Electrostatics

Electrostatics
There has been NO CHANGE in the latest Syllabus of JEE MAIN provided by NTA in this Chapter

1. ELECTRIC CHARGE 2. Charge is quantized. The smallest discrete value of

charge that can exist in nature is the charge on an
Charge is the property of matter that causes it to produce and
electron, given as
experience electrical and magnetic effects. The study of the
e = −1.6 × 10–19 C
electrical charges at rest is called electrostatics. When both
This is the charge attained by an electron and a
electrical and magnetic effects are present, the interaction
proton.
between charges is referred to as electromagnetic.
A charge q must be an integer multiple of this basis
There exist two types of charges in nature: positive and unit. That is,
negative. Like charges repel, and unlike charges attract, each Q = ± ne where n = 1, 2, …
other. Charge on a body can never be (1/2)e, (2/3)e, or 5.7e,
The SI Unit of charge is coulomb (C). In SI units, the etc.
current is a fundamental quantity, having a unit of ampere 3. Charge is invariant that is the numerical value of an
(A). The unit of charge is defined in terms of the unit of elementary charge is independent of velocity.
current. Thus, one coulomb is the charge transferred in one
If motion of changes had any effect on magnitude of
second across the section of a wire carrying a current of one
changes then we could not have exact cancellation of
ampere.
nuclear and electronic charge and atoms would not
As q = It, we have
be neutral.
1 C = (1 A) (1 s)
The dimensions of charge are [A T]. 2. CONDUCTORS AND INSULATORS
The excess or deficiency of electrons in a body gives In conductors charges are free to move throughout its
the concept of charge. If there is an excess of electrons in a volume whereas in insulators or dielectrics, the charges
body, it is negatively charged. And if there is deficiency of remain fixed at the places where they were initially
electrons, the body becomes positively charged. Whenever distributed.
addition or removal of electrons takes places, the body A body can be charged in the following two ways:
acquires a charge.
Charging by Friction
Illustration 1
When a glass rod is rubbed with silk, due to friction there
An object has a charge of + 32 mC. How many electrons is transfer of electrons from the rod to silk and hence rod
have been added or removed form it? becomes positively charged and silk negatively charged.
Solution:
Positive charge means electrons have been removed
Charging by Induction
from the system and the number of removed electrons is There are following steps to charge a metal sphere by
q 32 # 10 -6 induction:
n= e = = 2 # 10 +14
1.6 # 10 -19
+ –
Properties of Charge –
+ –
– + –
1. Charge is conserved, that is for an isolated system, –
–
the total charge remains constant. Charge is neither
–
–
created nor destroyed; it is transferred from one body
to the other.
Physics
(A) (B) constant. e r is also sometimes called dielectric constant, and
+ is represented by letter K.
+
+ + +
–
– +
–
–
–
–  If two charges q1 and q2 are kept in a medium of dielectric
constant K separated by a distance r. The net force on the
q1 q2
(C) (D) charge q2 (or q1) is . But the force on q2 due to
4rKf 0 r2
q1 q2
(A) The metal sphere is supported on an insulated stand. q1 or vice versa is equal to . This is due to the
4rf 0 r2
(B) A negatively charged rod is brought near to it, without
fact that force of interaction between two charges is not
touching, then the free electrons in the metal sphere dependent on presence of other charges.
are repelled by the excess of electrons of the rod and
the left surface accumulates positive charges.
Coulomb’s Law in Vector Form
(C) The surface of the sphere is earthed using a wire and
Suppose the position vectors of two
Y
the negative charged flows to earth. q2 F2
F1 q1
(D) The wire is disconnected and the rod is removed. charges q and q are rv1 and rv2 ,
1 2

then electric force on charge q due r 
Important Points 1
r2

to charge q is,
• The rod (inducing body) neither gains nor loses 2 X

charge.
3 ^r1 - r2 h
1 q1 q2
F1 = 4rf
• If the rod is negatively charged the sphere attains 0 r -r
1 2
Similarly,
positive charge i.e., nature of induced charge is
always opposite to that of inducing charge. Electric force on q due to charge q is
2 1

3 ^r2 - r1 h
1 q1 q2
• Induced charge can never be greater than the inducing F2 = 4rf
0 r -r
charge. 2 1

Here q and q are to be substituted with sign.

Charging by Conduction 1 2

When a charged body touches a neutral body, some charge Important Points
is transferred from charged body to neutral body. Even • The Coulomb force acts along the straight line
on removing the charged body, the neutral body remains connecting the points of location of the charges.
charged. The neutral body will have the same type of charge • The Coulomb force is central and spherically
as charged body after conduction. symmetric.
3. COULOMB’S LAW • The vector form of Coulomb’s Law is
The force of interaction of two stationary point charges kq1 q2
F = rt
in vacuum is directly proportional to the product of these r2
where rt is the unit vector has its origin at the source of
charges and inversely proportional to the square of their
the force.
separation,
kq1 q2 For example, to find the force on the charge q2, the origin
F=
r2 of r is placed at q1 as shown in fig.
where F is in Newton, q1 and q2 in Coulomb, r in metre, and
^
k is a constant given in SI units by q1
r
q2
1
K = 4re = 9 × 109 N m2 C−2 r
0

where e 0 = 8.85 × 10 −12 C2 N−1 m−2 and is called the Superposition Principle
permittivity of free space (vacuum or air). Consider N point charges 1, 2, ....., N with values q1, q2,........
For mediums other than air or vacuum, the electrostatic v v v v
qN having position vectors r1, r2, r3 .......rN , respectively, from
force between two charges becomes
some fixed origin O. The force exerted on the charge q located
1 qq qq
F = 4re d 1 2 2 n = 4re e d 1 2 2 n
1
r 0 r r at rv by these charges is obtained by summing vectorially the
forces by individual charges. So,
Here e = e 0 e r, is called the absolute permittivity or
permittivity of the medium, and e r (= e/e 0) is the relative
N
qi _ r - ri i
Fq = q / 4rf 3
0 r-r
permittivity of the medium which is a dimensionless i=1 i
 Electrostatics
Illustration 2 q3 = –8 nC – + q2 = 4 nC
Three point charges are placed at the following points y
on the x axis: q1 = + 2 µC at x = 0, q2 = – 3µC at x = 40
cm, q3 = – 5 µC at x = 120 cm. Find the force on the –3 3 cm

µC charge.
x

Solution: q4 = –12 nC – + q1 = 8 nC
4 cm
q1 q2 q3
x, cm Solution:
q3 – + q2
120
40

The force on q2 is the vector sum of two contributions, the 

3 cm
attractive force due to q1 (towards q1), and the repulsive force F13
due to q3 (also towards q1). The sum of these two forces, q4 – 
+ –q1
taken algebraically, since they are along the same line, is 
F14 
4 cm F12
- k q1 q2 k q2 q3
]0.40g2 ]0.80g2
F = F1 - F2 = -
Step 1. The force Fv12 (between q1 and q2) is repulsive, while
= – 9 × 109 the forces Fv13 (between q1 and q3) and (between q1
^2 # 10 -6 Ch^3 # 10 -6 Ch ^5 # 10 -6 Ch^3 # 10 -6 Ch
< F and q4) are attractive.
]0.40mg2 ]0.80mg2
-
Step 2. The magnitude of the forces Fv12 and Fv13 , and Fv14 are
= – 0.55 N or 0.55 N to left.
kq1 q2 ^9 # 109h^8 # 10 -9h^4 # 10 -9h
Fv12 = =
Illustration 3 r12
2
^3 # 10 -2h2
Two identical positive charges of magnitude Q are = 32 ×10–5 N
separated by a distance d. Another charge q is placed kq1 q3
F13 =
midway on the line joining the two. Identify the r132
magnitude and nature of the charge q so that the system ^9 # 109h^8 # 10 -9h^8 # 10 -9h
=
is in equilibrium. ^5 # 10 -2h2
+Q q +Q = 23 × 10–5 N
kq1 q4
d/2 d/2 F14 =
Solution: r142
^9 # 109h^8 # 10 -9h^12 # 10 -9h
Since the charge q is placed symmetrically between the =
^4 # 10 -2h2
two Q charges, therefore, it is in equilibrium.
= 54 × 10–5 N
Let us check the equilibrium of any of the + Q charges.
Step 3. v
F12 = (– 32 × 10–5) tj
kQ2 kQq Q
]d/2g2 = 0   or   q = - 4
Thus, 2 - Fv13 = (F13 cos q) it + (F13 sin q) tj
d
The charge q is a negative charge. = :- ]23gb 54 l it + ]23gb 53 ltjD # 10 -5
Illustration 4 = [−18.4 it + 13.8 tj ]10–5
v
F14 = (–54 × 10–5) it
Four point charges are located at the corners of a
rectangle, as shown in figure. Find the net force acting Step 4. The net force on q1 is Fv1 = Fv12 + Fv13 + Fv14
on the charge q1. = [– 72.4 it − 18.2 tj ] × 10–5 N

INTEXT EXERCISE: 1
1. When a glass rod is rubbed with silk, it acquires a through two strings of length 20 cm each, the separation
positive charge because between the suspension points being 5 cm. In equilibrium,
(a) protons are added to it the separation between the balls is 3 cm. The tension in the
(b) electrons are added to it
string is ______. The charge on each ball has a magnitude
(c) electrons are removed from it
(d) protons are removed from it 2.0 × 10–8 C.
2. Two identical pith balls are charged by rubbing against (a) 8 × 10–2 N (b) 6.2 × 10–2 N
each other. They are suspended from a horizontal rod (c) 8.2 × 10–3 N (d) 6.2 × 10–3 N
Physics
3. Two particles A and B, each having a charge Q, are
(a) - 9 # 109 <itd + n + vj + kvF
1 1
placed a distance d apart. Where should a particle of 3 3 8
charge q be placed on the perpendicular bisector of AB
vj + kv
(b) - 9 # 109 =itd + 4 n+ G
so that it experiences maximum force? 1 1
d d 3 3 3 3
(a) (b)
(c) - 9 # 109 <itd - n + vj + kvF
2 3 1 1
d d 3 3 8
(c) (d)
2 2 3 3 (d) None of these
4. A point charge q is placed at a distance r from one edge
of a line charge of length l and charge Q uniformly 8. Three charges, each of +q, are placed at the vertices
distributed over the whole length. The force on point of an equilateral triangle. The charge needed at the in
charge is centre of the triangle for the charges to be in equilibrium is
Qq Qq q q
4rf 0 ]r + l g r
(a) (b) (a) – (b) +
4rf 0 b r + 2 l
l 2 3 3
Qq 1
(c) 4rf ; 2 - ]r + l g2 E (d) None of these
1 3q
(c) + (d) – 3 q
0 r

5. Four equal charges, each +q are placed at the four 9. Three equal charges each +q are placed on the corners
corners of a square of side a. Then the coulomb force of an equilateral triangle of side a. Then the coulomb
experienced by one charge due to the rest of three is force experienced by one charge due to the rest is
^2 2 + 1 h Kq2 3Kq2 (a)
Kq2
(b)
2Kq2
(a) (b) a2 a2
2a 2
a2
2 2 Kq2 3 Kq2
(c) (d) zero (c) (d) zero
a2 a2
6. The vector form of Coulomb’s law is
1 q1 q2 10. Two identical pendulums, A and B, are suspended from
(a) F12 = 4rf
o r2 - r1 the same point. The bobs are given positive charges,
qq
^rv - rv h
1 with A having more charge than B. They diverge and
(b) F12 = 4rf 1 2
0 r
v1 - rv2 2 1 2 reach equilibrium, with A and B making angles θ1 and
3 ^r1 - r2 h
1 q1 q2 θ2 with the vertical respectively.
(c) F12 = 4rf
0 r -r
1 2 (a) θ1 > θ2
(d) None of these (b) θ1 < θ2
7. Three charges each of +1C are placed at (1, 0, 0), (c) θ1 = θ2
(2, 1, 1) and (3, 0, 0), then force on the charge placed
at (1, 0, 0) is (d) The tension in A is greater than that in B

Answer Key
1. (c) 2. (a) 3. (c) 4. (a) 5. (a) 6. (c) 7. (b) 8. (a) 9. (c) 10. (c)

4. ELECTRIC FIELD

The electric field strength Ev at a E

An electric charge influences its surroundings by creating point is defined as the force Fv per unit  P
r
an electric field. The second charges does not interact charge experienced by a vanishing small
directly with the first, rather, it responds to whatever field it positive test charge q0.
q
encounters. v
Ev = lim qF O
It is the space around a charge in which its influence q "0 0
0

Ev is also called the electric field intensity or simply the

(electric forces) can be experienced. In principle the electric
electric field.
field of a charge extends up to infinity.
Ev at a point is equal to the force experienced by a unit
Electric Field Intensity positive charge placed at that point. Obviously, if Ev is the
The electric field strength or intensity due to a point source electric field at any point, then the force experienced by a
charge q at that point is
charge q at any point P at a distance r from it is given by the
force acting on a unit positive test charge placed at that point. Fv = qEv
 Electrostatics
SI unit of E is Newton per Coulomb (N/C), which is same
Q
as Volt per metre (V/m): The dimensions of E are [MLT –3A–1]. E2 = (- cos 30°it + sin 30°tj)
a 2
4re 0 d n
The electric field created by a point charges q is given by 3
^+ tj h
kq Q
E = 2 rt E3 =
a 2
r 4re 0 d n
where the unit vector r has its origin at the source charge q. 3
The electric field due to a charge points away from the 6Q t
∴ E net = E1 + E2 + E3 = j
charge if it is positive and toward it if the charge is negative. 4re 0 a2
Illustration 5 4.1 ELECTRIC FIELD DUE TO FINITE ROD AT
A 5.0 µC point charge is placed at the point x = PERPENDICULAR DISTANCE X FROM THE
0.2 cm, y = 0.3 cm. Find the magnitude of E due to
it (a) at the origin at (b) at point A with coordinates WIRE
x = 1.0 m, y = 1.0 m. y = x tan φ
dy = x sec2 φ dφ
Solution:
x
(a) The direction of Ev at the origin is shown in figure. 
1 2
The distance of point O from the charge is 
P
y
q y
0.3

dy
r 

O 
0.2
Kmdy cos z
∴ dEx =
^ x sec zh2
E

r = ]0.2g + ]0.3g = 0.36m

2 2

Kmdy sin z
0.3 dEy =
tan q = tan i = 0.2 = 1.5
also ^ x sec zh2
⇒ θ = tan–1 1.5 = 56.3° Km
so Ex = x (sin q2 – sin q1)
The magnitude of electric intensity
Km
Kq Ey = x (cos q1 – cos q2)
          = = 3.5 × 106 N/C
r2 i 1 and i 2 are to be used with sign.
Illustration 6 r r
For eg in this figure q1 = 3 , i2 =- 4
Three point changes are placed at the corners of an
equilateral triangle as shown in figure. What is the
/3
value of electric field at the centre. /4
P

Solution:
Let us take origin at the centre and x and y-axis as shown (i) For semi infinite wire q1 = 0, q2 = – p/2
in figure. O
Km
–Q Ex = x
y
Km
Ey = x
E3

a E1 a 2 Km 
E= at 45° with OP.
E2
x
(ii) For infinite wire q1 = p/2, q2 = – p/2
o x
a x Km
+Q +Q
Ex = 2 x
E1, E2 and E3 are the electric field due to + Q, + Q and – Q
Ey = 0
Q
E1 = (cos 30°it + sin 30°tj) 4.2 ELECTRIC FIELD DUE TO ARC
a 2
4re 0 d n λ = charge distribution/unit length. Find field at the centre
3
due to arc. Arc subtends an angle f at center
Physics

Ev3 = Electric field due to wire 3 (Ring).

dE dE

Evnet = E1 + E2 + E3
r

°
 R In X axis : E1 cos 45° – E2 cos 45° = 0 {E1 = E2}
2km
d
In Y axis : E1 sin 45° + E2 sin 45° – R
Rd R d km km 2km
So only the horizontal components are cancelled and + R - R =0
vertical components are added. ⇒ Electric field at centre = 0
K ]mRdig cos i
z/2

∴ # dE = # R2 4.3 ELECTRIC FIELD DUE TO THE RING

0
2Km sin z/2 (at a point P lying a distance x from its center along the
∴ Evnet = R central axis perpendicular to the plane of the ring)
Illustration 7 dq
++
A thread carrying a uniform +
+ a +
charge l per unit length + + r

has the configuration +

+
 P
x
shown in figure a and b. O O +
+ dEx
+
Assuming a curvature r +
+ + dE
dE
r
radius r to be considerably (a) (b)
less than the length of the The magnitude of the electric field at P due to the segment
thread, find the magnitude of the electric field of charge dq is
dq
strength at the point O. dE = k 2
r
Solution: This field has an x component dEx = dE cos θ along the
x-axis and a component dE^ perpendicular to the x-axis.
The resultant field at P must lie along the x-axis because the
2


perpendicular components of the field created by any charge

 E3
E1

element is canceled by the perpendicular component created

by an element on the opposite side of the ring.

