) 10 volt
(c) 15 volt
(b) 5 volt
(d) 0
11. When charge of 3 coulomb is placed in a uniform electric
field , it experiences a force of 3000 newton, within this
field, potential difference between two points separated by
a distance of 1 cm is
(a) 10 Volt
(b) 90 Volt
(d) 3000 Volt
(c) 1000 Volt
12. The potential difference between points A and B in the
given uniform electric field is
a
C
B
E
(d) –23 volt/m
16. The electric field in a region is directed outward and is
proportional to the distance r from the origin. Taking
the electric potential at the origin to be zero, the electric
potential at a distance r
(a) Increases as one goes away from the origin
(b) Is proportional to r2
(c) Is proportional to r
(d) Is uniform in the region
17. The diagram shows three infinitely long uniform line
charges placed on the X, Y and Z axis. The work done
in moving a unit positive charge from (1, 1, 1) to
(0, 1, 1) is equal to
b
Y 3
A
E
(a) Ea
(b) E (a 2 + b 2 )
(c) Eb
(d) ( Eb / 2 )
13. The equation of an equipotential line in an electric
field is y = 2x, then the electric field strength vector at
(1, 2) may be
(a) 4iˆ + 3 ˆj
(b) 4iˆ + 8 ˆj
(c) 8iˆ + 4 ˆj
(d) –8iˆ + 4 ˆj
0.2 m
14. A, B, C, D, P and Q are points in a uniform electric
field. The potentials at these points are V(A) = 2 volt.
V(P) = V(B) = V(D) = 5 volt. V(C) = 8 volt. The electric
field at P is
B
C
P
A
(a)
(c)
(a) 10
along PQ
P Electrostatic Potential and Capacitance
W
2πε
( 3λn2 )
2πε 0
σq
(r2 − r1 )
3ε 0
(b)
σq
(r2 − r1 )
2ε 0
(c)
2σq
(r2 − r1 )
ε0 m
(d)
σq
(r2 − r1 )
ε0 m
(b) 15 2 Vm–1 along PA
(c) 5 Vm–1 along PC
(d) 5 Vm–1 along PA
( λn2 )
(a)
0.2 m
Vm–1
Z
(b)
( λln2 )
πε0
(d) None of these
18. A electric charge q, and mass m is initially at rest. It is at a
distance r1 from an infinite plane of positive charge having
surface charge density s. If charge is now released, find its
velocity at a distance r2 from the plane.
Q
D
X
2
99
19. Figure shows three circular arcs, each of radius R and total
charge as indicated. The net electric potential at the center
of curvature is
22. A point charge q is brought from infinity (slowly so that
heat developed in the shell is negligible) and is placed at the
center of a conducting neutral spherical shell of inner radius
a and outer radius b, then work done by external agent is
45º
b
30º
–2Q
q
a
+3Q
(a)
Q
2πε0 R
(b)
Q
4πε0 R
(a) 0
(c)
2Q
πε0 R
(d)
Q
πε0 R
(c)
20. Six charges of magnitude +q and –q are fixed at the corners
of a regular hexagon of edge length a as shown in the figure.
The electrostatic potential energy of the system of charged
particles is
+q
k q2 k q2
–
2b
2a
(b)
k q2
2b
(d)
k q2 k q2
–
2a
2b
23. Two short electric dipoles are placed as shown (r is the
distance between their centers). The energy of electric
interaction between these dipoles will be
P1
–q
r
–q
+q
C
+q
(a)
q 2  3 15 
− 

π ∈0 a  8
4
(b)
q2  3 9 
− 

π ∈0 a  2 4 
–q
(C is center of dipole of moment P2)
(a) 2 k P1 P2 cos θ
r3
(b) − 2 k P1 P2 cos θ
r3
(c) − 2 k P1 P2 sin θ
r3
q  3 15 
− 
(c)