E2 1
++
3
+
1 + +

Ev1 = field due to wire 1 (Infinitely long)

+
here +
+ 
Ev2 = field due to wire 2
+ dE2
+
+
Ev3 = field due to wire 3 +
+
dE1
+ +
So, Net Electric field : E net = E1 + E2 + E3 2
dq x
dEx = dE cos θ = d k n =
Ev1 = - Ev2 kx
dq
r2 r ^ x2 + a2h3/2
sin b 4 l
km
Evnet = Ev3 = 2R r
All elements of the ring make the same contribution to
the field at P because they are all equidistant from this point.
Thus, we can integrate to obtain the total field at P
2
# kx kx
Ex = dq = q
^ x2 + a2h3/2 ^ x2 + a2h3/2
1
2k
E3 =
Illustration 8
R
O
45° 45° A thin wire ring of radius “r” carries a charge q. Find
E2 E1 the magnitude of the electric field strength on the
3
axis of the ring as a function of distance L from the
Also here,
centre. Find the same for L >> r Find maximum field
Ev1 = Electric field due to wire 1.
strength and the corresponding distance L.
Ev2 = Electric field due to wire 2.
 Electrostatics
This result is valid for all values of x > 0 and x < 0. but for
(r2 + L2) x = 0 the answer is not valid as there is discontinuity at x = 0.
r
We can calculate the field close to the disk along the axis

O by assuming that R >> x; thus, the expression in bracket
L P E
reduces to unity to give us the near-field approximation.
v
Ex = 2πkσ = 2e
0
Solution:
This also gives the field due to an infinite sheet of charge
Due to a ring electric field strength at a distance “L” from
its centre on it can be given as with uniform charge density s. Note that the magnitude of
the field does not depend on the distance
qL
] g
4rf 0 ] L2 + r2g3 2
E= / " 1
Illustration 9
1 q Three large non-conducting parallel sheets are placed
For L >> r we have E = 4re 2
0 L at a finite distance from each other as shown in figure.
Thus the ring behaves like a point charge.
Find out electric field intensity at point A and B.
dE
For EMax, dL = 0. Q –2Q 3Q

From equation (1) we get y

q ]r2 + L2g3/2 - 3 ]r2 + L2g1/2 2L2

A B C D

4rf 0 > H= 0
dE 2
=
dL
]r2 + L2g3
(r2 + L2)3/2 = 2 ]r2 + L2g1/2 # 2L2
x
3

" ]2g
r Solution:
On solving we get L =
2
Substituting the value of “L” in equation (1) we get for point A
Q –2Q 3Q
1 q ^ r/ 2 h q
E = 4re # 2 2 3/2 =
0 ^ r + r /2 h 6 3 re 0 r2 E–2Q

EQ E3Q A

4.4 ELECTRIC FIELD DUE TO DISK

At a point P that lies along the central perpendicular axis of
the disk and at a distance x from the center of the disk. Q 3Q 2Q
E net = E Q + E 3Q + E -2Q =- 2Af it - 2Af it + 2Af it
Let disk has radius R has a uniform surface charge density σ. 0 0 0

Q t
dq = - Af i
R 0

r for point B
x P Q –2Q 3Q
dr

E–2Q

E3Q B EQ
The ring of radius r and width dr shown in has a surface
area equal to 2πrdr. The charge dq on this ring is 2πσr dr.
Using this result of field due to the ring
Q 3Q 2Q
dEx =
kx ]2rvr drg E net = E 3Q + E -2Q + E Q = 2Af it - 2Af it + 2Af it = 0
^ x2 + r2h3/2 0 0 0

To obtain the total field at P, we integrate this expression

4.5 ELECTRIC FIELD DUE TO UNIFORM SHELL
over the limits r = 0 to r = R. x is a constant
R For the homogeneous spherical shell electric field is given by
# 2r dr
Ex = kxπσ
^ x2 + r2h3/2 the expressions
0 KQ
E = - 2 rt r>a (i)
x
= 2rkv d1 - n
r
^ x2 + R2h1/2
E=0 r<a (ii)
Physics
Calculation of the electric field of a uniform shell E=
KQ
r>a (i)
The radius of the shell as a and let r be the distance from r2
E=0 r<a (ii)
any point P to the centre C of the sphere. We are interested
in obtaining the strength of the electric field at P. Consider
E
a
first the case when P is outside the shell (fig.). We may
divide the surface of the shell into narrow circular strips, KQ/r 2 ~1/r2
all with centers on the line AB. The radius of each strip is r
O
a sin θ and the width is a dθ. Therefore the area of the strip is
r=0

ad 
Outside sphere
Inside sphere
a
B Alternate proof of E = 0 at x < R
d
a sin  A2
 R
C A
P

P
2 a sin 
Area = length × width = (2πa sin θ) (a dθ) A1
= 2πa2 sin θ dθ. Consider a point p inside the shell field at P due to two
If Q is the total charge, uniformly distributed over the opposite elements
surface of the sphere, the charge per unit area is Q/4πa2 and A1 A2
the charge on the circular strip is E = ve 2 - 2 o
r1 r2
1
(Q/4πa2) (2πa2 sin θ dθ) = 2 Q sin θ dθ and A1 = Ωr12, A2 = Xr22
here Ω is the same solid angle formed by two cones cutting
on the shell.
B
a Electric Field due to Solid homogeneous sphere
C
q For the homogeneous solid sphere electric field is given by
R
A the expressions
KQ
a
r–a E= r≥a …(i)
r+a r P
r2
KQr
E= r<a …(ii)
(a) a3
In case of a solid homogeneous sphere, sphere may be
considered as composed of a series of thin spherical shells.
Therefore, the electric field at point outside is the same as if
B a all the charge were concentrated at the centre. i.e. E = KQ/r2.
C R

a P A
r a
a–r
a+r O P

(b)
Figure : Calculation of the electric field at a point P (a)
outside and (b) inside a mass distributed uniformly over a r
spherical shell. To obtain the field inside the homogeneous sphere,
R2 = a2 + r2 – 2ar cos θ consider a point P a distance r from the centre, with
Differentiating, since a and r are constant, we obtain r < a. We draw a sphere of radius r, (observe that those shells
RdR with radius larger than r do not contribute to the field at P,
2R dR = 2ar sin θ dθ or sin θ dθ = ar since P is inside them. All shells with a radius smaller than
Integrating after putting in the formula of electric field r produce a field. Let us call q′ the charge inside the dashed
due to ring we obtain sphere. By equation (i), the field at P will be
 Electrostatics
Kql If E
(i) A is the
Point A electric field intensity at PB due to charge +q
(ii) Point
E= 2 then
r (iii) Point C (iv) Centre of the sphere
The charge contained in the sphere of radius r is then 1 q
Solution: EA = z
4πε0 ( r + a )2
b 4rr l = 3
Q 3 Qr3
q′ =
^4ra /3h
3
3 a Method I :
Again, if EB is electric field intensity at P due to charge
Substituting this result in the preceding equation, we (i) For point A : We can consider the solid part of
– q then
finally obtain the field at a point inside the homogeneous sphere to be made of large number of spherical
shellsE which 1 have quniformly distributed charge
sphere as B=
KQr 4πε0Now
on its surface. )2 point A lies inside all
( r − asince
E= spherical shells so electric field intensity due to all
a3
So, shellsEaxial =E
will – EA{ a EB > EA}
beB zero.
Illustration 10
v q q
A solid non-conducting sphere of radius R and Eaxial = E A = 0 2 − 2
uniform volume charge density ρ has its centre at (ii) For point B4:πε 0 (the
All a)
r −spherical 4πε 0 ( r +for
shells a )which point
origin. Find out electric field intensity in vector form
1 make
B lies inside will 4rqaelectric field zero at point B.
at following positions : Eaxial =
R R
So electric 4field
πε0 will
( r 2 −bea 2due)2 to charge present from
(i) b 2 , 0, 0 l (ii) d , 0 n (iii) (R, R, 0)
R , radius r to OB.
2 2 Since p = q(2a)
vv
Solution: 4 ^ 3 3h
1 K3r OB - r t
2 pr
(i) at (R/2, 0, 0) : Distance of point from centre = So,   Eaxial =E B = 32 ; ForOBr >>> a
4πε0 ( r − OB
2
a2 )
] R/2g2 + 02 + 02 = R/2 < R, so point lies inside the t 6OB3 - r3@
sphere so      E B =1 3f2 p OB
Eaxial = 03 OB3
 trv 4πε0 r
: D
t Rt
E =
3f 0 = 3f 0 2 i (iii) For point C, similarly we can say that for all the
R R Electricshells Due Ctolies
Fieldpoint outside
Dipole at athe shellLying on the
Point
(ii) At d , , 0 n ; distance of point from centre = Equiatorial Line K : 3 r ]R - r3gD
4
2 2
3

If EA is electric field intensity at5P ?3 to +q

EC = OC
^ R/ 2 h + ^ R/ 2 h + 02 = R, so point lies at the OC
2 2
due
surface of sphere, therefore t R3 - r3
0 5OC ?=
E C = 31f q 3 OC 1 q
 EA =

KQ
E = 3 rv 4πε0 AP 2
4πε0 ( r + a 2 )
2
R Method : II We can consider that the spherical cavity is
 with charge density ρ and also –ρ, thereby making
filled
4
K 3 rR3 t R E A is represented both in magnitude and direction
R t
< jF
net
charge
 density zero after combining. We can consider
= it +
R3 2 2 two
by PD . concentric solid spheres one of radius R and charge
t R t R t density ρ and other Eof sin radius r and charge density –ρ.
= 3f < i+ jF 
Applying superposition principle.
A D
0 2 2

(iii) The point is outside the sphere P

– 2EA cos 
4 
K 3 rR3 t
 

6Rit + Rjt@
KQ r C  |EA| = |EA|
so   E = 3 rv =
R +
r ^ 2 Rh
3 EA sin  EB

6Rit + Rjt@
t r2 + a2 r2 + a2
t ]OA g 6- t ]OA g@
        = r
6 2 f0
(i) E A = E t + E - t = 3f 0 +  3f–q =0
+q 0
Illustration 11
t ]OB ga K : 3 rr ^- thD
B4
O
A 3
A Uniformly charged solid nonconducting sphere a
]OBg3
(ii) E B = E t + E - t = 3f + OB
of uniform volume charge density ρ and radius R If EB is electric field intensity
0
at P due to –q
is having a concentric spherical cavity of radius r. r3 t
3 OB = t ;1 - r 3 E OB
3
t
3f 0 ]OBEg B is represented
Find out electric field intensity at following points, as = 31f - q 3f 0
EB = 0 OB
both in magnitude
shown in the figure : 4πε0 ( r + a
2 2 )
(iii) E C = E t + E- t
and direction by PC clearly, EA = EB. Let us resolve EA and
K b 3 components
rR3 t l inK btwo (- t) l perpendicular
4 4 3
EB into two 3 rrmutually
OC + v
=
EB alongOCthe equitorial
R
r
directions components
OC 3
of E A andOCv 3

line cancel each 6R - r @ OC     

O A B C t other3but 3the components perpendicular to
] g3 added up because they act in same
=
3 f 0 OC
equitorial line get

(iv) E OSo
direction. =E t + E -t =
magnitude + 0 = 0 intensity E at P
of0resultant
Physics
5. ELECTRIC LINES OF FORCE
The electric lines of forces (field lines)are purely geometrical
construction which help in visualizing the nature of electric
field in a region. It has no physical existence. Lines of forces +q

are drawn in such a way that the tangent to a line of force

gives the direction of resultant electric field. The density
of field lines in any region is proportional to the magnitude
of the electric field in that region. Field lines originate on
positive charges and terminate on negative charges.
–q

1. There cannot be any closed line of force in an


electrostatic field E . Electric lines of force emerge
from positive charges and terminate on negative
Electric field lines for two positive
charges (or extends to infinity).
point-charges of equal magnitude.

+ –
+ +q

Fig. (A) Fig. (B) – –q

2. Crowded lines represent strong field, while spread out
lines represent weak field.
3. The number of lines originating or terminating on a
charge is proportional to the magnitude of the charge. Electric field lines for two point-charges of equal
magnitude but opposite sign an electric dipole.
In SI units, the number of electric field lines associated
with a unit charge (i.e., 1 coulomb) is taken as 1/e0. 6. If there is no electric field in a region of space, there
Thus, if a body encloses a charge q, total lines of force will be no lines of force. This is why inside a conductor
or at a neutral point (where resultant intensity is zero)
associated with it (also called flux) will be q/e0.
there cannot be any line of force.
7. If the lines of force are equidistant straight lines, the
+ +q field is uniform [Fig. (A)]. If either lines of force are
not equidistant or straight lines or both, the field will
be non-uniform [Fig. (B), (C)].
–
–2q

Fig. Electric field lines of two unequal charges of opposite sign

(A) Uniform field (B) Non-uniform field (C) Non-uniform field
Note that the total lines of force may be fractional as
6. ELECTRIC FLUX
lines of force are imaginary.
Consider some surface in an electric field E . Let us select
4. Lines of force can never cross each other. It is obvious, a small area element dS on this surface. The electric flux
because if they cross at a point the intensity at that of the field over the area element is given by z E = E : dS .
point will have two directions. Direction of dS is normal to the surface.