π ∈0 a  4
2
2
(d)
(d) − 4 k P1 P2 cos θ
r3
q 2  3 15 
− 

π ∈0 a  2
8
21. In the figure shown, A is a fixed charged. B (of mass m) is
given a velocity V perpendicular to line AB. At this moment
the radius of curvature of the resultant path of B is
24. For an infinite line of charge having charge density
λ lying along x-axis, the work required in moving charge
from C to A along arc CA is
B
C
a
V
A
+q
r
A
B
+q
a
++++++++++++
(a) 0
(b) ∞ (infinity)
(c)
100
2
4π ∈0 r mV
q2
(d) r
P2
2
(a)
qλ
log e 2
πε 0
(b)
qλ
log e 2
4πε 0
(c)
qλ
log e 2
4πε 0
(d)
qλ
1
log e
2πε 0
2
X
JEE (XII) Module-1 PW
25. Find the equivalent capacitance of circuit and charge on 5
µF capacitor
12µF
5µF
10µF
9µF
8µF
29. In the circuit shown, if the charge present in the first vertical
branch capacitor is equal to Q then what is the charge in the
Nth vertical branch capacitor of capacitance C?
2C
2C
C
C
+ 60 V –
(a) 4 µF, 50 µC
(c) 4 µF, 25 µC
(b) 8 µF, 25 µC
(d) 8 µF, 50 µC
26. The potential across a 3 µF capacitor is 12 V when it is not
connected to anything. It is then connected in parallel with
an uncharged 6 µF capacitor. At equilibrium, the charge q
on the 3 µF capacitor and the potential difference V across
it are
(a) q = 12 µC, V = 4 V
(b) q = 24 µC, V = 8 V
(c) q = 36 µC, V = 12 V (d) q = 12 µC, V = 6 V
27. Two identical capacitors in parallel are charged with charge
Q each. Separation of the plates in each capacitor is d0.
Suddenly, the left plate of the first capacitor and the right
plate of the second capacitor starts moving to the left with
speed v, then
2C
2C
2C
C
Observe carefully that the last vertical
capacitor is of 2C. All the remaining
vertical capacitors are of C.
(a) Q/2N
(b) Q/2N–1
(c) Q
(d) 2NQ
30. Both capacitors are initially uncharged and then connected as
shown and switch is closed. What is the potential difference
across the 3µF capacitor?
19 V
2 µF
9V
3 µF
15 V
(a) 30 V
(b) 10 V
(c) 25 V
(d) None of these
(a) Charge on the two capacitor as a function of time are
Q(d 0 − vt ) Q(d 0 + vt )
,
2d 0
2d 0
(b) Charge on the two capacitors as a function of time are
Qd 0
Qd 0
,
2(d 0 − vt ) 2(d 0 + vt )
31. Figure shows an arrangement of four identical rectangular
plates A, B, C and D each of area S. Ignore the separation
between the plates in comparison to the plate dimensions.
a
b
c
+Q1
+Q2
A
C
(c) Current in the circuit will increase as time passes
(d) Current in the circuit will be constant
28. Four capacitors and two sources of e.m.f. are connected as
shown in the figure. The p.d. in volts between the points a
and b is
3µF
b
13V
1µF
3µF
1µF
a
(a) zero
(b) 13
(c) 17
(d) 27
P Electrostatic Potential and Capacitance
W
B
D
(a) Potential difference between plates A and B is
independent of Q1
(b) Potential difference between plates C and D is
independent of Q1
(c) Potential difference between plates A and B is
independent of Q2
(d) Potential difference between plates C and D is
independent of Q2
101
32. Four uncharged capacitors c1, c2, c3 and c4 are connected as
shown in the figure. The points A, B, C and D are kept at
potential v1, v2, v3 and v4 respectively. Then the potential v0
at point O is
35. In the circuit, both capacitors are identical. Column-I
indicates action done on capacitor 1 and Column-II indicates
effect on capacitor 2. Select correct alternative.
(1)
(2)
v4 D
c4
A
v1
v3
C
O
c1
c3
c2
v2 B
(a) v1 + v2 + v3 + v4
v1c1 + c2 v2 + c3 v3 + c4 v4
c1 + c2 + c3 + c4
(b)
1 c1v1 + c2 v2 + c3 v3 + c4 v4
1
(d)
(v1 + v2 + v3 + v4 )
2
c1 + c2 + c3 + c4
4
33. In the circuit shown in figure initially K1 is closed and K2
is open for a long time. Then K1 was opened and K2 was
closed (order is important), what will be the charge on C2
& C3 now? [C = 1 µF]
(c)
C1 = 6C
E = 9V
K1
K2
•
•
C2 = 3C
C3 = 3C
Column-I
Column-II
A. Plates are moved further p. Amount of charge on left
apart.
plate increases
B. Area increased
q. Potential
difference
increases
C. Left plate is earthed
r. Amount of charge on
right plate decreases
It’s
Plates
are
short
D.
s. None of the above effects
circuited
(a) A-(r); B-(p,q); C-(s); D-(p,q)
(b) A-(s), B-(p,q); C-(r), D-(p,q)
(c) A-(p,q); B-(r), C-(s), D-(p,q)
(d) A-(s), B-(r), C-(p,q); D-(p,q)
36. In the network shown we have three identical capacitors.
Each of them can withstand a maximum 100 V p.d. What
maximum voltage can be applied across A and B so that no
capacitor gets spoiled?
C
(a) 18 µC, 0 µC
(b) 0 µC, 18 µC
(c) 9 µC, 9 µC
(d) 4.5 µC, 4.5 µC
34. In figure shown, a capacitor is partially filled with 2
dielectric slabs. The capacitor is connected to a battery of
voltage V. Column-I indicates the region shown in figure and
Column-II indicates the electric field in that region. Match
the appropriate entries.
C
A
(a) 150 V
B
(b) 120 V
d
A/2
R K=2
B. Q
C. R
D. S
d/2
Column-II
p. 8V
5d
q. 4V
3d
r. 2V
5d
(d) A-(p), B-(q), C-(r), D-(s)
4µF
2µF
2µF
4µF
(a) 400/3 µC (b) 100 µC (c) 50 µC
(d) 100/3 µC
38. Two identical capacitors are connected in series as shown
in the figure. A dielectric slab (k > 1) is placed between the
plates of the capacitor B and the battery remains connected.
Which of the following statement(s) is/are correct following
the insertion of the dielectric?
Before
A C
s. 2V
3d
(a) A-(p), B-(q), C-(s), D-(r)
(b) A-(q), B-(p), C-(s), D-(r)
(c) A-(q), B-(p), C-(r), D-(s)
102
(d) 200 V
50 V
Q A/2
K=4 S
Column-I
A. P
(c) 180 V
37. The circuit was in the shown state from a long time. Now the
switch S is closed. The charge that flows through the switch is
A
P
C
V
B
(a)
(b)
(c)
(d)
C
C
V
C
The charge supplied by the battery increases
The capacitance of the system increases
The electric field in the capacitor B increases
The electrostatic potential energy decreases
JEE (XII) Module-1 PW
Exercise-3 (JEE Advanced Level)
MULTIPLE CORRECT TYPE QUESTIONS
1. Potential at a point A is 3 volt and at a point B is 7 volt, an
electron is moving towards A from B
(a) It must have some K.E. at B to reach A
(b) It need not have any K.E. at B to reach A
(c) To reach A it must have more than or equal to 4 eV KE
at B
(d) When it will reach A, it will have K.E. more than or at
least equal to 4 eV if it was released from rest at B
2. Three points charges are placed at the corners of an
equilateral triangle of side L as shown in the figure
+q
–
O
+
axis
–
(a) The potential at all the points on the axis will be zero
(c) The direction of electric field at all points on the axis
will be along the axis
L
L
+
(b) The electric field at all the points on the axis will be
zero
–2q
L
6. The figure shows a nonconducting ring which has positive
and negative charge non uniformly distributed on it such that
the total charge is zero. Which of the following statements
is false?
+q
(a) The potential at the centroid of the triangle is zero
(b) The electric field at the centroid of the triangle is zero
(c) The dipole moment of the system is
2 qL
(d) If the ring is placed inside a uniform external electric
field then net torque and force acting on the ring would
be zero
7. Which of the following is true for the figure showing
electric lines of force? (E is electrical field, V is potential)
(d) The dipole moment of the system is
3 qL
3. Which of the following quantities depend on the choice of
zero potential or zero potential energy?
(a) Potential at a particular point
(b) Change in potential energy of a two-charge system
(c) Potential energy of a two - charge system
(d) Potential difference between two points
4. At distance of 5 cm and 10 cm outwards from the surface
of a uniformly charged solid sphere, the potentials are 100V
and 75V respectively. Then
(a) Potential at its surface is 150V
(b) The charge on the sphere is (5/3) × 10–9 C
(c) The electric field on the surface is 1500 V/m
(d) The electric potential at its center is 225 V
5. The electric potential decreases uniformly from 180
V to 20 V as one moves on the X-axis from x = – 2 cm
to x = + 2 cm. The electric field at the origin
(a) Must be equal to 40 V/cm
(b) May be equal to 40 V/cm
(c) May be greater than 40 V/cm
(d) May be less than 40 V/cm
P Electrostatic Potential and Capacitance
W
B
A
(a) EA > EB
(c) VA > VB
(b) EB > EA
(d) VB > VA
8. A and B are two conducting concentric spherical shells. A is
given a charge Q while B is uncharged. If now B is earthed
as shown in figure. Then
B
+
++ + +
+
+ A +
+
+
+ + ++
(a) The charge appearing on inner surface of B is –Q
(b) The field inside and outside A is zero
(c) The field between A and B is not zero
(d) The charge appearing on outer surface of B is zero
103
9. The plates of a parallel plate capacitor with no dielectric are
connected to a voltage source. Now a dielectric of dielectric
constant K is inserted to fill the whole space between the
plates with voltage source remaining connected to the
capacitor.
(a) The energy stored in the capacitor will become K-times
(b) The electric field inside the capacitor will decrease to
K-times
(c) The force of attraction between the plates will increase
to K2-times
(d) The charge on the capacitor will increase to K-times
10. 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
(c) The net potential difference across them is different
from the sum of the individual initial potential
differences
(d) The net energy stored in the two capacitors is less than
the sum of the initial individual energies
14. Charge Q coulombs is uniformly distributed throughout the
volume of a solid hemisphere of radius R metres. Then the
potential at center O of the hemisphere in volts is
R
Q1 + Q2
C
(b)
Q3
C
(d)
Q2
C
1
[(Q1 + Q2 ) − (Q3 − Q4 )]
C
11. Rows of capacitors containing 1, 2, 4, 8, ..........∞ capacitors,
each of capacitance 2F, are connected in parallel as shown
in figure. The potential difference across AB = 10 volt,
then
A
B
Row1
Row2
Row3
(a) Total capacitance across AB is 4F
(b) Charge of each capacitor will be same
(c) Charge on the capacitor in the first row is more than on
any other capacitor
(d) Energy of all the capacitors is 50 J
12. The two plates X and Y of a parallel plate capacitor of
capacitance C are given a charge of amount Q each. X is
now joined to the positive terminal and Y to the negative
terminal of a cell of emf E = Q/C.
(a) Charge of amount Q will flow from the negative
terminal to the positive terminal of the cell inside it
(b) The total charge on the plate X will be 2Q
(c) The total charge on the plate Y will be zero
(d) The cell will supply CE2 amount of energy
104
(b) Net charge on connected plates equals the sum of
initial charges
O
Q4
Q2
(c)
(a) Net charge on connected plates is less than the sum of
initial individual charges
Q3
Q1
(a)
13. When two identical capacitors are charged individually to
different potentials and connected parallel to each other, after
disconnecting them from the source
(a)
1 3Q
4πε0 2 R
(b)
1 3Q
4πε0 4 R
(c)
1 Q
4πε0 4 R
(d)
1 Q
4πε 0 8 R
COMPREHENSION BASED QUESTIONS
Comprehension (Q. 15 to 17): A leaf electroscope is a simple
apparatus to detect any charge on a body. It consists of two metal
leaves OA and OB, free to rotate about O. Initially both are very
slightly separated. When a charged object is touched to the metal
knob at the top of the conducting rod, charge flows from knob
to the leaves through the conducting rod. As the leaves are now
charged similarly, they start repelling each other and get separated,
(deflected by certain angle).
Metal knob
Metal rod
+
+++
Rubber
O
Glass
window
Gold leaves
+ ++
+
+
+
++
A B
O
A
B
The angle of deflection in static equilibrium is an indicator of the
amount of charge on the charged body.
15. When a + 20 C rod is touched to the knob, the deflection
of leaves was 5°, and when an identical rod of – 40 C is
touched, the deflection was found to be 9°. If an identical
rod of +30 C is touched, then the deflection may be
(a) 0
(b) 2°
(c) 7°
(d) 11°
JEE (XII) Module-1 PW
16. If we perform these steps one by one.
(ii)
(a)
5
2
+ +++
++
r
0
–5
Then the
positively
charged rod is
touched to the
knob
(b)
(iii) Now the +vely
charged rod is
removed, and
a negatively
charged rod of
same magnitude
is brought closer
at same distance
+ +++
++
(c)
––––––
In which case, the leaves will converge (come closer), as
compared to the previous state?
(a)
(b)
(c)
(d)
5
(10 Nm /C)
A positively
charged rod is
brought closer
to initially
uncharged knob
(i)
(i)
(i) and (iii)
Only (iii)
In all cases, the leaves will diverge
17. In an electroscope, both leaves are hinged at the top point O.
Each leaf has mass m, length l and gets charge q. Assuming
the charge to be concentrated at ends A and B only, the
small angle of deviation (θ) between the leaves in static
equilibrium, is equal to
1/3
1/3
 4 kq 2 
(a) 
 l 2 mg 


 k q2 
(b) 
 l 2 mg 


1/3
1/3
 2k q 2 
(c)  2

 l mg 
 64 k q 2 
(d)  2

 l mg 
Comprehension (Q. 18 to 20): A charged particle is suspended
at the center of two thin concentric spherical charged shells, made
of non conducting material. Figure A shows cross section of the
arrangement. Figure B gives the net flux f through a Gaussian
sphere centered on the particle, as a function of the radius r of the
sphere.
shell A
charged
particle
shell B
O
rA
rB
Figure A
P Electrostatic Potential and Capacitance
W
Figure B
18. What is the charge on the central particle?
(a) 0.2 mC
(b) 2 mC
(c) 1.77 mC
(d) 3.4 mC
19. What is the charge on shell A?
(a) 5.31 × 10–6 C
(b) –5.31 × 10–6 C
–6
(c) –3.54 × 10 C
(d) –1.77 × 10–6 C
20. In which range of the values of r is the electric field zero?
(a) 0 to rA
(b) rA to rB
(c) For r > rB
(d) For no range of r, electric field is zero
Comprehension (Q. 21 to 23): A solid conducting sphere of
radius ‘a’ is surrounded by a thin uncharged concentric conducting
shell of radius 2a. A point charge q is placed at a distance 4a from
common center of conducting sphere and shell. The inner sphere
is then grounded.
2a
4a
a
q
21. The charge on solid sphere is
(a) −
q
2
(b) − q
4
(c) −
q
8
(d) −
q
16
22. Pick up the correct statement
(a) Charge on surface of inner sphere is non-uniformly
distributed
(b) Charge on inner surface of outer shell is non-uniformly
distributed
(c) Charge on outer surface of outer shell is non-uniformly
distributed
(d) All the above statements are false
23. The potential of outer shell is
(a)
q
32πε0 a
(b)
q
16πε0 a
(c)
q
8πε0 a
(d)
q
4πε0 a
105
Comprehension (Q. 24 to 27): The figure shows a diagonal
symmetric arrangement of capacitors and a battery.
24. Identify the correct statements
MATCH THE COLUMN TYPE QUESTIONS
28. Column-I gives a situation in which two dipoles of dipole
moment piˆ and 3 p ˆj are
y
placed at origin. A circle of
radius R with center at origin
(a) Both the 4µF capacitors carry equal charges in
opposite sense
(b) Both the 4µF capacitors carry equal charges in same
sense
(c) VB – VD > 0
A.
25. If the potential of C is zero, then
(a) VA = + 20V
C.
(b) 4(VA – VB) + 2(VD – VB) = 2VB
(c) 2(VA – VD) + 2(VB – VD) = 4VD
(d) VA = VB + VD
26. The potential of the point B and D are
(a) VB = 8V
D.
(b) VB = 12V
(c) VD = 8V
(d) VD = 12V
27. The value of charge q1, q2 and q3 as shown in the figure are
q2
q1
B
+ –
+ –
+
q3
–
+ – C
+
A
D
q1
q2
(d) q1 = 3 µC ; q2 = 4 µC ; q3 = + 2 µC
Column-Ι
shell I
106
The coordinates of point
on circle where potential
is maximum
The coordinates of point
on circle where potential
is zero
The coordinates of point
on circle where magnitude
of electric field intensity is
1 4p
4πε0 R3
p.
The coordinates of
point on circle where
magnitude of electric
field intensity is
1 2p
4πε0 R3
s.
q.
r.
R 3 R
 ,