5. Lines of force have tendency to contract longitudinally 
dS
E
(like a stretched elastic string) and repel each other

laterally. This concept, like the lines of force, is also
imaginary. It helps in understanding how attraction is
produced between opposite charges, and how repulsion is
produced between similar charges.
 Electrostatics
The electric flux over the whole area is Solid Angle
given by Recall how a (plane) angle is defined in two-
# E $ dS

fE = 
A

dimensions. Let a small transverse line r
S E
element ∆l be placed at a distance r from a  l

point O. Then the angle subtended by ∆l at

O can be approximated as ∆θ = ∆l/r.
If the electric field is uniform over that area then s
Likewise, in three-dimensions the solid
fE = E $ S
angle subtended by a small perpendicular r
Illustration 12 plane area ∆S, at a distance r, can be written 
x as ∆Ω = ∆S/r2.
The electric field in a region is given by E = E0 L it.
TS
Find the electric flux through cubical volume ∆Ω = 2
r
bounded by the surface x = 0, x = L, y = 0, y = L,
The total solid angle subtended by a sphere at the center
z = 0 and z = L.
A 4 rR 2
is Ω = 2 = = 4r . However, the same result is to be
Solution: R R2
At x = 0, E = 0 and at x = L, E = E0 it expected whether or not the closed surface is spherical.
Solid angle included in a cone with semi vertical
The direction of the field is along the x-axis, so it
angle q is given by Ω = 2p(1 – cos q)
will cross the yz-face of the cube. The flux of this field
y


Eo
Illustration 13
x Find the flux x through a closed surface when a
z charge q is kept (a) inside the surface (b) outside the
f = fleft face + fright face surface.
= 0 + E0L2 = E0L2 Solution: dA
Note that the flux through remaining four surfaces is zero
because the field and the area vector are perpendicular to (a) Inside the surface
r
each other. Kq d
Ev = 2
r q
Physical Meaning
The electric flux through a surface inside an electric field v = EdA = Kq dA = Kq dW
df = Ev $ dA 2
r
represents the total number of electric field lines crossing the f= # d z = # KqdX = Kq # dX = 4rKq
surface in a direction normal to the surface. S S S
For a closed surface, outward flux is taken to be positive q
f= f
while inward flux is taken as negative. 0

E Note: The result does not depend on the shape of surface

n
or position of charge.
Surface Surface (b) Outside the surface
Total flux through the intercepted area

Negative-flux dA2
Positive-flux dA
r1 1
The flux through a closed surface kept in a uniform field r2
is zero. q
n

E
Kq Kq
df = E 1 $ dA + E 2 $ dA2 = – dA + dA2 = 0
E E r12 1 r22
n n
dA1 dA2
Since dW = = 2 =0
r12 r2
E = –R2E E = 0 E = +R2E
Cylinder in a uniform field
Therefore the net flux through the surface is zero.
Physics
INTEXT EXERCISE: 2
1. Four point charges q, –q, 2Q and Q are placed in order
+
at the corners of A, B, C and D of a square. If the field
q +
at the midpoint of CD is zero, the value of Q is
5 +
x 
(a) 1 (b) 
2 +
2 2 5 5
(c) 5 (d) 2
+

(a) 4rf x ^sin a + sin bh it + 4rf x ^cos a - cos bhtj

2. A charged sphere of diameter d, density r is immersed m m
in oil of density s. There is uniform electric field E o o

(b) 4rf x 6^cos a + cos bh it + ^sin a - sin bhtj@

directed vertically upward such that the sphere is m
suspended in oil. Charge on the sphere is _____ . o

Assume viscous force is absent. (c) 4rf x 6^sin a - sin bh it + ^cos a + cos bhtj@
m
rd3 ^t - vh g rd3 ^t - vh g
o

(d) 4rf x 6^cos a - cos bh it + ^sin a + sin bhtj@

(a) 3E (b) 2E m
rd3 ^t - vh g rd3 ^t - vh g o
(c) 6E (d) 12E 8. 1 mC charge is placed at point (1, 2, 4). The electric
field at a point P(0, –4, 3) is ______ N/C.
3. Two point charges q1 and A

q2 of magnitude + 10 C
-8
(a) - 38.42 it - 230.52 tj - 38.42 kt
-8
and 10 C , respectively, (b) + 38.42 it + 38.42 tj - 230.52 kt
0.1m 0.1m
are placed 0.1 m apart.
(c) - 38.42 it - 38.42 tj + 230.52 kt
Calculate the magnitude
(d) None of these
of electric field at point A, q1 q2

shown in figure. 0.05m 0.05m 9. Two point charges q A = 3 µC and qB =− 3 µC are

3
(a) 9 # 10 N/C (b) 5 # 10 N/C
3 located 20 cm apart in vacuum. What is the electric
3 3 field at the midpiont O of the line AB joining the two
(c) 3 # 10 N/C (d) 11 # 10 N/C
charges ?
4. Two rings of radii 2 cm and 5 cm, charged with
6 6
charge density of –5 C/m and 16 C/m are placed (a) 1.2 # 10 N/C (b) 5.4 # 10 N/C
co-axially, centered at same point. The distance of 6 6
(c) 6.9 # 10 N/C (d) 9.9 # 10 N/C
point from centre, along the axis where | Ev | = 0 is
(a) 3 (b) 3 10. An electric line of force in x – y plane is given by
(c) 5 (d) none of these x2 + y2 = 1. A particle with unit positive charge, initially
5. Charge Q is distributed uniformly on length ‘l’ of a at rest at the point x = 1, y = 0 in the x – y plane:
wire. It is bent in form of semicircle. The electric field (a) will not move at all
strength near the centre of arc is (b) will move along the straight line
Qr Q (c) will move along the circular line of force
(a) (b)
4f o l 2
4rf o l2 (d) information is insufficient to draw any conclusion
Q Q
(c) (d) 11. An electric charge of + 5 mC is placed at a distance of
2rf o l2 2f o l2
2m from another charge of – 5 mC. The electric field at
6. A block of mass ‘m’ and charge ‘q’ is placed a smooth
the mid-point of the line joining them will be
horizontal table which terminates in a vertical non
conducting wall at a distance ‘d’ from the block. A (a) zero (b) 9000 V m–1
(c) 90000 V m –1 (d) 4500 V m–1
horizontal electric field E towards right is switched on.
Assuming elastic collision, the time period of motion is 12. A pendulum bob of mass 10 mg and carrying
2dm 4dm 9.8 × 10–9 C is suspended in a horizontal uniform
(a) Eq (b) Eq electric field of 1000 Vm–1. The string of the pendulum
lies inclined to the vertical at an angle
8dm 10dm
(c) qE (d) Eq (take g = 9.8 m/s2)

7. For a finite line of charge with linear charge density l, (a) 30° (b) 45o
the electric field at a point shown is given by (c) 60o (d) 0o
 Electrostatics
Answer Key
1. (d) 2. (c) 3. (a) 4. (b) 5. (d) 6. (c) 7. (a) 8. (a) 9. (b) 10. (d)
11. (c) 12. (b)

7. GAUSS’S LAW Important points

Consider charges q1, q2, q3, ........ qn inside a closed surface (i) Flux through Gaussian surface is independent of its
and charges Q1, Q2 ,......Qn outside that surface. such a shape.
(ii) Flux through Gaussian surface depends only on total
surface under consideration is also called a Gaussian surface.
charge present inside Gaussian surface.
Consider a point P on the surface. Let E 1, E 2, ...... E n be (iii) Flux through Gaussian surface is independent of
the fields produced by q1, q2, q3,.......qn at P and E 11, E 12, ...... E 1n position of charges inside Gaussian surface.
be the fields produced by the charges Q1, Q2, .........Qn at P (iv) Electric field intensity at the Gaussian surface is due
Q1 to all the charges present inside as well as outside the
Gaussian surface.
E q1
P q3

q2  (a) If a closed surface does not enclose any charge, then

# E .ds 1
= e (q) = 0
Q2 Q3
Now the resultant electric field at P is given by 0

E = ] E 1 + E 2 + ... + E ng + _ E 11 + E 2 + ... + E n i
1 1 i.e, if a closed body (not enclosing any charge) is placed
in an electric field (either uniform on non‑uniform),
The flux of resultant electric field through the closed total flux linked with it will be zero.
surface is

f= # E .ds = %o# E 1 .ds + E 2 .ds + .... + E n .ds /

+ %o# E 11 .ds + E 12 .ds + .... + o# E 1n .ds /

Sphere

Here # E 11 .ds is the flux due to q1 which is q1/∈0 and

E = 0 E = 0
# E 11 .ds is the flux due to Q1 which is zero, as it is not (b) If a closed body encloses a charge q, total flux
linked with the body will be # E .ds = e ^q h
enclosed by the surface. 1
Similarly the flux due to the other charges also can be o

written. Now we can write

Illustration 14
# E .ds = ( eq10 + eq20 + ..... + eqn0 2 + !0 + 0.... + A hemisphere is kept in a uniform electric field as
shown in the figure. find the flux of the electric field
q1 + q2 + .... + qn
= d e0
n through the curved surface.
E
Rqenclosed
⇒ # E .ds =
e0
Here Rqenclosed is the sum of all enclosed charges which
can be positive, negative or zero. R

Statement of gauss’s law,

n
(B)
Solution:
The surface integral of the electric field intensity over any
1 Considering the hemispherical body as a closed body
closed hypothetical surface (Gaussian surface) is f times
0 with a curved surface and a plane base (cross-section),
that of net charge enclosed by that surface. Here, e0 is the the flux linked with the body will be zero as it does not
permittivity of free space. encloses any charge i.e.,
n
If S is the Gaussian surface and / qi is the total charge f = fCS + fPS = 0 ...(1)
i=1 As field is perpendicular to base, the flux linked with base
enclosed by the Gaussian surface, then according to Gauss’s
fPS = E × pR2 cos 180° = – pR2E
law,
1
n So substituting this value of fPS in Eq.(1), we get fCS
f = # E .dS = f / qi = pR2E
0 i=1
Physics
Illustration 15 Total flux emerges from the system (Two cubes) is
A charge is placed at the centre of a cylindrical Q
ftotal = e
surface. Find total flux passing through lateral 0 Q
Flux through the given cube is fcube = 2e
curved surface. 0

2R (c) If a point charge is kept at the corner of a cube

R 45° q

Solution:
Total flux linked can be divided into parts
(i) Flux through lateral surface fL
(ii) Flux through end caps fA
(A )
If end cap substends solid angle Ω at centre. (B)
q For enclosing the charge completely, seven more
fL + 2fA =  identical cubes are required. So total flux linked with
f0
q Q
× b 4r l
X the 8 cube system is ftotal = e .
fA = 0
f0 Q
Note: Solid angle is given by Ω = 2π (1 – cos θ) \ Flux through the given cube fcube = 8e .
0
and Flux through one face opposite to the charge, of
2r d1 - n q d1 - n
1 1
q 2 2 Q/8e 0 Q
= f = the given cube is fface = 3 = 24e 0 (Because flux
0 4r 2f 0
q through other three faces is zero).
fL =
2 f0
7.1 APPLICATION OF GAUSS’S LAW
Illustration 16 Field Due to an Infinite Line Charge
Find the flux linked with a cube if a charge Q is kept Assume cylindrical gaussian surface of length L and radius r.
(a) at center of the cube The charge enclosed by the cylinder is Q = lL.
(b) at the center of one of the faces Applying Gauss’s law to the curved surface, we have
(c) at one of the vertices
mL
E # dS = E(2prL) = f
Solution: 0
m
(a) If a point charge is kept at the centre of a cube, then or E = 2rf r
0
find the total flux linked with the cube is

+ dS
+ 
S2 + E
+
+
+
+
Q + r
 
E + E

+ S1 
dS + dS
+
+
1
ftotal = e 0 (Q ) ;
+
+
+ 
1 S3 + E
Flux linked with each face of the cube is fface = 6e (Q) ; +
+ 
0
+ dS
(b) If a point charge is kept at the centre of a face of the cube, Field Due to an Infinite Charged Plane Sheet Having
the first we should enclose the charge by assuming a Surface Charge Density (s):
Gaussian surface (an identical imaginar cube) Assume cylindrical Gaussian surface of cross sectional
area A.
The flux through the two plane ends of cylindrical
Q
Gaussian surface
   fE = # E .dS + # E .dS
(A ) (B )
     = EA + EA = 2EA
 Electrostatics
Qr3 1 Q tr
E(4pr2) = 3 or E = 4rf 3 r = 3f
+ +
+ ++
R f0 0 R
+
++ + + + 0
++
 P
+ + + +
P  The field increases linearly with distance from the centre
E + + ++ + E
+
(b) At an external point (r > R): To find the
+ 

+ + dS
+ +
+ ++ + + +
dS + electric field outside the charged sphere, we
use a spherical Gaussian surface of radius
The charge enclosed by the Gaussian surface q = sA
(r > R). This surface encloses the entire charged sphere.
Applying Gauss’s law, we have
So from Gauss’s law, we have
vA v
2EA = f & E = 2f     Q
0 0 E(4pr2) = f
Electric Field Due To A Uniformly Charged Spherical
0
1 Q
Shell : or, E = 4rf 2
0 r
(a) At an external point (r > R)
 The field at points outside the sphere is same as that of
E
 a point charge at the centre.
+ + + dS
Illustration 17
+ +

Charge is distributed throughout a spherical region

+ + +

r Gaussian
R of space in such a manner that its volume charge
+

surface
density is given by
+ + +

ρ = ar2, 0 ≤ r ≤ R
Q where a s is a constant. Find the field at distance r
E # dS = E ^4rr2h = f from the center.
0
Therefore, Y
1 Q
E = 4rf 2
0 r
R
For points outside the charged conducting sphere or the dr

charged spherical shell, the field is same as that of a point r

dv = 4r2dr
charge at the centre.
(b) At an Internal Point (r < R): The field still has the Z X

same symmetry and so we again pick a spherical

Gaussian surface, but now with radius r less than R. Solution:
Since the charge enclosed is zero, from Gauss’s law Since charge distribution is volumetric we choose a
we have small volume element dV within the sphere. Because the
E(4pr2) = 0 \ E = 0
charge distribution is spherically symmetric, we choose
Thus, we conclude that E = 0 at all points inside a uniformly
a thin spherical shell of radius r and thickness dr. We
charged conducting sphere or the charged spherical shell.
can think of entire spherical charge to be made up of
Electric Field Due To A Uniformly Charged Non- concentric shells.
Conducting Sphere Having Charge ‘Q’:
Area of sphere is 4πr2.
(a) At an internal point (r < R): Applying Gauss’s Law
Therefore, dV = 4πr2dr
Ql
   # E .dS = E ^4rr2h = f r
0
q= # tdV = # ^ar2h^4rr2 drh
Q Qr3
Here, Q′ = b 3 rr3 l t = b 3 rr3 l # 4
4 4 0
= 3 4
R = 5 rar5
3 rR
3

q
By Gauss law: E(4pr2) = f
+ + 0

E = 4re b 5 l (πar3)
+
+
+ 1 4
+
+ r Gaussian 0
+ + surface
+ + +
+
+
+ +
+
Field inside a spherical cavity
+
+R + A sphere of radius R has a uniform volume density ρ. A
+
spherical cavity of radius b whose center lies at rv = av is
Where r is volume charge density Therefore removed from the sphere.
Physics
Similarly, the electric field formed by the charge density
P –ρ inside the cavity is
^- th sv

 s
E2 = 3f 0 ; s = r - a
r
v v v

O a

s

The field within the cavity or outside is the superposition r
of the field due to the original uncut sphere, plus the field 
due to a sphere of the size of the cavity but with a uniform a
negative charge density. Electric field Ev1 caused by the 
O
charge distribution +ρ at a point rv , inside the spherical s is the radius vector from cavity center to the point P.
- t ]rv - avg
cavity.
tr trv \ E2 =
Ev1 = 3f rt = 3f ; where r̂ is a unit vector in radial 3f 0
0 0
direction. The resultant electric field inside the cavity is therefore
s given by the superposition of E 1 and E 2
 - t ]rv - avg
r
trv
E = E 1 + E 2 = 3f + < F
a O +
O O
R
0 3f 0
(1) Charge density (2) Charge density tav
of big sphere of smaller sphere = + 3f = constant
is  is (–) 0

INTEXT EXERCISE: 3
1. An electric dipole of dipole moment P is placed at the q q
(a) f0 (b) 2f0
centre of a sphere of radius ‘R’, then the flux passing 0 0
through sphere is 2q0
P P (c) f0 (d) None of these
(a) (b) Rf
R2 f 0 0

P 6. The electron is projected from a P u

(c) f (d) zero distance d and with initial velocity d
0

2. A point charge ‘q’ is placed at centre of a cube with u parallel to a uniformly charged
x
side a. The flux linked with each face of cube is flat non‑conducting plate as shown. l

q q It strikes the plate after traveling a

(a) af (b) 6f distance l along the x-direction. The
0 0
q qa surface charge density of the conducting plate is equal to:
(c) f (d) f
0 0
4df 0 mu2 2df 0 mu
3. The electric field in a region is given by Ev = ayit . Here (a) (b) el
el2
a is a constant of proper dimensions. Find the total flux
df 0 mu2 df 0 mu
passing through a cube bounded by the surfaces, x = l, (c) el (d) el
x = 2 l, y = 0, y = l, z = 0, z = l.
(a) 6al2 (b) al3 7. Electric charge is uniformly distributed
(c) 3al3 (d) zero along a long straight wire of radius 1 mm.
The charge per cm length of the wire is Q
4. A hemisphere body of radius R is placed in a uniform
electric field E. What is the flux linked with the curved coulomb. Another cylindrical surface of
surface, if the field is parallel to the base radius 50 cm and length 1 m symmetrically
(a) pR2E (b) 4pR2E encloses the wire as shown in the figure.
(c) 2pR E 2 (d) Zero The flux through the cylindrical surface is
Q 100Q
5. A charge q0 is distributed uniformly on a ring of radius (a) f (b) f0
R. A sphere of equal radius R is constructed with its 0

centre on the circumference of the ring. The electric 10Q 100Q

^rf 0 h ^rf 0 h
(c) (d)
flux through the surface of the sphere is
 Electrostatics
8. Given: Ev = ^10it + 7tj h Vm -1 . The electric flux (a) 600 mC (b) 60 mC
through 1 m2 area in XZ plane is (c) 7 mC (d) 6 mC
(a) 10 Vm (b) 7 Vm 10. In a region of space, the electric field is in the
(c) 100 Vm (d) 49 Vm x-direction and proportional to x, i.e., Ev = E0 xit .
Consider an imaginary cubical volume of edge a, with
9. The electric field in space is
Y
0.1 m its edges parallel to the axes of coordinates. The charge
(see figure) Ex = 600x1/2 × inside this volume is
106 V/m, Ey = 0, Ez = 0. The
O
X (a) zero (b) e0E0a3
charge inside the cube is 1 1
(c) f E0 a3 (d) 6 f 0 E0 a2
0.1 m
nearly Z
0

Answer Key
1. (d) 2. (b) 3. (d) 4. (d) 5. (d) 6. (a) 7. (b) 8. (b) 9. (c) 10. (b)

8. ELECTRIC POTENTIAL From definition of potential

r

In electrostatic field the electric potential (due to some source - # _q0 E i $ drv
Wext^3 " ph
charges) at a point P is defined as the work done by external V= = 3 q
q0 0
agent in taking a point unit positive charge from a reference r
KQ KQ
point (generally taken at infinity) to that point P without V = – # 2 dr = r
r
acceleration. 3

Illustration 18
If (W∞P)ext is the work required in moving a point charge q
When charge 10 µC is shifted from infinity to a point
from infinity to a point P, the electric potential of the point
in an electric field, it is found that work done by
P is electrostatic forces is –10 µJ. If the charge is doubled
F
W3P) ext and taken again from infinity to the same point
VP =
q acc = 0 without accelerating it, then find the amount of work
Note: done by electric field and against electric field.
(i) Potential is a scalar quantity and proper signs of both Solution:
W and q must be used. Wext)∞p = –wel)∞p = wel)p∞ = 10 µJ
(ii) If we know the potential at some point (interms of
Wext) 3p 10nJ
numerical value or interms of formula) then we can Vp = = 10nC = 1 V
q
find out the work done by electric force when charge
moves from point ‘P’ to ∞ by the formula So if now the charge is doubled and taken from infinity then
WP3) electric = qVP wext h3p
1 = 20nC ⇒ wext)∞p = 20 µJ
(iii) Electric field is conservative. In an electric field work
⇒ Wel) ∞P = –20 µJ
is path independent and work done in moving a point
charge q between two fixed points having a potential
Illustration 19
difference DV is equal to,
A charge 3 µC is released at rest from a point P where
WAB = q(VB – VA)= qDV
electric potential is 20 V then its kinetic energy when
Hence in moving a charged particle in an electric field it reaches to infinity is:
work is always done unless the points are at same potential Solution:
as shown in Fig. below. Wel = ∆K = Kf – 0
B
A
Wel)P → ∞ = qVP = 60 µJ
so, Kf = 60 µJ
I
II I +Q
II
I
II A B Illustration 20
+Q
A +q L L B Two point charges 2 µC and – 4 µC are situated
(A) (B) (C) at points (–2 m, 0 m) and (2 m, 0 m) respectively.
Electric potential due to a point charge Find out potential at point C(4 m, 0 m) and
D (0 m, 5 m).
Q r P
Physics
D Electric Potential due to a charged ring
dq
A B C
R 2
q1 = 2 C q2 = – 4 C R
+ X2
(–2, 0) (2, 0)
Solution: x P

Potential at point C
VC = Vq + Vq Consider an element of charge dq on the ring.
K ^2 nCh K ^- 4 nCh
1 2

Potential at point P due to charge dq will be

K ^dq h
= Vq + Vq =
1 26 + 2
dV =
9 # 10 # 2 # 10
9 -6
9 # 109 # 4 # 10 -6 R2 + x2
= 6 - 2
Net potential at point P due to all such element will be
= –15000 V. KQ
Similarly, VD = Vq + Vq V = # dV =
1 2 R2 + x2
K ^2 nCh K ^- 4 nCh KQ
= + at centre x = 0 VC = R
^ 5h + 2 ^ 5 h + 22
2 2 2

Illustration 22
K ^2 nCh K ^- 4 nCh
= + A point charge q0 is placed on the axis at the distance
3 3
R from the centre of uniformly charged ring of total
= – 6000 V. charge Q and radius R. If the point charge is released
then find out its speed when it reaches to a large
Illustration 21 distance.
Four charges + q, + q, – q, – q are fixed respectively at Solution:
the corners of A, B, C and D of a square of side Only electric force is acting on q0
‘a’ arranged in the given order. 1
A +q +q
B \ Wel = ∆K = 2 mv2 – 0
What will be the work done by
KQ
external agent in carrying a a O +Q DWel = –Wext = – q0(V∞ – VP) = –q0 e 0 - o
charge Q slowly from O to E and
E
2R
from O to F? Kq0 Q 1 2 Kq0 Q
\ = 2 mv2 ⇒ v = mR
2R
D C –q –q
F
Solution:
KJK Kq ONO KJK Kq ONO Electric potential due to a uniformly charged disc
Potential at the point O = 2 KK a OO - 2 KK a OO = 0 V
Figure shows a uniformly disc of radius R with surface
KK OO KK OO
L 2P L 2 P charge density s coul/m2. To find electric potential at point
q q P we consider an elemental ring of radius y and width dy,
charge on this elemental ring is dq = σ2πydy. Due to this
ring, the electric potential at point P can be given as
a 5a/2
a kdq kv2rydy
dV = =
x +y
2 2
x2 + y2
s
–q a/2 F a/2 –q dy
  VE = 0 (from diagram symmetry)
2kq 2kq 4kq 1
- a/2 = a < - 1F
y x
VF = P
5
c 5 a m
2
⇒ Wext (O → E) = Q(VE – V0) = Q(0 – 0) = 0 J
Net electric potential at Point P due to whole disc can be
Wext (O → F) = Q(VF – V0) given as R v y dy
V = # dV = # 2e
        = Q < a d - 1 n - 0F J
4kq 1 0 0 x2 + y2
5 v 7 2
x + y2 A0
R
= 2e
4kqQ 1
a < 5
- 1F J
0
        = = 2e 7 x2 + R2 - xA
v
0
 Electrostatics
Electric Potential Due to Various Charge Distributions
Name/Type Formula Note Graph
Point charge Kq • q is source charge.
V= r • r is the distance of the point from the
V

point charge.
r

Ring (uniform/nonuniform KQ • Q is source charge. V

charge distribution) at centre V = R
• x is the distance of the point from centre.
KQ
at the axis V =
R2 + x2
r

Uniformly charged shell kQ • R is radius of shell V

for r ≥ R V = r • r is the distance from centre of shell to KQ/R
KQ the point.
for r ≤ R V = R
• Q is total charge = s4pR2. R
r

Uniformly charged solid kQ • R is radius of sphere. V

nonconducting sphere for r ≥ R V = r • r is distance from centre to the point.
KQ ^3R2 - r2h
3KQ/2R
(insulating material) 3
for r ≤ R, V = • Vcentre = 2 Vsurface.
KQ/R
3
2R
4
• Q is total charge = r 3 pR3.
= 6f ^3R2 - r2h
t r
R
0 • Inside sphere potential varies
parabolically.
• Outside potential varies hyperbolically.
Line charge Not defined • Absolute potential is not defined.
• Potential difference between two points
is given by formula
r
VB – VA = - 2Km ln d rB n
A

Infinite nonconducting thin Not defined • Absolute potential is not defined.

sheet • Potential difference between two point
is given by formula
VB – VA = – 2f ^rB - rAh
v
0

Infinite charged conducting Not defined • Absolute potential is not defined.

thin sheet • Potential difference between two point
is given by formula.
VB – VA = – f ^rB - rAh
v
0

Illustration 23
Solution:
Two concentric spherical shells of radius R1 and R2 KQ1 KQ2 KQ1 KQ2
(R2 > R1) are having uniformly distributed charges (i) VA = R1 + R2 (ii) VB = R1 + R2
Q1 and Q2 respectively. Find out potential KQ KQ
C (iii) VC = R 1 + R 2
2 2
KQ1 KQ2
R2 B (iv) for r ≤ R1       V = R1 + R2
KQ KQ
A R1 (v) for R1 ≤ r ≤ R2  V = r 1 + R 2
2
Q1 Q2 KQ1 KQ2
(vi) for r ≥ R2    V = r + r

(i) at point A Relation between electric field and electric potential

(ii) at surface of smaller shell (i.e. at point B) (a) Potential difference in a uniform electric field :
(iii) at surface of larger shell (i.e. at point C) VB – VA = – E $ AB
(iv) at r ≤ R1 ⇒ VB – VA = – | E | | AB | cos θ
(v) at R1 ≤ r ≤ R2 = – | E | d = – Ed
(vi) at r ≥ R2 d = effective distance between A and B
along electric field.
Physics
TV
or we can also say that E = Td because In the direction of electric field potential
B  always decreases.
(ii) | DVBC | = Ed = 20 × 2 × 10–2 = 0.4 so, VB – VC = 0.4 V
E

(iii) | DVCA | = Ed = 20 × 4 × 10–2 = 0.8 so, VC – VA = – 0.8 V

A because In the direction of electric field potential
d always decreases.
Special Cases: (iv) |DVDC | = Ed = 20 × 0 = 0 so, VD – VC = 0
Case 1. Line AB is parallel to electric field. because the effective distance between D and C is
⇒ VA – VB = Ed zero.
d (v) | DVAD | = Ed = 20 × 4 × 10–2 = 0.8
A B so, VA – VD = 0.8 V
E because In the direction of electric field potential
always decreases.
Case 2. Line AB is perpendicular to electric field. (vi) The order of potential VA > VB > VC = VD.
⇒ VA – VB = 0 ⇒ VA = VB
Equipotential Surface
A
d E If potential of a surface (imaginary or physically existing) is
B same throughout then such surface is known as a equipotential
surface.
(i) Properties of equipotential surfaces:
 In the direction of electric field potential always (a) When a charge is shifted from one point to another
decreases. point on an equipotential surface then work done
against electrostatic forces is zero.
Illustration 24 (b) Electric field is always perpendicular to
A uniform electric field is present in the positive equipotential surfaces.
x-direction. Ιf the intensity of the field is 5 N/C then (c) Two equipotential surfaces do not cross each other.
find the potential difference (VB – VA) between two (ii) Examples of equipotential surfaces: V2
points A (0 m, 2 m) and B (5 m, 3 m) V1
(a) Point charge: Equipotential
Solution:  q
surfaces are concentric and
VB – VA = – E . AB = – (5 iˆ ) . (5 iˆ + ĵ )
R1

spherical as shown in figure. In

R2
= –25 V.
figure we can see that sphere of
Illustration 25 radius R1 has potential V1 throughout its surface
Find out following and similarly for other concentric sphere potential
(i) VA – VB (ii) VB – VC is same.
(iii) VC – VA (iv) VD – VC (b) Line charge: Equipotential surfaces have curved
(v) VA – VD surfaces as that of coaxial cylinders of different radii.
(vi) Arrange the order of potential for points A, B, C
and D. V2
D
uniform electric V1
field E = 20 N/C
C
2 cm
A B
2 cm

Solution:
(i) | DVAB | = 20 × 2 × 10–2 = 0.4 (c)
Uniformly charged large conducting/non
so, VA – VB = 0.4 V conducting sheets Equipotential surfaces are
parallel planes.
 Electrostatics
+
V1 V2 V3 Potential difference in a Non uniform electric field
2V 2V 2V
Ex = – 2x , Ey = – 2y , Ez = – 2z
+
+

E = Ex iˆ + Ey ĵ + Ez k̂
+
+ ⇒

= – ;it 2x V + tj 2y V + kt 2z V E
+
+ 2 2 2
+

= – ;it 2x + tj 2y + kt 2z E V
+
2 2 2
Note: In uniform electric field equipotential surfaces are
always parallel planes. = – ∇V = –grad V
Illustration 26 2V
Where 2x = derivative of V with respect to x (keeping y
Some equipotential surfaces are shown in figure. What and z constant)
can you say about the magnitude and the direction of the 2V
electric field ? 2y = derivative of V with respect to y (keeping z and x
y (cm)
constant)
2V
2z = derivative of V with respect to z (keeping x and y
10 V 20 V 30 V 40 V

0
30° 30° 30° 30°
constant)
10 20 30 40 x (cm)
If electric potential and electric field depends only on one
coordinate, say r:
 2V
Solution: (i) E = – 2r rt
Here we can say that the electric field will be
where r̂ is a unit vector along increasing r.
perpendicular to equipotential surfaces.

Also | E | =
TV (ii) # dV =– # E $ dr
Td rB
where ∆V = potential difference between two ⇒ VB – VA = –
# v
E $ dr
equipotential surfaces. rA
∆d = perpendicular distance between two equipotential v is along the increasing direction of r.
dr
surfaces.
10 Illustration 28
So | Ev | =
]10 sin 30°g # 10 -2
= 200 V/m
V = x2 + y , Find E .
Now there are two perpendicular directions. We know that
in the direction of electric field electric potential decreases Solution:
2V 2V 2V
so the correct direction makes an angle 120° with the 2x = 2x, 2y = 1 and 2z = 0
positive x-axis. 
E = – cit 22Vx + tj 22Vy + kt 22Vz m
Illustration 27
= –(2x iˆ + ĵ )
Figure shows some equi-potential surfaces produce
by some charges. At which point the value of electric Electric field is nonuniform.
field is greatest?
Illustration 29
(50 V)

(40 V)
(30 V)

(20 V)

For given E = 2xit + 3yjt find the potential at (x, y) if

B
V at origin is 5 volts.