2 2 
 R
3 R
− ,−

2 
 2

3 R R
, 
−
2 2

 3 R R
,− 

2
 2
(a) A-(p); B-(r,s); C-(p,q) D-(r,s)
(b) A-(p); B-(r,q); C-(p,r) D-(r,s)
29. Column-I gives certain situations involving two thin
conducting shells connected by a conducting wire via a key K.
In all situations one sphere has net charge +q and other sphere
has no net charge. After the key K is pressed, column-II gives
some resulting effect. Match the figures in Column-I with
the statements in Column-II.
(c) q1 = 32 µC ; q2 = 24 µC ; q3 = + 8 µC
+q
Column-II
(d) A-(r); B-(p,q); C-(p,r) D-(r,s)
(b) q1 = 48 µC ; q2 = 16 µC ; q3 = + 8 µC
initially no
net charge
R
pi^
(c) A-(q); B-(r,s); C-(p,q); D-(r,q)
+ –
E = 20 V
(a) q1 = 32 µC ; q2 = 24 µC ; q3 = – 8 µC
A.
x
Column-I
B.
(d) VD – VB > 0
3pjˆ
is drawn as shown in figure.
Column-II gives coordinates
of certain positions on the
circle. Match the statements in
Column-I with the statements
in Column-II.
p.
Column-II
Charge flows through connecting wire.
K
shell II
JEE (XII) Module-1 PW
B.
+q
q.
Potential energy of system of spheres decreases.
r.
No heat is produced.
s.
The sphere I has no charge after equilibrium is reached.
K
shell II
shell I
C.
initially no
net charge
+q
K
shell I
shell II
D.
+q
initially no
net charge
K
shell I
shell II
(a) A-(p,q); B-(p,q); C-(p,q,s); D-(r,s)
(b) A-(p,q); B-(p,q,s); C-(r,s); D-(q,r)
(c) A-(q,r); B-(p,q); C-(p,r); D-(p,q,r)
(d) A-(q,r); B-(p,q,r); C-(r); D-(r,s)
30. In each situation of Column-I, some charge distributions are given with all details explained. In Column-II The electrostatic
potential energy and its nature is given situation in Column-II. Then match situation in Column-I with the corresponding results
in Column-II.
A.
Column-Ι
A thin shell of radius a and having a charge –Q
uniformly distributed over its surface as shown.
Column-II
–Q
p.
1 Q2
in magnitude
8π ∈0 a
q.
3 Q2
in magnitude
20π ∈0 a
r.
2 Q2
in a magnitude
5π ∈0 a
s.
Positive in sign
a
B.
C.
5a
and having a charge –Q
2
uniformly distributed over its surface and a point
charge –Q placed at its center as shown.
–Q
A solid sphere of radius a and having a charge –Q
uniformly distributed throughout its volume as
shown.
–Q
A thin shell of radius
–Q
5a
2
a
D.
A solid sphere of radius a and having a charge –Q
uniformly distributed throughout its volume. The
solid sphere is surrounded by a concentric thin
uniformly charged spherical shell of radius 2a and
carrying charge –Q as shown.
–Q
–Q
a
2a
(a) A-(p,q); B-(p,q); C-(p,q,s); D-(r,s)
(b) A-(p,s); B-(q,s); C-(q,s); D-(s)
(c) A-(q,r); B-(p,q); C-(p,r); D-(p,q,r)
(d) A-(q,r); B-(p,q,r); C-(r); D-(r,s)
P Electrostatic Potential and Capacitance
W
107
31. On a capacitor of capacitance C 0 following steps are
performed in the order as given in column-I.
(i) Capacitor is charged by connecting it across a battery
33. In the shown fig. points A and B are 4 cm apart along the
lines of a uniform field E = 600 ˆi V/m . Find the change in
potential |VB – VA| in volt.
of EMF V0
E
(ii) Dielectric of dielectric constant K and thickness d is
inserted
(iii) Capacitor is disconnected from battery
A
B
(iv) Separation between plates is doubled
Column-I
(Steps performed)
Column-II
Final value of Quantity
(Symbols have usual
meaning)
A. (i) (iv) (iii) (ii)
p.
B. (iv) (i) (iii) (ii)
q.
C. (ii) (i) (iii) (iv)
r.
D. (i) (ii) (iv) (iii)
s.
Q=
C0V0
2
Q=
KC0V0
K +1
C=
KC0
K +1
V=
V0 ( K + 1)
2K
34. Four point charges of 0.6 µC, 2.2 µC, –3.6 µC and
+ 4.8 µC are placed at the corners of a square of side
10 cm. What is the work done to bring a charge of
–5 µC from infinity to the center of the square in Joule
(Assume speed of –5µC charge is kept constant).
35. An electric field is given as =
E 2 xiˆ − 3 y 2 ˆj N /C . Find the
change in potential in volt from the position r = ˆi − 2jˆ m
A
to rˆB = 2 iˆ + ˆj + 3 kˆ m.
36. A uniform electric field of 400 V/m is directed at 37° below
the x-axis, as shown in figure. Find the changes in potential
VB – VA in volt.
(a) A-(p,r,s); B-(p,r,s); C-(r); D-(q,r)
(b) A-(p,r,s); B-(p,q,s); C-(r,s); D-(q,r)
(c) A-(q,r); B-(p,r,s); C-(p,r), D-(p,q,r)
(d) A-(q,r); B-(p,q,r); C-(r); D-(p,r,s)
A 3cm B 37º
x
NUMERICAL TYPE QUESTIONS
32. Figure shows two equipotential (dashed) surfaces such that
VA = –5V and VB = –15 V. The external work needed to
move a –2 µC charge at constant speed from A to B along
the indicated path in Joule is given by x × 10–5 then x is
B
A
108
37. A 10 µF capacitor is fully charged across a 12 volt battery.
The capacitor is then disconnected from the battery and
connected across an initially uncharged capacitor, C. The
voltage across each capacitor is now 3 volts. What is the
unknown capacitance C (in µF)?
38. 2 conducting objects one with charge of +Q and another with
–Q are kept on x-axis at x = –3 and x = + 4 respectively. The
4