Solution:
C V ^ x, yh x y

A # dV = - # Ev $ dr = – # Ex dx – # Ey dy
5 ^0, 0h 0 0

2x2 3y
2
Solution: ⇒ V–5=– 2 - 2
E is larger where equipotential surfaces are closer. In the
figure we can see that for point B they are closer so E at 2x2 3y
2
⇒ V = – 2 - 2 + 5.
point B is maximum.
Physics
INTEXT EXERCISE: 4
1. An electric field E = ^20it + 30tj h N/C exists in the (a) 2.4 V (b) 1.2 V
space. If the potential at the origin is taken to be zero, (c) 3.6 V (d) 4.8 V
the potential at (2 m, 2 m) is _________
8. The electric field Ev in the given situation is constant
(a) 100 V (b) –100 V
in both magnitude and direction. Consider a path of
(c) 50 V (d) –50 V length d at an angle q = 60° with respect to field lines
2. An electric field E = Ax it exists in the space, where shown in figure. The potential difference between
A = 10 V/m2. Take the potential at (10 m, 20 m) to be points 1 and 2 is:
zero. The potential at the origin is ________
(a) 100 V (b) 200 V 2

(c) 400 V (d) 500 V

d 
E
1 60°
3. A charge q = 10 mC is distributed uniformly over
the circumference of a ring of radius 3 m placed on 3

x – y plane with its centre at origin. Find the electric E

potential at a point P(0, 0, 4 m). (a) (b) Ed cos 60°
d cos 60 o
(a) 1.8 kV (b) 18 kV Ed E
(c) (d) d cos 60 o
(c) 2.8 kV (d) 28 kV cos 60 o

4. Angle between equipotential surface and lines of force is 9. Choose the correct relation regarding potential. Here
(a) Zero (b) 180° A, B, C and D all are at equal distance from point O.
(c) 90° (d) 45° then:
C
5. There is an electric field E in X-direction. If the work
done by electric field on moving a charge 0.2 C through
a distance of 2m along a line making an angle 60° with A –q O +q B
the X-axis is 4.0 J, what is the value of E.
D
(a) 3 N/C (b) 4 N/C (a) | VA | = | VB | > | VC | = | VD|
(c) 5 N/C (d) 20 N/C (b) | VC | = | VD | > | VA | = | VB |
6. A spherical distribution of charge consists of uniform (c) | VA | > | VC | = | VD | > | VB |
a (d) | VB | > | VC | = | VD | > | VA |
charge density, r1 from r = 0 to r = 2 and a uniform
a
charge density r2 from r = 2 and r = a. The potential 10. Figure shows three spherical and C

a
B
at r = 2 is ________ equipotential surfaces A, B and C A
around a point charge q. The
(a) 24f ^t1 + 2t2h (b) 24f ^2t1 + 9t2h
a2 a2 t1
0 0
potential difference VA – VB = VB t2
– VC. If t1 and t2 are the distances
(c) 8f ^t1 + 3t2h
a 2
(d) None of these
0 between them then:
7. Eight mercury droplets having a radius of 1 mm (a) t1 = t2 (b) t1 > t2
and a charge of 0.066 pC each merge to form one
droplet. Its potential is: (c) t1 < t2 (d) t1 ≤ t2

Answer Key
1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (b) 7. (a) 8. (b) 9. (a) 10. (c)

Electrostatic Potential Energy If the source of the potential is a point charge Q, the
Consider a point charge q placed at position where the potential at a distance r from Q is V = kQ/r. Therefore, the
potential is V. The potential energy associated with the potential energy shared by two charges q and Q separated
interaction of this single charge with the charges that created by r is
V is kqQ
U= r
U = qV
 Electrostatics
Implicit in Equation is the choice U = 0 at r = ∞, which
allows the following interpretation 1
mu =mv + mv, i.e., v = 2 u
The potential energy of the system of two charges is the
And by conservation of energy’
external work needed to bring the charges from infinity to the
1 2 1 2 1 2 1 e2
separation r without a change in kinetic energy. 2 mu = 2 mv + 2 mv + 4rf 0 r
When both charges have the same sign, their potential 1 2 – m bul =
2
1 e2 : uD
⇒ 2 mu 2 4rf 0 r as v = 2
energy is positive:
When the charges have opposite signs, the external work 1 2 e2
⇒ 4 mu = 4rf 0 r
is negative. Negative potential energy means that external
work is required to separate the charges. e2
⇒ r=
rmf 0 u2
Electric Potential Energy of a System of Point Charges:
Illustration 31
The electric potential energy of such a system is the work
Three equal charges q are placed at A
done in assembling this system starting from infinite
the corners of an equilateral triangle
separation between any two-point charges. of side a.
For a system of point charges q1, q2, ..... qn, the potential
(i) Find out potential energy of B C
energy is
n n q1 q j charge system.
1
U = 2 / / 4rf r (ii) Calculate work required to decrease the side of
i=1 j=1 0 ij
triangle to a/2.
It simply means that we have to consider all the pairs that
(iii) If the charges are released from the shown
are possible.
position and each of them has same mass m then
Important points: find the speed of each particle when they lie on
1. The potential energy of the system of charges depends triangle of side 2a
essentially on the seperation between the charges and Solution:
is independent of their location in space. (i) U = U12 + U13 + U23
2. As electrostatic force is conservative, work done Kq2 Kq2 Kq2 3Kq2
by external agency is given by    = a + a + a = a
(ii) Work required to decrease the sides A q
Wex = U(final) – U(intial)
= q(Vfinal – Vinitial) 3Kq2 3Kq2
W = Uf – Ui = a/2 – a
3. When free to move a positive charge moves
3Kq2 B C
from higher potential point(higher potential    =
q q
a a
energy) to lower potential point(lower potential (iii) Work done by electrostatic forces
energy); A negative charge moves from lower = change is kinetic energy of particles.
potential(higher potential energy) point to higher Ui – Uf = Kf – Ki
potential point(lower potential energy). 3Kq2 3Kq2
b1 l
– 2a = 3 2 mv – 0
2
a
Illustration 30
Kq 2
A proton moves from a large distance with a speed ⇒ v= am
u m/s directly towards a free proton originally at rest.
Find the distance of closet approach for the two protons 9. SELF ENERGY
in terms of mass of proton m and its charge e. (i) For hollow/solid uniformly charged conducting
Solution: sphere or hollow uniformly charged non‑conducting
sphere :
As here the particle at rest is free to move, when one
KQ2 Q2
particle approaches the other, due to electrostatic Uself = 2R = 8rf R
repulsion other will also start moving and so the velocity 0

of first particle will decrease while of other will increase Q : Charge on sphere, R : Radius of sphere.
and at closest approach both will move with same (ii) For uniformly charged solid nonconducting
velocity. So if v is the common velocity of each particle sphere:
at closest approach, then by ‘conservation of momentum’ 3KQ2 3Q2
of the two protons system. Uself = 5R = 20rf 0 R
Physics
Illustration 32 Electrostatic Energy Density
A spherical shell of radius R with uniform charge q Energy density is defined as energy stored in unit volume in
is expanded to a radius 2R. Find the work performed
any electric field.
by the electric forces and external agent against
1
electric forces in this process. Energy density = 2 εE2
Solution: where     E = electric field intensity at a point
q2 q2 q2
Wext = Uf – Ui = 16rf R - 8rf R =- 16rf R
0 0 0        ε =ε0εr electric permittivity of medium

Illustration 33 Illustration 34
Two concentric spherical shells of radius R1 and R2 (R2 Find out energy stored in an imaginary cubical volume
> R1) are having uniformly distributed charges Q1 and of side a in front of a infinitely large nonconducting
Q2 respectively. Find out total energy of the system. sheet of uniform charge density σ.

R2
Solution:
Energy stored
R1
# 1
U =  2 f 0 E dV where dV is small volume
2

Q1 Q2
1 # dV
= 2 f 0 E2 a E is constant
Solution: 1 v2
Utotal = Uself 1 + Uself 2 + UInteraction = 2 f 0 2 (a3)
4f 0
Q2 Q2 Q1 Q2 v2 a3
= 8rf1 R + 8rf2 R + 4rf = 8f
0 1 0 2 0 R2 0

INTEXT EXERCISE: 5
1. Four charges q1 = 1 mC, q2 = 2 mC, q3 = –3 mC and 1
(a) Q2 (b)
q4 = 4 mC are kept on the vertices of a square of side v2
1 m. The electric potential energy of this system of 1 1
(c) v (d) m
charges is _______
4. The electric field strength at a distance r from
(a) –7.62 × 10–2 J (b) –8.62 × 10–2 J the centre of a charged sphere of radius R is E.
(c) –7.62 × 10–4 J (d) –8.62 × 10–4 J If r > R, how much work will be done in bringing a test
2. Two point charges q1 = q2 = 2 mC are fixed at charge q0 from infinity to that point
x1 = + 3 m and x2 = –3m as shown in figure. A third 1
(a) q0RE (b) 2 q0RE
particle of mass 1 g and charge q3 = –4 mC is released 1
from rest at y = 4.0 m. Find the speed of the particle as (c) q0rE (d) 2 q0rE
it reaches the origin. 5. Point charge q moves from point P to point S along
the path PQRS in a uniform electric field Ev pointing
q3 y=4m parallel to the positive direction of x–axis. The
coordinates of the point P, Q, R and S are (a, b, 0) (2a,
0, 0), (0, –b, 0) and (0, 0, 0) respectively. The work
q2 q1
done by the field in the above process is given by
(a) qEa (b) – qEa
x2 = – 3 m x1 = 3 m Eqm
(c) qEa 2 (d) 2t
(a) 3.2 m/s (b) 4.2 m/s
6. A particle A has charge +q and particle B has charge
(c) 5.2 m/s (d) 6.2 m/s +4q with each of them having the same mass m. When
3. A particle A of mass m and charge Q moves directly allowed to fall from rest through the same electric
v
towards a fixed particle B, which has charge Q. potential difference, the ratio of their speed v A will
B
The speed of A is v when it is far away from B. The become
minimum separation between the particles is NOT (a) 1 : 2 (b) 2 : 1
proportional to (c) 1 : 4 (d) 4 : 1
 Electrostatics
7. Figure shows three points. X, Y and Z forming an 8. A particle of mass 0.002 kg and a charge 1 mC is held
equilateral triangle of side ‘s’ in an uniform electric at rest on a frictionless horizontal surface at a distance
field of strength E. A unit positive test charge is moved of 1 m from a fixed charge of 1 mC. If the particle is
from X to Y, from Y to Z, and from Z back to X. Which released, it will be repelled. The speed of the particle
one of the following correctly gives the work done when it is at a distance of 10 m from the fixed charge is
against electrical forces in moving the charge along the (a) 60 ms–1 (b) 75 ms–1
various parts of this path? (c) 90 ms–1 (d) 100 ms–1
s 9. Two identical thin rings, each of radius R metre, are co-
X
axially placed at a distance R metre apart Q1 coulomb
60° Y
and Q2 coulomb are the charges uniformly spread on
the two rings. The work done in moving a charge q
from the centre of one ring to that of the other is
q ^Q1 - Q2 h^ 2 - 1 h
Z
E (a) zero (b)
4 2 rf 0 R
2 q ^Q1 + Q2 h q ^Q1 + Q2 h^ 2 + 1 h
X to Y Y to Z Z to X
(a) + Es + Es cos 60° – Es cos 60° (c) 4rf 0 R (d)
4 2 rf 0 R
(b) 0 – Es sin 60° + Es sin 60° 10. In bringing an electron towards a second electron, the
(c) + Es + Es cos 60° – Es cos 60° electrostatic potential energy of the system
(d) 0 + Es sin 60° – Es sin 60° (a) increases (b) decreases
(c) remains the same (d) becomes zero
Answer Key
1. (a) 2. (d) 3. (c) 4. (c) 5. (b) 6. (a) 7. (b) 8. (c) 9. (b) 10. (a)

10. ELECTRIC DIPOLE Solution:

Two equal and opposite charges separated by a distance
Given diagram is:
together constitute a dipole. Q
Dipole moment ^ pv h is defined as the simple product of
magnitude of either charge and the distance of separation
between the two charges.
 
p = q( 2a ) –2Q
a
Q

Equitorial Line It can be shown in the form of dipoles as

Q
+q –q
O Axial Line
P
a a
–Q 60°
–Q Q
Dipole moment pv always points from – q to + q. Its SI P
where, Pv = dipole moment of single dipole
unit is Coulomb metre (Cm).
So, net dipole moment is :
Illustration 35 Pnet = P2 + P2 + 2P $ P cos 60 o
Three charges are arranged on the vertices of an or Pnet = 3 P = 3 Qa
equilateral triangle as shown in figure. Find the and direction is along the bisector of the angle at –2Q,
dipole moment of the combination. towards the triangle.
Q
Electric Field Due To A Dipole At A Point Lying On
Axial Line
Consider an electric dipole consisting of two point charges
Q
– q and +q separated by some distance 2a. Let P be an
–2Q
a observation point on axial line such that its distance from
centre of the dipole is r.
Physics
If EA is the electric field intensity at P due to charge +q then Resolving pv along OP and perpendicular to OP, we get p
1 q
EA = 4rf cos q and p sin q respectively. Point P is on the axial line of
0 ]r + ag
2
dipole of dipole moment p cos q. Let Ev1 be the electric field
Again, if EB is electric field intensity at P due to charge – q then
q intensity at P due to p cos q.
1
0 ]r - ag
EB = 4rf 2

B E 
So, Eaxial = EB – EA{ a EB > EA}
A

q q
4rf 0 ]r - ag 4rf 0 ]r + ag2
P
Eaxial = 2 -  r
os
pc 
1 4rqa
Eaxial = 4rf 2
0 ^r - a h
q  q
2 2
O p
p sin 
Since p = q(2a)
1 2pr 1 2p cos i
Eaxial = 4rf 2 For r >>> a Then E1 = 4rf along PA
2 2 ;
0 ^r - a h
0 r3
v
Let E2 be the electric field intensity at P due to p sin q
1 2p
Eaxial = 4rf 3 1 p sin i
0 r Then Ev2 = 4rf along PB
0 r3
Electric Field Due to Dipole at a Point Lying on the If E is the magnitude of the resultant electric intensity Ev, then
Equiatorial Line
1 p
E2 = E21 + E22 = < 4rf 3 F (4 cos2 q + sin2 q)
2

If EA is electric field intensity at P due to +q 0 r

1 q 1 q 1 p
EA = 4rf or E = 4rf 3 3 cos i + cos2 i + sin2 i
^ a2h
2
2 = 4rf
0 AP 0 r 2
+ 0 r

Ev A is represented both in magnitude and direction 1 p

or E = 4rf 3 3 cos2 i + 1
0 r
by PD .
EA sin  If b is the angle which Ev makes with Ev1 , then
D
E 1 p sin i # 4rf 0 r
3
tan i
tan b = E2 = 4rf = 2
P

2EA cos  1 0 r3 2p cos i
b = tan–1 b 2 tan i l
   1
|EA| = |EA| or
C
EA sin  EB
r2 + a2 r2 + a2 Electric Potential due to an Electric Dipole at a
Point on the Axial Line of the Electric Dipole
r

+q
 
–q Let P be the observation point on the axial line of the electric
A
O
B dipole AB. Let r be the distance of the observation point P
a a from the mid point O of the electric dipole. Potential, due to
If EB is electric field intensity at P due to –q 1 q
charge +q, at P = 4rf r - a .
1 ^- q h
1 q v 0
EB = 4rf 2 2 E B is represented both in magnitude
0 ^r + a h Potential, due to charge – q at P = 4rf r + a
0
and direction by PC clearly, EA = EB. Let us resolve EA Potential at P due to electric dipole,
and EB into two components in two mutually perpendicular r–a
directions components of Ev A and Ev B along the equitorial B O A P

line cancel each other but the components perpendicular

–q +q
a a
to equitorial line get added up because they act in same r
direction. So magnitude of resultant intensity Ev at P
r+a

2q
Potential at a point on axial line
a
E = 2EA cos q =
4rf 0 r2 + a2h r2 + a2
^ 1 q 1 q
p V = 4rf r - a - 4rf r + a
1
Eequatorial = 4rf 2 ; For r >> a 0 0
0 ^r + a h
= 4rf q : r - a - r + a D
2 3/2
1 1 1
1 p
Eequatorial = 4rf 3 0
0 r
= 4rf q ; E = 4rf
1 r+a-r+a 1 q # 2a
Electric Field Intensity at a General Point due 0 r2 - a2 0 r -a
2 2

1 p
to Short Electric Dipole = 4rf 2
0 r -a
2  { q × 2a = p}
Let P be the general point. Consider a short electric dipole 1 p
If r >> a, then V = 4rf 2
of dipole moment pv placed in vacuum. Let O be the mid 0 r
Kpv $ rv
point of the dipole. Let the line OP make an angle q with pv. In vector form it can also be written as V =
r3
 Electrostatics
Illustration 36 It is convenient to express the torque in vector form as the
 
Find out the magnitude of electric field intensity cross product of the vectors p and E , so vectorially,
and potential due to a dipole of dipole moment xv = pv # Ev
Pv = it + 3 tj kept at origin at (–1, 3 , 0)
Potential Energy of a Dipole Placed in a
Solution: Uniform Electric Field
ˆj
Work must be done by an external agent to rotate the dipole
2 through a given angle in the field. This work done is then
stored as potential energy in the system, that is, the dipole
and the external field. The work dW required to rotate the
° 3
60
dipole through an angle dq is given by
P ˆi
(0, 0, 0)
ˆk dW = t dq
Since, t = pE sin q
Given :( Pv = it + 3 tj )
This work is transformed into potential energy U. We find
Pv = 1 + 3 = 2 this for a rotation from q0 to q. So,
q is the angle between the dipole moment pv and the i i i i

position vector of the point rv = # xdi = # PE sin idi = # PE sin idi = pE # sin idi
i0 i0 i0 i0
pv $ rv
⇒   U = pE ]- cos ig = pE ^cos i 0 - cos ih
cos θ = pr i
i0

^it + 3 tj h $ ^- it + 3 tj + 0kth 1 The term involving cos q0 is a constant that depends on

= 2#2 = 2 the initial orientation of the dipole. It is convenient to choose
KP q0 = 90°, so that cos q0 = cos 90° = 0. In this case, we can
\ E= 3 cos2 i + 1
r3 express U as
K ]2g 3 K 7 U = –pEcos q
= 4 +1 = 8
23 This is equivalent to the dot product of the vectors pv and
pv $ rv Ev . So, U = - pv.Ev
V= K 3
r Force on an electric dipole in Non–uniform
^it + 3 tj h $ ^- it + 3 tj + 0kth electric field
=K
_ ]- 1g2 + ^ 3 h2 + ]0g2 i
3
If in a non–uniform electric field dipole is placed at a point
2 K where electric field is E, the interaction energy of dipole
\ V= K 8 = 4
at this point U = – - pv.Ev . Now the force on dipole due to
TU
11. DIPOLE IN AN EXTERNAL FIELD electric field F = – Tr . If dipole is placed in the direction
dE
Torque on a Dipole Placed in a Uniform Electric Field of electric field then F = – p dx

+q Illustration 37
qE
Calculate force on a dipole
 2a sin  in the surrounding of a long
+ + + + + + + + + + +

charged wire as shown in the

figure.
p
–q –q +q
qE E r

Suppose an electric dipole is placed in a uniform external

electric field Ev where the dipole moment makes an angle q
with the field. The forces on the two charges are equal and 
opposite each having a magnitude F = qE.
Solution:
Thus the net force on the dipole is zero.
In the situation shown in figure, the electric field strength
However, the two forces produce a net torque on the 2km
dipole, and the dipole tends to rotate such that its axis gets due to the wire, at the position of dipole as E = r
aligned with the field. Thus force on dipole is
2kpm
     F = – p. dr = – p ;- 2 E =
t = 2Fa sin q dE 2km
r r2
Since F = qE and p = 2aq
Here –ve charge of dipole is close to wire hence net
⇒ t = 2aqE sin q
force an dipole due to wire will be attractive.
⇒ t = pE sin q
Physics
Force between two short dipoles Illustration 38
(a) When dipoles are placed on the same axis: Suppose Prove that the frequency of oscillation of an electric
the dipoles are placed at a separation r. The P.E. of the dipole of moment p and rotational inertia I for
dipole Pv2 in the field of Pv1 is small amplitudes about its equilibrium position in a
uniform electric field strength E is 2r b l
2P P2 1 pE
U = – P2E1 cos 0° = - c 4rf m 13
1
0 r I
P1 P2
Solution:
Let an electric dipole (charge q and –q at a distance
E2 E1 2a apart) placed in a uniform external electric field of
r

The force between them is given by strength E.
2P1 P2
F = - dr = - dr <- 4rf F
dU d 1 +q 
0 r3 F
6P P2
= - c 4rf m 14
1
0 r 2a  

Here minus sign of F indicates attractive force.
E

P
(b) When dipoles are placed parallel to each other: 
Suppose dipoles are placed at a separation r. The P.E. F –q
+ +
of dipole P2 in the field of dipole P1 is
Restoring torque on dipole t = –pE sin q = –pEq
PP
U = – P2E1 cos 180° = c 4rf m 1 3 2
1
(as q is small)
0 r
Here – ve sign shows the restoring tendency of torque.
r x PE
t = Ia \ angular acceleration = a = I = I i
P1 P2
E2 E1

The force between them is given by pE
1 PP For SHM a = – w2q comparing we get w =
F = dr = - dr < 4rf 1 3 2 F
- dU d I
0 r ~
Thus frequency of oscillations of dipole n = 2r
3P1 P2
= c 4rf m 4
1
c m
r 1 pE
0
= 2r

As the force is positive, so it is of repulsive nature. I

INTEXT EXERCISE: 6
1. The electric potential at a point situated at a distance r (a) force = 0, torque may not be zero
on the axis of a short electric dipole of moment ‘P’ is (b) force may not be zero
1
4rf 0 times _________ (c) force and torque both may not be zero
P P (d) force = 0, torque = 0
(a) 3 (b) 2
r r 5. An electric dipole is placed in a non–uniform electric
P field. Then net
(c) r (d) None of these
(a) force experienced is zero while torque is not zero
2. An electric dipole of moment P is kept along an
(b) force experienced is zero and torque is also zero
electric field E. The work done in rotating it from an
equilibrium position by an angle q is (c) both force and torque may NOT be zero
(a) PE(1 – cos q) (b) PE(1 – sin q) (d) force experienced is not zero while torque is zero
(c) PE cos q (d) PE sin q 6. A molecule of HCl is placed in a electric field of
3. If the magnitude of intensity of electric field at a 2.5 × 104 N/C. The dipole moment of molecule is 3.4 ×
distance x on axial line and at a distance y on equatorial 10–30. The maximum torque that can act on molecule is
line of a given dipole are equal, then x : y is (a) 6.5 × 10–26 N-m (b) 7.5 × 10–26 N-m
(a) 1 : 1 (b) 1 : 2 (c) 8.5 × 10–26 N-m (d) 1.5 × 10–26 N-m
(c) 1 : 2 (d) 3
2 :1 7. For short dipole, electric field magnitude at a point
4. An electric dipole is placed in a uniform field, the net which is at a distance r from it and making an angle ‘q’
force and net torque on the dipole is with its axis, is
 Electrostatics
p sin i p cos i (a) 90c and 180c (b) 0c and 90c
(a) 4rf (b) 4rf
0 r3 0 r3 (c) 90c and 0c (d) 0c and 180c
p
(c) 1 + 3 cos2 i (d) None of these 10. An electric dipole is placed along the x-axis at the or-
4rf 0 r3
8. The dipole moment of a dipole is 10tj . This is kept in igin O. A point P is at a distance of 20 cm from this
π
a uniform electric field Ev = 3it + 4tj , then the torque origin such that OP makes an angle 3 with the X-ax-
acting on it is ______ is. If the electric field at P makes an angle θ with the
(a) 30kt (b) - 30kt X-axis, the value of θ would be.
(c) 40kt (d) - 40kt
(b) 3 + tan c
π π -1 3m
9. The potential at a point due to an electric dipole will be (a) 3
2
maximum and minimum when the angles between the
(d) tan c
axis of the dipole and the line joining the point to the 2π -1 3m
(c) 3 2
dipole are respectively
Answer Key
1. (b) 2. (a) 3. (d) 4. (a) 5. (c) 6. (c) 7. (c) 8. (b) 9. (d) 10. (b)

12. ELECTROSTATIC PRESSURE Solution:

Take a small charge element on the surface of sphere (charge We know that electrostatic force on a charged conductor is
density s). dF v2
given by ds = 2f
E1 is the field at nearby point due to the charge element. 0
v2
E2 is the field at the point due to remaining charge on sphere. So the conductor will break by this force if, 2f
0

> Breaking strength i.e., s2 > 2 × 9 × 10–12 × 3.5 × 108

E2
E1

i.e., smin = ^3 7 h × 10–2 = 7.94 × 10–2 (C/m2)

E1 E2

Now as the charge on an electron is 1.6 × 10–19 C, the

+

excess electrons per m2 is 4.96 × 1017

Inside E1 – E2 = 0      ⇒ E1 = E2 Further as in case of a conductor near its surface
v
Just outside E = E1 + E2 = 2E2⇒ E2 = 2e v 7.94 # 10 -2
0 E= f = = 8.8 × 109 V/m
Force due to electrostatic pressure is directed normally 0 9 # 10 -12
outwards to the surface.
+ + ++
13. CONDUCTOR AND IT’S PROPERTIES
+ + ds [FOR ELECTROSTATIC CONDITION]
+
+ + (i) Conductors are materials which contain large number
+ + of free electrons which can move freely inside the
+ +
+ + + conductor.
Force on small element of area ds of charged conductor (ii) Ιn electrostatics, conductors are always equipotential
dF = (Charge on ds ) × Electric field surfaces.
v v2 (iii) Charge always resides on outer surface of conductor.
= (σ ds) 2e = 2e ds
0 0 (iv) Ιf there is a cavity inside the conductor having no
The electric force acting per unit area of charged surface charge then charge will always reside only on outer
is defined as electrostatic pressure. surface of conductor.
dF v2 (v) Electric field is always perpendicular to conducting
Pelectrostatic = dS = 2e surface.
0
(vi) Electric lines of force never enter into conductors.
Illustration 39
Brass has a tensile strength 3.5 × 108 N/m2. What (vii) Electric field intensity near the A

charge density on this material will be enough to conducting surface is given by

break it by electrostatic force of repulsion? How v
formula Ev = f nt
C
0
be? What is the value of intensity just out side the
B
v v v
surface? Ev A = f nt; Ev B = fB nt ; and EvC = fC nt
A
0 0 0
Physics
(viii) When a conductor is grounded its
Solution:
V=0
potential becomes zero.
Consider a Gaussian surface as shown in figure. Two
(ix) When an isolated conductor is grounded then its faces of this closed surface lie completely inside the
conductor where the electric field is zero. The flux
charge becomes zero.
through these faces is, therefore, zero. The other parts
(x) When two conductors are connected there will be of the closed surface which are outside the conductor
charge flow till their potential becomes equal are parallel to the electric field and hence the flux on
Illustration 40 these parts is also zero. The total flux of the electric
field through the closed surface is, therefore zero. From
Prove that if an isolated (isolated means no charges
Gauss’s law, the total charge inside this closed surface
are near the sheet) large conducting sheet is given a should be zero. The charge on the inner surface of A
charge then the charge distributes equally on its two should be equal and opposite to that on the inner surface of B.
surfaces. E=0
Q1 A
Solution: E
Let there is x charge on left side of sheet and Q – x charge Q2 B
on right side of sheet. E=0

x Q–x Q1 – q P
A
+q

P –q
Q–x x
B
2A0 2A0 Q2 + q

To find the value of q, consider the field at a point

P inside the plate A. Suppose, the surface area
of the plate (one side) is A. Using the equation
Since point P lies inside the conductor so EP = 0 E = σ/(2ε0), the electric field at P
Q-x Q Q -q
x 2x due to the charge Q1 – q = 21Af (downward); due to
2Af 0 - 2Af 0 = 0 ⇒ 2Af 0 = 2Af 0 q 0

Q Q the charge + q = 2Af (upward),

⇒ x= 2 Q–x= 2 0
q
due to the charge – q = 2Af (downward), and due to
So charge is equally distributed on both sides 0
Q +q
Illustration 41 the charge Q2 + q = 22Af (upward).
0

If an isolated infinite sheet contains charge Q1 on The net electric field at P due to all the four charged
its one surface and charge Q2 on its other surface surfaces is (in the downward direction)
then prove that electric field intensity at a point in Q1 - q q q Q2 + q
Ep = 2Af 0 - 2Af 0 + 2Af 0 - 2Af 0
Q
front of sheet will be 2Af , where Q = Q1 + Q2. As the point P is inside the conductor, this field should be
0
Q -Q
Solution: zero. Hence, Q1 – q – q + q – Q2 – q = 0 ⇒ q = 1 2 2
Electric field at point P:
Q1 Q2 Q1 + Q2 Q Illustration 43
2Af 0 + 2Af 0 n = 2Af 0 n = 2Af 0 n
t t t
Figure shows three large –Q 3Q Q
Q1 Q2 metallic plates with charges
P Q1 Q2 – Q, 3Q and Q respectively.
2A0
+
2A0 Determine the final charges
on all the surfaces.
This shows that the resultant field due to a sheet depends Solution:
only on the total charge of the sheet and not on the
We assume that charge on surface 2 is x. Following
distribution of charge on individual surfaces.
conservation of charge, we see that surfaces 1 has charge
Illustration 42 (– Q – x). The electric field inside the metal plate is zero
so fields at P is zero.
Two conducting plates A and B are placed parallel to 3Q Q
each other. A is given a charge Q1 and B a charge Q2. 1 2 3 4 5 6

Prove that the charges on the inner facing surfaces –Q – x x

are of equal magnitude and opposite sign. Also find P

the charges on inner & outer surfaces.
 Electrostatics
-Q - x x + 3Q + Q (ii) If a charge q is kept inside the +q + Q

Resultant field at P : EP = 0 ⇒ 2Af = 2Af 0 cavity of a conductor and conductor

0
- 5Q is given a charge Q then – q charge –q
⇒ – Q – x = x + 4Q ⇒ x = 2 q
will be induced on inner surface
We see that charges on the facing surfaces of the plates and total charge on the outer surface
are of equal magnitude and opposite sign.
will be q + Q.
Thus the final charge distribution on all the surfaces is:
(iii) Resultant field, due to q (which B
S2
C
is inside the cavity) and induced S1

charge on S1, at any point outside

q q+Q
A

S1 (like B, C) is zero. Resultant

–q
+ 3Q –5Q 5Q +Q –Q +3Q
2 2 2 2 2 2
field due to q + Q on S2 and any
other charge outside S2 , at any point inside of
surface S2 (like A, B) is zero
(iv) Resultant field in a charge free
14. SOME OTHER IMPORTANT RESULTS FOR cavity in a closed conductor is
A CLOSED CONDUCTOR zero. There can be charges No

outside the conductor and on the

+q charge
(i) Ιf a charge q is kept in the cavity
surface also. Then also this result
then – q will be induced on the
is true. No charge will be induced
inner surface and + q will be –q
q on the inner most surface of the conductor.
induced on the outer surface of
the conductor (v) Charge distribution for different types of cavities in
conductors
S2 S2
S2

q q
S1 q
(A) (B) (C)
S1 S1
C C
C

charge is at the common centre charge is not at the common centre charge is at the centre of S2
(S1, S2  spherical) (S1, S2  spherical) (S2  spherical)

S2 S2 S2

S1 C
q
q
(D) S1 C
(E) C (F) S1 q

charge is not at the centre of S2

charge is at the centre of charge is not at the centre of
(S2  spherical) S1(Spherical) S1(Spherical)


Using the result that Eres in the conducting material should be zero and using result (iii) We can show that

Case A B C D E F
S1 Uniform Non Uniform Non Uniform Non Uniform Uniform Non Uniform
S2 Uniform Uniform Uniform Uniform Non Uniform Non Uniform