electric field on the x-axis is given by 3Q  x 2 +  . What
3

–1
is the value of 1/C (in F ) of this configuration of objects,
where C is capacitance?
JEE (XII) Module-1 PW
SUBJECTIVE TYPE QUESTIONS
39. A positive charge is distributed in a spherical region
with charge density r = r 0 r for r ≤ R (where r 0 is
a positive constant and r is the distance from center).
Find out electric potential and electric field at following
locations.
C
41. The electric potential varies in space according to the relation
V = 3x + 4y. A particle of mass 10 kg starts from rest from
point (2, 3.2) under the influence of this field. Find the
velocity of the particle when it crosses the x-axis. The charge
on the particle is +1C. Assume V and (x, y) are in S.I. units.
42. A small dipole of dipole moment P= (iˆ + ˆj ) µCm is placed
at the origin. Find out electric potential at a point having
position vector =
r (4iˆ + 3 ˆj ) m .
(a) At a distance r from center inside the sphere.
(b) At a distance r from center outside the sphere.
40. A solid sphere of radius ‘R’ is uniformly charged with charge
R
density ρ in its volume. A spherical cavity of radius
is
2
made in the sphere as shown in the figure.
Find the electric potential at the center of the sphere.
Exercise-4 (Past Year Questions)
6. In an electrical circuit, a battery is connected to pass 20 C
of charge through it in a certain given time. The potential
difference between two plates of the battery is maintained
at 15 V. The work done by the battery is ___________ J.
JEE MAIN
1. 27 identical drops are charged at 22V each. They combine
to form a bigger drop. The potential of the bigger drop will
be _________ V.
(2022)
2. Sixty four conducting drops each of radius 0.02 m and each
carrying a charge of 5µC are combined to form a bigger drop.
The ratio of surface density of bigger drop to the smaller
drop will be
(2022)
(a) 1 : 4
(b) 4 : 1
(c) 1 : 8
(d) 8 : 1
3. If the electric potential at any point (x, y, z) m in space is
given by V = 3x2 volt. The electric field at the point (1, 0, 3)
m will be
(2022)
(a)
(b)
(c)
(d)
3 Vm–1, directed along positive x-axis
3 Vm–1, directed along negative x-axis
6 Vm–1, directed along positive x-axis
6 Vm–1, directed along negative x-axis
4. 512 identical drops of mercury are charged to a potential of
2V each. The drops are joined to form a single drop. The
potential of this drop is ___________ V.
(2021)
5. Two identical conducting spheres with negligible volume
have 2.1nC and –0.1 nC charges, respectively. They are
brought into contact and then separated by a distance of
0.5 m. The electrostatic force acting between the spheres is
1

___________ × 10–9 N Given = 4π ∈0=
SI Unit 
9
9 × 10


(2021)
P Electrostatic Potential and Capacitance
W
(2021)
7. Given below are two statements
(2021)
Statement-I: An electric dipole is placed at the center of a
hollow sphere. The flux of electric field through the sphere is
zero but the electric field is not zero anywhere in the sphere.
Statement-II: If R is the radius of a solid metallic sphere
and Q be the total charge on it.
The electric field at any point on the spherical surface of
radius r(<R) is zero but the electric flux passing through this
closed spherical surface of radius r is not zero.
In the light of the above statements. Choose the correct
answer from the option given below
(a) Statement-I is true but Statement-II is false
(b) Statement-I is false but Statement-II is true
(c) Both Statement-I and Statement-II are true
(d) Both Statement-I and Statement-II are false
8. Consider two charged metallic spheres S1 and S2 of radii R1
and R2, respectively. The electric fields E1 (on S1) and E2
(on S2) on their surfaces are such that E1/E2 = R1/R2. Then
the ratio V1(on S1) / V2 (on S2) of the electrostatic potentials
on each sphere is
(2020)
R 
(a)  1 
 R2 
3
(b)
R1
R2
R 
(c)  2 
 R1 
(d)  R1 
 R2 
2
109
9. A particle of charge q and mass m is subjected to an electric
field E = E0 (1 – ax2) in the x-direction, where a and E0 are
constants. Initially the particle was at rest at x = 0. Other
than the initial position the kinetic energy of the particle
becomes zero when the distance of the particle from the
origin is
(2020)
(a)
3
a
(b)
(c)
1
a
(d) a
2
a
10. A charge Q is distributed over two concentric conducting
thin spherical shells radii r and R (R > r). If the surface
charge densities on the two shells are equal, the electric
potential at the common center is
(2020)
r
R
1
(R + r)
Q
(c)
4πε0 2( R 2 + r 2 )
1 (2 R + r )
Q
4πε0 ( R 2 + r 2 )
11. Concentric metallic hollow spheres of radii R and 4R hold
charges Q1 and Q2 respectively. Given that surface charge
densities of the concentric spheres are equal, the potential
difference V(R) – V(4R) is
(2020)
3Q2
(a)
4πε0 R
3Q1
16πε0 R
(c)
Q2
4πε0 R
(d)
3Q1
4πε0 R
110
(c) Increase by
x-axis
–q
q2
4πε0 d
2q 2
3πε0 d
(b) Decrease by
4q 2
3πε0 d
(d) Increase by
3q 2
4πε0 d
14. Ten charges are placed on the circumference of a circle of
radius R with constant angular separation between successive
charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each,
while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and
the electric field E at the center of the circle are respectively.
(Take V = 0 at infinity)
(2020)
10q
10q
;E
=
(b) V =
4πε0 R
4πε0 R 2
10q
=E=0
4πε0 R
(d) V = 0; E =
10q
4πε0 R 2
15. A solid sphere of radius R carries a charge Q+ q distributed
uniformly over its volume. A very small point like piece of
it of mass m gets detached from the bottom of the sphere
and falls down vertically under gravity. This piece carries
charge q. If it acquires a speed v when it has fallen through
a vertical height y (see figure), then
(Assume the remaining portion to be spherical). (2020)
Q
R
q
y
v


qQ
(a) v2 = y 
+ g
 4πε0 R( R + y )m

12. Hydrogen ion and singly ionized helium atom are accelerated,
from rest, through the same potential difference. The ratio
of final speeds of hydrogen and helium ions is close to
(2020)
(a) 2 : 1
(c) 10 : 7
(a) Decrease by
(c) V =
1 ( R + 2r )Q
(b)
4πε0 2( R 2 + r 2 )
(b)
4q
(a) V = 0; E = 0
1 (R + r)
(a)
Q
4πε0 ( R 2 + r 2 )
(d)
13. A two point charges 4q and –q are fixed on the x-axis at x
= –d/2 and x = d/2, respectively. If a third point charge ‘q’ is
taken from the origin to x = d along the semicircle as shown
in the figure, the energy of the charge will
(2020)
y-axis
(b) 1 : 2
(d) 5 : 7