Note: In all cases charge on inner surface S1 = – q and on outer surface S2 = q. The distribution of charge on S1 will not change
even if some charges are kept outside the conductor (i.e. outside the surface S2). But the charge distribution on S2 may change
if some charges(s) is/are kept outside the conductor.
Physics
Illustration 44 Kq
(ii) VA = R
An uncharged conductor of inner radius R1 and outer 2
radius R2 contains a point charge q at the centre as Kq
(iii) VB = CB
shown in figure
(iv) EA = 0 (point is inside metallic conductor)
t
Kq \
S2
(v) EB = CB
S1 q R1 CB2
KQq \
C
R2
(vi) FQ =
O
CB
A CB2
B
(vi) Sharing of charges: Two conducting hollow spherical
(i) Find E v and V at points A, B and C. shells of radii R1 and R2 having charges Q1 and Q2
(ii) If a point charge Q is kept out side the sphere at respectively and separated by large distance, are joined
a distance ‘r’ (>> R2) from centre then find out by a conducting wire. Let final charges on spheres are
resultant force on charge Q and charge q. q1 and q2 respectively.
q1 q2
Solution: R1 R2
Kq Kq K (- q) Kq
At point A: VA = OA + R + R , EA = OA
2 1 OA3
Kq K ^- qh Kq Kq
Potential on both spherical shell become equal after
At point B: VB = OB + OB + R = R , EB = 0; joining, therefore
2 2
Kq1 Kq2 q1 R1
Kq Kq R1 = R2 ; q2 = R2 ...(i)
At point C: VC = OC , EvC = OC
OC3 and q1 + q2 = Q1 + Q2...(ii)
+q
from (i) and (ii)
^Q1 + Q2 h R1 ^Q1 + Q2 h R2
–q q1= R1 + R2 ; q2 = R1 + R2
q
q R v 4rR12 R
ratio of charges q1 = R1 ; 1 = R1
2 2 v2 4rR22 2

KqQ v R
(ii) Force on point charge Q: FvQ = rt (r = distance ratio of surface charge densities v1 = R2
r2 2 1
of ‘Q’ from centre ‘O’) Force on point charge q: Illustration 46
Fvq = 0 (using result (iii) & charge on S1 uniform)
The two conducting spherical shells
Illustration 45 are joined by a conducting wire and Q
–3Q
cut after some time when charge R
An uncharged conductor of inner radius R1 and outer stops flowing. Find out the charge
radius R2 contains a point charge q placed at point on each sphere after that. 2R
P (not at the centre) as shown in figure? Find out the
Solution:
following:
B After cutting the wire, the potential of both the shells is
equal. Thus, potential of inner shell
Kx K ^- 2Q - xh K ^ x - 2Qh
A

R1
P Vin = R + 2R = 2R
C q
S1 R2 and potential of outer shell
Kx K ^- 2Q - xh - KQ
D
S2 Vout = 2R + 2R = R

(i) VC (ii) VA –2Q – x

(iii) VB (iv) EA x

(v) EB
K ^ x - 2Q h
(vi) force on charge Q if it is placed at B
KQ
As  Vout = Vin ⇒ - R = 2R
Solution:
⇒  – 2Q = x – 2Q ⇒ x = 0
Kq K ^- qh Kq So charge on inner spherical shell = 0 and outer spherical
(i) VC = CP + R + R
1 2 shell = – 2Q.
 Electrostatics
Illustration 47 3Q
Find charge on each spherical shell after joining the
inner most shell and outer most shell by a conducting
–Q

wire. Also find charges on each surface. R

5Q
–2Q
3
2 Q 2R
R
1
Solution:
2R
3R
When inner shell is grounded to the Earth then the potential
of inner shell will become zero because potential of the
Kx K3Q
Solution: Earth is taken to be zero. R + 2R = 0
Let the charge on the innermost sphere be x. 3Q
6Q – x
x
–2Q
3
2 x R
R
1

2R 2R
3R
- 3Q
Finally potential of shell 1 = Potential of shell 3 x= 2 ,
Kx K ^- 2Qh K ^6Q - xh KQ k ^- 2qh k ^5Qh - 3Q
R + 2R + 3R = 3R + 3R + 3R the charge that has increase = 2 – (– Q)
Q -Q Q
3x – 3Q + 6Q – x = 4Q; 2x = Q; x = 2 = 2 hence charge flows into the Earth = 2
Q
Charge on innermost shell = 2 , charge on outermost
Illustration 49
5Q
shell = 2 middle shell = – 2Q An isolated conducting sphere of charge Q and radius
R is connected to a similar uncharged sphere (kept
Q
+3Q/2
–3Q/2
–Q/2
at a large distance) by using a high resistance wire.
Q/2 After a long time what is the amount of heat loss ?
Solution:
When two conducting spheres of equal radius are
connected charge is equally distributed on them.
So we can say that heat loss of system ∆H = Ui – Uf
Illustration 48
Q2
=d n-d n = 16rf R
Two conducting hollow spherical shells of radii R and Q2 Q 2 /4 Q2 /4
8rf 0 R - +
8rf 0 R 8rf 0 R
2R carry charges – Q and 3Q respectively. How much 0

charge will flow into the earth if inner shell is grounded?

INTEXT EXERCISE: 7
1. Electric field just outside the spherical conducting shell (a) zero
v
(b) 2f
whose surface charge density is s is ______ 0
v v v
(a) 2f (b) f (c) f (d) None of these
0 0 0
2v v 3. A charge q is distributed uniformly
(c) f (d) 3f
0 0
on the surface of a sphere of radius
2. A positively charged sphere R. It is covered by a concentric R q

suspended with a silk thread is + + hollow conducting sphere of radius

+
slowly pushed in a metal bucket. +
+ 2R
++ + 2R. Find the charge on outer surface
After its insertion the lid is closed. +
What will be the electric field of hollow sphere if it is earthed.
intensity inside when the sphere q q
(a) 2 (b) - 2
has touched the bucket? s is the
surface charge density of sphere. (c) –q (d) zero
Physics
4. In the above problem if the thickness of the outer (a) a potential difference appears between the two
sphere is considerable, find the charge on outer surface cylinders when a charge density is given to the
of hollow sphere if it is earthed. inner cylinder.
q q (b) a potential difference appears between the two
(a) 2 (b) - 2 cylinders when a charge density is given to the
(c) –q (d) zero outer cylinder.
(c) no potential difference appears between the two
5. For the isolated charged conductor shown in figure, the cylinders when a uniform line charge is kept along
potentials at points A, B, C and D are VA, VB, VC and VD the axis of the cylinders.
respectively, then: (d) no potential difference appears between the two
+
+ + D+ cylinders when same density is given to both the
+
+
+
+ cylinders.
+ +
+
++
+
+ 8. Two conducting spheres of radii R1 and R2 are charged
+ +
++
+
+ with charges Q1 and Q2, respectively. On bringing
+ C + them in contact, there is
+ A B ++
+
(a) no change in the energy of the system
(b) an increase in the energy of the system if
+
+ + +

(a) VA = VB > VC > VD (b) VD > VC > VB = VA Q1R2 ≠ Q2R1

(c) always a decrease in the energy of the system
(c) VD > VC > VB > VA (d) VD = VC = VB = VA (d) a decrease in the energy of the system if
6. A solid conducting sphere having a charge Q is Q1R2 ≠ Q2R1
surrounded by an uncharged concentric conducing 9. If conducting sphere-1 with a charge q touches an
hollow spherical shell. Let the potential difference uncharged conducting sphere-2, of half the radius of 1,
between the surface of the solid sphere and that charges on them are
of the outer surface of the hollow shell be V. If the q q q 2q
(a) 2 , 2 (b) 3 , 3
shell is now given a charge of –3Q, the new potential
2q q q 4q
difference between the same two surfaces is: (c) 3 , 3 (d) 5 , 5
(a) V (b) 2 V 10. The net charge given to an isolated conducting solid
(c) 4 V (d) –2 V
sphere:
7. A long, hollow conducting cylinder is kept coaxially (a) must be distributed uniformly on the surface
inside another long, hollow conducting cylinder of (b) may be distributed uniformly on the surface
larger radius, both the cylinders are initially electrically (c) must be distributed uniformly in the volume
neutral. (d) may be distributed uniformly in the volume.
Answer Key
1. (b) 2. (a) 3. (d) 4. (d) 5. (d) 6. (a) 7. (a) 8. (d) 9. (c) 10. (a)
 Electrostatics
Unsolved Exercises
EXERCISE - 1

Charges and Coulomb’s Law 7. The maximum electric field intensity on the axis of a
uniformly charged ring of charge q and radius R will be:
1. A charged particle q1 is at position (2, –1, 3). The q
1 1 2q
electrostatic force on another charged particle q2 at (0, (a) 4rf (b) 4rf
0 3 3R 0 3R
2 2
0, 0) is:
1 2q 1 3q
(a) 561rf2 ^2it - tj + 3kth
qq (c) 4rf (d) 4rf
0 3 3R 0 2 3R
2 2
0

^2it - tj + 3kth
q1 q2
(b) 8. A charged particle of charge q and mass m is released
56 14 rf 0
from rest in a uniform electric field E. Neglecting the
(c) 561re2 ^ tj - 2it - 3kth
qq
effect of gravity, the kinetic energy of the charged
0
particle after time ‘t’ seconds is
^ tj - 2it - 3kth
q1 q2
(d)
56 14 rf 0 Eqm E2 q2 t2
(a) t (b) 2m
2. Three charges +4q, Q and q are placed in a straight
2E t 2 2 Eq 2
m
line of length , at points at distance 0, , /2 and , (c) mq (d)
2t 2
respectively from one end of line. What should be the
value of Q in order to make the net force on q to be zero? 9. A positively charged pendulum is oscillating in a
(a) – q (b) – 2q uniform electric field as shown in Figure. Its time
(c) – q/2 (d) 4q period of SHM as compared to that when it was
uncharged. (mg > qE)
3. Two similar very small conducting spheres having
charges 40 µC and – 20 µC are some distance apart.
Now they are touched and kept at the same distance.
The ratio of the initial to the final force between them is: E
(a) 8 : 1 (b) 4 : 1
(c) 1 : 8 (d) 1 : 1 +
4. Two point charges placed at a distance r in air exert a
force F on each other. The value of distance R at which
they experience force 4F when placed in a medium of (a) Will increase
dielectric constant K = 16 is: (b) Will decrease
(a) r (b) r/4 (c) Will not change
(c) r/8 (d) 2r (d) Will first increase then decrease
5. Mark out the correct option. 10. The electric field above a uniformly charged
(a) The total charge of the universe is constant. nonconducting sheet is E. If the nonconducting sheet
(b) The total positive charge of the universe is
is now replaced by a conducting sheet, with the charge
constant.
same as before, the new electric field at the same point is:
(c) The total negative charge of the universe is
constant (a) 2E (b) E
(d) The total number of charged particles in the E
(c) 2 (d) None of these
universe is constant.
Electric Field Electric lines of force, Gauss’s law
6. Charges 2Q and – Q are placed –Q +2Q
11. If electric field is uniform, then the electric lines of
as shown in figure. The point at – +
forces are:
which electric field intensity is
A B

zero will be: (a) Divergent (b) Convergent

(a) Somewhere between – Q and 2Q (c) Circular (d) Parallel
(b) Somewhere on the left of – Q 12. The figure shows the electric lines of force emerging
(c) Somewhere on the right of 2Q from a charged body. If the electric fields at A and B
(d) Somewhere on the perpendicular bisector of line are EA and EB respectively and if the distance between
joining – Q and 2Q A and B is r, then
Physics
q q
(a) 24e (b) 12e
0 0
q q
A (c) 6e (d) 8e
0 0

B
Electric potential, potential energy
18. At a certain distance from a point charge, the electric
field is 500 V/m and the potential is 3000 V. What is
the distance?
(a) EA < EB (b) EA > EB (a) 6 m (b) 12 m
E E (c) 36 m (d) 144 m
(c) E A = rB (d) E A = 2B
r 19. Figure represents a square +q P +q
13. Select the correct statement: carrying charges +q,
(a) The electric lines of force are always closed curves + q, – q, – q at its four corners as A C
(b) Electric lines of force are parallel to equipotential shown. Then the potential will B

surface be zero at points: (A, C, P and Q –q

are mid points of sides)
–q
(c) Electric lines of force are perpendicular to Q

equipotential surface (a) A, B, C, P and Q (b) A, B and C

(d) Electric line of force is always the path of a (c) A, P, C and Q (d) P, B and Q
positively charged particle.
20. A semicircular ring of radius 0.5 m is uniformly
14. If the electric flux entering and leaving a closed surface charged with a total charge of 1.5 × 10–9 coul. The
are respectively of magnitude φ1 and φ2, then the electric potential at the centre of this ring is:
electric charge inside the surface will be: (a) 27 V (b) 13.5 V
z 2 - z1 (c) 54 V (d) 45.5 V
(a) f0 (b) (φ1 – φ2)e0
21. When a charge of 3 coul is placed in a uniform electric
(c) e0(φ2 – φ1) (d) e0(φ2 + φ1)
field, it experiences a force of 3000 newton. The
15. Electric charges are distributed in a small volume. The potential difference between two points separated by a
flux of the electric field through a spherical surface of distance of 1 cm along field within this field is:
radius 20 cm surrounding the total charge is 50 V-m. The (a) 10 volt (b) 90 volt
flux over a concentric sphere of radius 40 cm will be: (c) 1000 volt (d) 3000 volt
(a) 50 V-m (b) 75 V-m 22. A 5 coulomb charge experiences a constant force of
(c) 100 V-m (d) 200 V-m 2000 N when moved between two points separated
by a distance of 2 cm in a uniform electric field. The
16. An imaginary closed surface P is potential difference between these two points is:
constructed around a neutral
(a) 8 V (b) 200 V
conducting wire connected to a
(c) 800 V (d) 20,000 V
battery and a switch as shown in
figure. As the switch is closed, P battery 23. The kinetic energy which an electron acquires when
the free electrons in the wire accelerated (from rest) through a potential difference
start moving along the wire. In of 1 volt is called:
any time interval, the number of electrons entering the (a) 1 joule (b) 1 electron volt
closed surface P is equal to the number of electrons (c) 1 erg (d) 1 watt
leaving it. On closing the switch, the flux of the electric 24. The potential difference between points A and B in the
field through the closed surface: given uniform electric field is:
(a) remains unchanged (b) remains zero a
C B
(c) is increased (d) is decreased
E
17. Eight point charges (can be assumed
q q
b
as uniformly charged small q
D q C
B
spheres and their centres at the
A

corner of the cube) having value q q A

E
q
each are fixed at vertices of a cube. q
The electric flux through square q (a) Ea (b) E ^a2 + b2h
surface ABCD of the cube is (c) Eb (d) ^ Eb/ 2 h
 Electrostatics
25. An equipotential surface and an electric line of force: 33. Six charges of magnitude + q +q –q