qQ R
(b) v2 = 2y 
+ g
3
 4πε0 ( R + y ) m



qQ
(c) v2 = y 
+ g
2
 4πε0 R ym



qQ
(d) v2 = 2y 
+ g
 4πε0 R( R + y )m

JEE (XII) Module-1 PW
16. A parallel plate capacitor is made of two square plates of side
a, separated by a distance d(d << a). The lower triangular
portion is filled with a dielectric of dielectric constant K, as
shown in the figure. Capacitance of this capacitor is (2019)
d
K
(a)
(b)
a
K ε0 a 2
(a)
ln K
d ( K − 1)
(c)
2
(b)
K ε0 a
ln K
d
(c)
K ε0 a 2
2d ( K + 1)
(d)
1 K ε0 a 2
2 d
(d)
K1
K2
L/2
K3
K4
L/2
d/2
d/2
( K1 + K 2 ) ( K3 + K 4 )
K=
2 ( K1 + K 2 + K 3 + K 4 )
(b) K =
(c) K =
2R
2 −1
2R
2 +1
R
2 −1
R
2 +1
19. A charge Q is distributed over three concentric spherical
shells of radii a, b, c(a < b < c) such that their surface charge
densities are equal to one another. The total potential at a
point at distance r from their common center, where r < a,
would be
(2019)
17. A parallel plate capacitor with square plates is filled with four
dielectrics of dielectric constants K1, K2, K3, K4 arranged as
shown in the figure. The effective dielectric constant K will
be
(2019)
(a)
18. Two electric dipoles, A, B with respective dipole moments
d A = −4qaiˆ and d B = −2qaiˆ are placed on the x-axis with
a separation R, as shown in the figure
(2019)
R
X
B
A
The distance from A at which both of them produce the same
potential is
( K1 + K 2 ) ( K3 + K 4 )
K1 + K 2 + K 3 + K 4
( K1 + K3 ) ( K 2 + K 4 )
K1 + K 2 + K 3 + K 4
K K
K1 K 2
(d) K
=
+ 3 4
K1 + K 2 K3 + K 4
P Electrostatic Potential and Capacitance
W
(a)
(c)
Q (a + b + c)
(
2
2
4πε0 a + b + c
(
Q a 2 + b2 + c2
(
3
3
2
)
4πε0 a + b + c
3
)
)
(b)
Q
4πε0 ( a + b + c )
Q ab + bc + ca
(d) 12πε
abc
0
20. A parallel plate capacitor is of area 6 cm2 and a separation
3 mm. The gap is filled with three dielectric materials
of equal thickness (see figure) with dielectric constants
K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a
material which when fully inserted in above capacitor, gives
same capacitance would be
(2019)
K1
K2
K3
3mm
(a) 36
(b) 14
(c) 12
(d) 4
21. Four equal point charges Q each are placed in the xy plane
at (0, 2), (4, 2), (4, –2) and (0, – 2). The work required to
put a fifth charge Q at the origin of the coordinate system
will be
(2019)
(a)
Q2
2 2πε 0
(b)
Q2 
1 
1+


4πε 0 
3
(c)
Q2
4πε 0
(d)
Q2 
1 
1+


4πε 0 
5
111
22. The given graph shows variation (with distance r from center)
of
(2019)
r0
(a)
(b)
(c)
(d)
r0
r
Potential of a uniformly charged spherical shell
Electric field of a uniformly charged sphere
Electric field of uniformly charged spherical shell
Potential of a uniformly charged sphere
23. Three charges Q + q and +q are placed at the vertices of
a right-angle isosceles triangles as shown below. The net
electrostatic energy of the configuration is zero, if the value
of Q is
(2019)
Q
+q
+q
(a)
(c)
− 2q
(b) + q
2 +1
−2q
(d)
−q
1+ 2
24. In the figure shown below, the charge on the left plate of
the 10 mF capacitor is –3 mC. The charge on the right plate
of the 6 mF capacitor is
(2019)
26. In the figure shown, after the switch ‘S’ is turned from
position ‘A’ to position ‘B’, the energy dissipated in the
circuit in terms of capacitance ‘C’ and total charge ‘Q’ is
(2019)
A
B
S
(a)
3 Q2
8 C
(b)
1 Q2
8 C
(c)
5 Q2
8 C
(d)
3 Q2
4 C
3C
C
27. In the circuit shown, find C if the effective capacitance of
the whole circuit is to be 0.5 mF. All values in the circuit are
in mF.
(2019)
C
2
A
2
2
1
2
6F
2
2
B
2F
10F
4F
(a) + 18 mC
(c) + 12 mC
(b) –12 mC
(d) –18 mC
25. There is a uniform spherically symmetric surface charge
density at a distance R 0 from the origin. The charge
distribution is initially at rest and starts expanding because
of mutual repulsion. The figure that represents best the speed
V(R(t)) of the distribution as a function of its instantaneous
radius R(t) is
(2019)
V(R(t))
V(R(t))
(b)
(a)
R0
R0
R(t)
V(R(t))
V0
V(R(t))
(c)
(d)
R0
112
R(t)
R(t)
R0
R(t)
(a)
7
µF
11
(b) 4µF
(c)
6
µF
5
(d)
7
µF
10
28. A solid conducting sphere, having a charge Q, is surrounded
by an uncharged conducting hollow spherical shell. Let the
potential difference between the surface of the solid sphere
and that of the outer surface of the hollow shell be V. If
the shell is now given a charge of –4Q, the new potential
difference between the same two surfaces is
(2019)
(a) –2V
(b) V
(c) 2V
(d) 4V
(a) 180 V
(b) –520 V
(c) 320 V
(d) –48 V
E ( Ax + B ) iˆ,
29. The electric field in a region is given by =
where E is in NC–1 and x is in metres. The values of constants
are A = 20 SI unit and B = 10 SI unit. If the potential at
x = 1 is V1 and that at x = – 5 is V2, then V1 – V2 is:
(2019)
JEE (XII) Module-1 PW
30. A positive point charge is released from rest at a distance r0
from a positive line charge with uniform density. The speed
(v) of the point charge, as a function of instantaneous distance
r from line charge, is proportional to
(2019)
33. Figure shows charge (q) versus voltage (V) graph for series
and parallel combination of two given capacitors. The
capacitances are
(2019)
q(mc)
A
500
r0
B
80
10V
(a) 40mF and 10mF
(c) 60mF and 40mF
r
(a) v ∝ ln  
 r0 
(b) v ∝ e
+
34. A point dipole p = − p0 x̂ is kept at the origin. The potential
r
r0
and electric field due to this dipole on the y-axis at a distance
d are, respectively
r
(c) v ∝ ln  
 r0 
r
(d) v ∝  r 
 0
31. A system of three charges are placed as shown in the figure
If D >> d,the potential energy of the system is best given by
(2019)
D
+Q
–Q
Q
d
(a)
1  q 2 qQd 
− − 2 
4πε0  d
D 
(b)
1
4πε0
(c)
1  q 2 2qQd 
− +

4πε0  d
D2 
(d)
1
4πε0
 q 2 qQd 
+ 2 
+
D 
 d
(2019)
35. Two identical parallel plate capacitors, of capacitance C
each, have plates of area A, separated by a distance d. The
space between the plates of the two capacitors, is filled with
three dielectrics, of equal thickness and dielectric constants
K1, K2 and K3. The first capacitor is filled as shown in fig.
I, and the second one is filled as shown in fig II.
If these two modified capacitors are charged by the same
potential V, the ratio of the energy stored in the two, would
be (E1 refers to capacitor (I) and E2 to capacitor (II))
 q 2 qQd 
− −
2 
 d 2D 
( n + 1)V
( K + n)
(c) V
(Take V = 0 at infinity)
p
p
,
(a)
4πε0 d 2 4πε0 d 3
p
−p
,
(b)
4πε0 d 2 4πε0 d 3
−p
(c) 0,
4πε0 d 3
p
0,
(d)
4πε0 d 3
(2019)
K1
K2
32. The parallel combination of two air filled parallel plate
capacitors of capacitance C and nC is connected to a battery
of voltage, V. When the capacitors are fully charged, the
battery is removed and after that a dielectric material of
dielectric constant K is placed between the two plates of the
first capacitor. The new potential difference of the combined
system is
(2019)
(a)
V(volt)
(b) 20mF and 30mF
(d) 50mF and 30mF
K1
K2
K3
K3
(I)
(II)
(a)
E1 ( K1 + K 2 + K 3 )( K 2 K 3 + K 3 K1 + K1 K 2 )
=
E2
K1 K 2 K 3
(b)
E1 ( K1 + K 2 + K 3 )( K 2 K 3 + K 3 K1 + K1 K 2 )
=
E2
9 K1 K 2 K 3
(b)
V
K +n
(c)
9K1 K 2 K3
E1
=
E2 ( K1 + K 2 + K 3 )( K 2 K 3 + K 3 K1 + K1 K 2 )
(d)
nV
K +n
(d)
K1 K 2 K 3
E1
=
E2 ( K1 + K 2 + K 3 )( K 2 K 3 + K 3 K1 + K1 K 2 )
P Electrostatic Potential and Capacitance
W
113
JEE ADVANCED
36. Six charges are placed around a regular hexagon of side
length a as shown in the figure. Five of them have charge q,
and the remaining one has charge x. The perpendicular from
each charge of the nearest hexagon side passes through the
center O of the hexagon and is bisected by the side.(2022)
q
q
(b) If rB = 3 , then the electric potential just outside B is
2
k
∈
(c) If rB = 2, then the total charge of the configuration is
15πk
5
, then the magnitude of the electric field just
2
13πk
outside B is
∈0
(d) If rB =
90º
q
38. A disk of radius R with uniform positive charge density σ
is placed on the xy plane with its center at the origin. The
Coulomb potential along the z-axis is
O
a
q
q
=
V ( z)
x
Which of the following statements is (are) correct in SI
units?
(a) When x = q, the magnitude of the electric field at O is
zero
(b) When x = –q, the magnitude of the electric field at O is
q
6π ∈0 a 2
σ
2 ∈0
7q
origin
4 3π ∈0 a
(d) When x = – 3q, the potential at O is –
3q
4 3π ∈0 a
37. In the figure, the inner (shaded) region A represents a
sphere of radius rA=1, within which the electrostatic charge
density varies with the radial distance r from the center
as ρA= kr, where k is positive. In the spherical shell B of
outer radius rB, the electrostatic charge density varies as
2k
ρ B = . Assume that dimensions are taken care of. All
r
physical quantities are in their SI units.
(2022)
R2 + z 2 − z
)
A particle of positive charge q is placed initially at rest at
a point on the z axis with z = z0 and z0 > 0. In addition to
the Coulomb force, the particle experiences a vertical force
2c ∈0
. Which of the following
F = −ckˆ with c > 0. Let β =
qσ
statements(s) is (are) correct?
(2022)
(a) =
For β
(c) When x = 2q, the potential at O is
(
(b) =
For β
origin
1
25
=
and z0
R , the particle reaches the
4
7
1
3
=
and z0
R , the particle reaches the
4
7
1
(c)=
For β
=
and z0
4
z = z0
R
3
, the particle returns back to
(d) For β > 1 and z0 > 0 , the particle always reaches the
origin
39. In the following circuit C1 = 12 µF, C2 = C3 = 4 µF and C4
= C5 = 2 µF. The charge stored in C3 is _______ µC.
(2022)
C1
rA
A
B
6V
outside B
114
C3
C4
C5
2V
rB
Which of the following statement(s) is(are) correct?
(a) If rB =
C2
3
, then the electric field is zero everywhere
2
40. A medium having dielectric constant K >1 fills the space
between the plates of a parallel plate capacitor. The plates
have large area, and the distance between them is d. The
capacitor is connected to a battery of voltage V. as shown
in Figure (a). Now, both the plates are moved by a distance
of d/2 from their original positions, as shown in figure (b).
JEE (XII) Module-1 PW
In the process of going from the configuration depicted
in Figure (a) to that in Figure (b), which of the following
statement(s) is(are) correct?
(2022)
Switch
d
0.01m
d/2
d/2
200 V
d
V
Figure (b)
V
Figure (a)
(a) The electric field inside the dielectric material is
reduced by a factor of 2K
1
.
(b) The capacitance is decreased by a factor of
K +1
(c) The voltage between the capacitor plates is increased
by a factor of (K + 1)
(d) The work done in the process DOES NOT depend on
the presence of the dielectric material
41. Two point charges –Q and +Q/ 3 are placed in the
xy-plane at the origin (0, 0) and a point (2, 0) respectively,
as shown in the figure. This results in an equipotential circle
of radius R and potential V = 0 in the xy-plane with its center
at (b, 0). All lengths are measured in meters.
(2021)
y
–Q
+Q / 3
x
(a) The value of R is ____ meter.
(b) The value of b is ______meter.
42. Two large circular discs separated by a distance of
0.01 m are connected to a battery via a switch as shown
in the figure. Charged oil drops of density 900 kg m–3
are released through a tiny hole at the center of the top
disc. Once some oil drops achieve terminal velocity, the
switch is closed to apply a voltage of 200 V across the
discs. As a result, an oil drop of radius 8×10–7 m stops
moving vertically and floats between the discs. The
number of electrons present in this oil drop is ________.
(Neglect the buoyancy force, take acceleration due to gravity
=10 ms–2 and charge on an electron
(e) = 1.6 × 10–19 C)
(2020)
P Electrostatic Potential and Capacitance
W
43. Two capacitors with capacitance values C1 = 2000 ± 10 pF
and C2 = 3000 ± 15 pF are connected in series. The voltage
applied across this combination is V = 5.00 ± 0.02 V. The
percentage error in the calculation of the energy stored in
this combination of capacitors is _______.
(2020)
44. A thin spherical insulating shell of radius R carries a
uniformly distributed charge such that the potential
at its surface is V 0 . A hole with a small area a4pR 2
(a < < 1) is made on the shell without affecting the rest of
the shell. Which one of the following statements is correct?
(2019)
(a) The ratio of the potential at the center of the shell to
that of the point at 1/2 R from center towards the hole
1− α
will be
1 − 2α
(b) The magnitude of electric field at the center of the shell
αV0
is reduced by
2R
(c) The magnitude of electric field at a point, located on
a line passing through the hole and shell's center on a
distance 2R from the center of the spherical shell will
αV0
be reduced by
2R
(d) The potential at the center of the shell is reduced by
2aV0
45. Parallel plate capacitor of capacitance C has spacing d
between two plates having area A. The region between the
plates is filled with N dielectric layers, parallel to its plates,
d
each with thickness δ = . The dielectric constant of the
N
m

K m K 1 +  . For a very large N (> 103),
mth layer is =
N

 K ∈0 a 
the capacitance C is α 
 . The value of α will be
 d n 2 
_________.
(2019)
115
ANSWER KEY
CONCEPT APPLICATION
1.
11.
21.
31.
(b)
(a)
(c)
(c)
2.
12.
22.
32.
(a)
(c)
(d)
(d)
3.
13.
23.
33.
(b)
(a)
(b)
(d)
4.
14.
24.
34.
(a)
(a)
(a)
(d)
5.
15.
25.
35.
(c)
6. (b)
(b)
16. (a,c)
(a)
26. (d)
(a,b,c) 36. (d)
7.
17.
27.
37.
(b)
(d)
(a)
(a)
8.
18.
28.
38.
(a)
(c)
(a)
(a)
9.
19.
29.
39.
(a)
(b)
(d)
(a)
10. (a)
20. (b)
30. (c)
4.
14.
24.
34.
44.
54.
64.
(c)
(c)
(c)
(a)
(a)
(d)
(a)
5.
15.
25.
35.
45.
55.
65.
(c)
(d)
(d)
(c)
(d)
(c)
(d)
6.
16.
26.
36.
46.
56.
66.
(d)
(d)
(c)
(d)
(b)
(b)
(c)
7.
17.
27.
37.
47.
57.
67.
(a)
(b)
(a,c,d)
(c)
(d)
(b)
(b)
8.
18.
28.
38.
48.
58.
68.
(d)
(b)
(a)
(c)
(a)
(c)
(b)
9.
19.
29.
39.
49.
59.
69.
(a)
(c)
(b)
(a)
(c)
(c)
(a)
10.
20.
30.
40.
50.
60.
(b)
(b)
(a)
(b)
5.
15.
25.
35.
(b)
(a)
(a)
(a)
6.
16.
26.
36.
(b)
(b)
(a)
(a)
7.
17.
27.
37.
(b)
(b)
(d)
(c)
8.
18.
28.
38.
(b)
(d)
(c)
(a,b)
9. (b)
19. (a)
29. (b)
(b,c)
(c)
(a,b,c,d)
[0006]
6.
16.
26.
36.
(b,c,d) 7. (a,d)
(c)
17. (a)
(b,c)
27. (c)
[– 9.6] 37. [30]
8.
18.
28.
38.
(a,c,d) 9. (a,c,d)
(c)
19. (b)
(a)
29. (a)
[0119]
EXERCISE-1 (TOPICWISE)
1.
11.
21.
31.
41.
51.
61.
(b)
(b)
(d)
(d)
(a)
(d)
(d)
2.
12.
22.
32.
42.
52.
62.
(d)
(d)
(a)
(a)
(b)
(b)
(b)
3.
13.
23.
33.
43.
53.
63.
(b)
(c)
(b)
(a)
(a)
(d)
(b)
(b)
(a)
(b)
(a)
(c)
(d)
EXERCISE-2 (LEARNING PLUS)
1.
11.
21.
31.
(c)
(a)
(c)
(a,b)
2.
12.
22.
32.
(b)
(c)
(c)
(b)
3.
13.
23.
33.
(d)
(d)
(b)
(c)
4.
14.
24.
34.
10. (a)
20. (d)
30. (b)
EXERCISE-3 (JEE ADVANCED LEVEL)
1.
11.
21.
31.
(a,c)
(a,c)
(b)
(a)
2.
12.
22.
32.
(a,d)
3.
(a,b,c,d) 13.
(c)
23.
[0002] 33.
(a,c)
4. (a,b,c,d) 5.
(b,c,d) 14. (d)
15.
(a)
24. (b, c) 25.
[0024] 34. [2.55] 35.
2
2
ρ0 [4 R 3 − r 3 ] (b) ρ0 R 4
ρ0 R 4 40. 5ρR
0r
39.
(a) E ρ=
ˆ
=
rˆ ;V
=
E =
r
V
;
12 ∈0
4ε0 r
4ε0
12ε0
4ε0 r 2
10. (b,c)
20. (d)
30. (b)
7
× 10 –6 volt
500πε 0
41. −1.2 iˆ − 1.6 ˆj
42.
7. (a)
17. (d)
27. (a)
8. (d)
18. (b)
28. (b)
9. (a)
19. (a)
29. (a)
43. [1.30]
44. (a)
EXERCISE-4 (PAST YEAR QUESTIONS)
JEE Main
1.
11.
21.
31.
[198]
(b)
(d)
(a)
2.
12.
22.
32.
(b)
(a)
(a)
(a)
3.
13.
23.
33.
(d)
(b)
(a)
(a)
4.
14.
24.
34.
[128]
(a)
(a)
(c)
5.
15.
25.
35.
[36]
(d)
(d)
(c)
6. [300]
16. (a)
26. (a)
10. (a)
20. (c)
30. (a)
JEE Advanced
36. (a,b,c) 37. (b)
45. [0.99 to 1.01]
116
38. (a,c,d) 39. [8]
40. (b)
41. [1.73,3] 42. [6]
JEE (XII) Module-1 PW