(a) never intersect each other (b) intersect at 45º and – q are fixed at the
(c) intersect at 60º (d) intersect at 90º corners of a regular hexagon
of edge length a as shown in –q +q
26. A particle of charge Q and mass m travels through a
the figure. The electrostatic
potential difference V from rest. The final momentum
potential energy of the
of the particle is:
system of charged particles +q –q
mV is:
(a) Q (b) 2Q mV
q2 q2
(a) re a ;
3 15 E
(b) re a ;
3 9E
2QV 8 - 4 2 -4
(c) 2mQV (d) m
0 0

q2 q2
(d) re a ;
3 15 E
(c) re a ;
3 15 E
27. If a uniformly charged spherical shell of radius 10 cm 0 4 - 2 0 2 - 8
has a potential V at a point distant 5 cm from its centre,
then the potential at a point distant 15 cm from the 34. You are given an arrangement of three point charges
centre will be: q, 2q and xq separated by equal finite distances so that
V 2V electric potential energy of the system is zero. Then the
(a) 3 (b) 3 value of x is:
3 2 1
(c) 2 V (d) 3V (a) - 3 (b) - 3
2 3
28. A hollow uniformly charged sphere has radius r. If the (c) 3 (d) 2
potential difference between its surface and a point at
35. The variation of potential with distance r from a
distance 3r from the centre is V, then the electric field
fixed point is shown in Figure. The electric field at
intensity at a distance 3r from the centre is:
r = 5 cm, is:
(a) V/6r (b) V/4r
(c) V/3r (d) V/2r 5
29. A hollow sphere of radius 5 cm is uniformly charged
V in volt

such that the potential on its surface is 10 Volt then

potential at centre of sphere will be:
(a) Zero
(b) 10 Volt
(c) Same as at a point 5 cm away from the surface r in cm
0 2 4 6
(d) Same as at a point 25 cm away from the centre (a) (2.5) V/cm (b) (– 2.5) V/cm
30. If a charge is shifted from a high potential region to (c) (– 2/5) cm (d) (2/5) V/cm
low potential region, the electrical potential energy:
36. In the above question, the electric force acting on a
(a) Increases
point charge of 2 C placed at the origin will be:
(b) Decreases
(a) 2 N (b) 500 N
(c) May increase or decrease.
(c) – 5 N (d) – 500 N
(d) Remains constant
31. A flat circular fixed disc has a charge + Q uniformly 37. The electric potential V as a function of distance x (in
distributed on the disc. A charge + q is thrown with metre) is given by
kinetic energy K, towards the disc along its axis. The V = (5x2 + 10x – 9) volt.
charge q: The value of electric field at x = 1 m would be:
(a) may hit the disc at the centre (a) – 20 volt/m (b) 6 volt/m
(b) may return back along its path after touching the (c) 11 volt/m (d) – 23 volt/m
disc 38. A uniform electric field having a magnitude E0 and
(c) may return back along its path without touching direction along positive x-axis exists. If the electric
the disc potential V is zero at x = 0, then its value at x = + x will be:
(d) any of the above three situations is possible (a) Vx = xE0 (b) Vx = – xE0
depending on the magnitude of K (c) Vx = x2E0 (d) Vx = – x2E0
32. When the separation between two charges is decreased, 39. Let E be the electric field and V, the electric potential at
the electric potential energy of the charges a point.
(a) increases (a) If E ≠ 0, V cannot be zero
(b) decreases (b) If E = 0, V must be zero
(c) may increase or decrease (c) If V = 0, E must be zero
(d) remains the same (d) None of these
Physics
40. The electric field in a region is directed outward and is Y
proportional to the distance r from the origin. Taking
Q
the electric potential at the origin to be zero, the electric
potential at a distance r: Equatorial
(a) increases as one goes away from the origin. –q +q P
X

(b) is proportional to r2
(a) at P only (b) at Q only
(c) is proportional to r
(c) both at P and at Q (d) neither at P nor at Q
(d) is uniform in the region
41. A particle A has charge + q and particle B has charge 47. An electric dipole of dipole moment pv is placed at the
+ 4q with each of them having the same mass m. origin along the x-axis. The angle made by electric field
When allowed to fall from rest through same electrical with x-axis at a point P, whose position vector makes
1
potential difference, the ratio of their momenta will be: an angle q with x-axis, is: (where, tan α = 2 tan i )
(a) 2 : 1 (b) 1 : 2 (a) α (b) q
(c) 4 : 1 (d) 1 : 4 (c) q + α (d) q + 2α

42. There is a uniform electric field in X-direction. If the 48. An electric dipole consists of two opposite charges
work done by external agent in moving a charge of 0.2 each of magnitude 1.0 µC, separated by a distance of
C through a distance of 2 metre slowly along the line 2.0 cm. The dipole is placed in an external electric field
making an angle of 60º with X-direction is 4 joule, then of 1.0 × 105 N/C. The maximum torque on the dipole is:
the magnitude of E is: (a) 0.2 × 10–3 N-m (b) 1.0 × 10–3 N-m
–3
(c) 2.0 × 10 N-m (d) 4.0 × 10–3 N-m
(a) 3 N/C (b) 4 N/C
(c) 5 N/C (d) 20 N/C 49. A dipole of electric dipole moment P is placed in a
uniform electric field of strength E. If q is the angle
43. In an electron gun, electrons are accelerated through a between positive directions of P and E, then the
potential difference of V volt. Taking electronic charge potential energy of the electric dipole is largest when q is:
and mass to be respectively e and m, the maximum (a) zero (b) p/2
velocity attained by them is: (c) p (d) p/4
2eV 2eV 50. Two opposite and equal charges of magnitude
(a) m (b) m 4 × 10–8 coulomb each when placed 2 × 10–2 cm apart
(c) 2m/eV (d) (V 2/2em) form a dipole. If this dipole is placed in an external
44. In a cathode ray tube, if V is the potential difference electric field of 4 × 108 N/C, the value of maximum
between the cathode and anode, the speed of the torque and the work required in rotating it through
electrons, when they reach the anode is proportional 180º from its initial orientation which is along electric
to: (Assume initial velocity = 0) field will be: (Assume rotation of dipole about an axis
(a) V (b) 1/V passing through centre of the dipole):
(c) V (d) (V 2/2em) (a) 64 × 10–4 N-m and 44 × 10–4 J
(b) 32 × 10–4 N-m and 32 × 10–4 J
45. Three charges Q , + q and + q are placed at the vertices (c) 64 × 10–4 N-m and 32 × 10–4 J
of a right-angled isosceles triangle as shown. The net (d) 32 × 10–4 N-m and 64 × 10–4 J
electrostatic energy of the configuration is zero if Q is
51. At a point on the axis (but not inside the dipole and not
equal to: at infinity) of an electric dipole
Q
(a) The electric field is zero
(b) The electric potential is zero
(c) Neither the electric field nor the electric potential
is zero
(d) The electric field is directed perpendicular to the
+q +q
axis of the dipole
a
-q - 2q 52. An electric dipole is placed at the centre of a sphere.
(a) (b) Mark the correct options.
1+ 2 2+ 2
(a) The electric field is zero at every point of the sphere.
(c) – 2q (d) + q
(b) The flux of the electric field through the sphere is
Dipoles non-zero.
46. Due to an electric dipole shown in fig., the electric field (c) The electric field is zero on a circle on the sphere.
intensity is parallel to dipole axis: (d) The electic field is not zero anywhere on the sphere.
 Electrostatics
Self energy, Energy density 60. An uncharged sphere of metal is placed in a uniform
electric field produced by two large conducting parallel
53. A uniformly charged sphere of radius 1 cm has plates having equal and opposite charges, then lines of
potential of 8000 V at surface. The energy density near force look like:
the surface of sphere will be: + + + ++ + + + + + + +
(a) 64 × 105 J/m3 (b) 8 × 103 J/m3
(c) 32 J/m3 (d) 2.83 J/m3 (a) (b)

54. If ‘n’ identical water drops (assumed spherical each)

each charged to a potential energy U coalesce to form – – – – – – – – – – – –
a single drop, the potential energy of the single drop + + + + + + + + + +
is(Assume that drops are uniformly charged):
(a) n1/3 U (b) n2/3 U (c) (d)
(c) n4/3 U (d) n5/3 U
– – – – – – – – – –
Conductors 61. You are travelling in a car during a thunder storm. In
E
55. A metallic solid sphere is order to protect yourself from lightening, would you
placed in a uniform electric A
prefer to:
field. The lines of force
B
C (a) Remain in the car
follow the path(s) shown D (b) Take shelter under a tree
in figure as:
(c) Get out and be flat on the ground
(a) A (b) B (d) Touch the nearest electrical pole
(c) C (d) D
62. A positively charged body ‘A’ has been brought near
56. A neutral spherical metallic object A is placed near a neutral brass sphere B mounted on a glass stand as
a finite metal plate B carrying a positive charge. The
shown in the figure. The potential of B will be:
electric force on the object will be:
(a) away from the plate B (b) towards the plate B
(c) parallel to the plate B (d) zero ×
×
B
×

57. A positive point charge q is brought near a neutral ×

×

×
×

×
×

metal sphere.
×
×

×
×

(a) The sphere becomes negatively charged.

×

A
×

(b) The sphere becomes positively charged.

(a) Zero (b) Negative
(c) The interior remains neutral and the surface gets
non-uniform charge distribution. (c) Positive (d) Infinite
(d) The interior becomes positively charged and the 63. The amount of work done by electric field in joules in
surface becomes negatively charged. carrying a charge + q along the closed path PQRSP
58. Three concentric conducting spherical shells carry between the oppositely charged metal plates is: (where,
charges as follows: + 4Q on the inner shell, – 2Q on the E is electric field between the plates)
middle shell and – 5Q on the outer shell. The charge on –
+
the inner surface of the outer shell is:
S R
(a) 0 (b) 4Q + –

(c) – Q (d) – 2Q + –

59. Figure shows a thick metallic sphere. If it is given a + –

charge + Q, then electric field will be present in the + –

region +P
Q–
+Q

(a) zero (b) q

R1 R2 (c) qE (PQ + QR + SR + SP) (d) q/e0
64. Figure shows a closed surface which intersects a
conducting sphere. If a positive charge is placed at the
(a) r < R1 only (b) r > R1 and R1 < r < R2 point P, the flux of the electric field through the closed
(c) r ≥ R2 only (d) r ≤ R2 only surface:
Physics
65. A solid metallic sphere has a charge +3Q. Concentric
with this sphere is a conducting spherical shell having
•P charge – Q. The radius of the sphere is a and that of the
spherical shell is b(>a). What is the electric field at a
closed
distance r(a < r < b) from the centre?
conducting
surface sphere
(a) will become positive 1 2Q 1 5Q
(a) 4rf (b) 4rf
0 r 0 r
2 2
(b) will remain zero
(c) will become undefined 1 3Q 1 Q
(c) 4rf (d) 4rf 2
0 r 0 r
2
(d) will become negative
EXERCISE - 2

Charges and Coulomb’s Law M

1. Five balls, numbered 1 to 5, are suspended using A B

separate threads. Pairs (1, 2), (2, 4), (4, 1) show
electrostatic attraction, while pairs (2, 3) and (4, 5)
show repulsion. Therefore ball 1: (a) M attracts A (b) A repels B
(a) Must be positively charged
(b) Must be negatively charged (c) M attracts B (d) B attracts A
(c) May be neutral 6. Three charges – q1, + q2 and – q3 are placed as shown
(d) Must be made of metal in the figure. The x-component of the force on – q1 is
proportional to:
2. Two point charges of same +Q –Q
y
–q3
magnitude and opposite sign are A B
fixed at points A and B. A third small point charge is to
be balanced at point P by the electrostatic force due to
a 
these two charges. The point P: b
(a) lies on the perpendicular bisector of line AB x
(b) is at the mid point of line AB –q1 +q2
q2 q3 q2 q3
(c) lies to the left of A (a) - cos i (b) + sin i
b2 a2 b2 a2
(d) none of these. q q
q q
3. A total charge of 20 µC is divided into two parts and (c) 22 + 32 cos i (d) 22 - 32 sin i
b a b a
placed at some distance apart. If the charges experience
7. Mid way between the two equal and similar charges,
maximum coulombian repulsion, the charges should be:
we placed the third equal and similar charge. Which
(a) 5 mC, 15 mC (b) 10 mC, 10 mC of the following statements is correct, concerned to the
40 20 equilibrium along the line joining the charges ?
(c) 12 mC, 8 mC (d)3 nC, 3 nC
(a) The third charge experienced a net force inclined
4. The magnitude of electric force on 2 µC charge
to the line joining the charges
placed at the centre O of two equilateral triangles
each of side 10 cm, as shown in figure is P. If charge (b) The third charge is in stable equilibrium
A, B, C, D, E and F are 2 µC, 2 µC, 2 µC, – 2 µC, (c) The third charge is in unstable equilibrium
– 2 µC, – 2 µC respectively, then P is: (d) The third charge experiences a net force
F perpendicular to the line joining the charges

E A Electric Field
O 8. A simple pendulum has a length l
D B and mass of bob m. The bob is
given a charge q coulomb. The
C pendulum is suspended in a l

(a) 21.6 N (b) 64.8 N uniform horizontal electric field

(c) 0 (d) 43.2 N of strength E as shown in figure,
then calculate the time period of
5. A large nonconducting sheet M is given a uniform oscillation when the bob is
charge density. Two uncharged small metal spheres A slightly displaced from its mean
and B are placed near the sheet as shown in figure. position.
 Electrostatics
l 12. In the above question, if Q′ is removed then which
(a) 2r
l
g (b) 2r qE 4 * option is correct:
g+ m 
l
(c) 2r * qE 4 l
(d) 2r E
g- m
g +c
2 qE m2 (a) (b)
m
9. The linear charge density on upper half of a segment
r r
of ring is l and at lower half, it is – l. The direction of
R R
 
electric field at centre O of ring is: E
E
A
+
+ (c) (d)
+
+ D B
O r
– R r R
–
––
C
13. A solid sphere of radius R has a volume charge density ρ
= ρ0 r2 ( Where ρ0 is a constant and r is the distance from
(a) along OA (b) along OB
centre). At a distance x from its centre (for x < R), the
(c) along OC (d) along OD electric field is directly proportional to:
10. The particle of mass m and charge q will touch the (a) 1/x2 (b) 1/x
infinitely large plate of uniform charge density s if its (c) x3 (d) x2
velocity v is more than (sq > 0) 14. 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

v field strength at points between a and b and distance x
q, m from a. E is taken positive if it is along the line joining
d from a to b. From the graph, it can be decided that
........

2v q d
(a) 0 (b) me 0 a
x
b

vqd (a) a is positive, b is negative

(c) me 0 (d) none of these
(b) a and b both are positive
11. A charge ‘q’ is placed at the centre of a conducting (c) a and b both are negative
spherical shell of radius R, which is given a charge Q. (d) a is negative, b is positive
An external charge Q′ is also present at distance R′
15. A point charge q is placed at origin. Let Ev A, Ev B
(R′ > R) from ‘q’. Then the resultant field will be best
and EvC be the electric field at three points
represented for region r < R by:
A (1, 2, 3), B (1, 1, – 1) and C (2, 2, 2) due to charge q.
Q
Then [i] Ev A = Ev B [ii] Ev B = 4 EvC select the correct
alternative
q R Q (a) only [i] is correct (b) only [ii] is correct
R (c) both [i] and [ii] are correct
(d) both [i] and [ii] are wrong
 16. The direction (q) of Ev at point P due to uniformly
E charged finite rod will be
(a) (b) Ey y

d . 
R r + 0  P Ex x
R r 90
  +
+
E E +
(c) (d) +
300
+
+
R r R r
Physics
(a) at angle 30° from x-axis (b) 45° from x-axis mq 2mq
(c) 60° from x-axis (d) none of these (a) 2re 0 m (b) re 0 m
17. An equilateral triangle wire frame of side L having 3 mq 3mq
(c) re 0 m (d) re 0 m
point charges at its vertices is kept in x‑y plane as shown.
Component of electric field due to the configuration in 22. A charge Q is placed at a distance of 4R above the
z direction at (0, 0, L) is [origin is centroid of triangle] centre of a disc of radius R. The magnitude of flux
q
y
through the disc is f. Now a hemispherical shell of
radius R is placed over the disc such that it forms a
closed surface. The flux through the curved surface
x (taking direction of area vector along outward normal
as positive), is
q –2q Q
9 3 kq
(a) (b) zero
8L2 4R
9kq
(c) (d) None R
8L2
18. A point charge 50 mC is located in the XY plane at (a) zero (b) f
the point of position vector rv0 = 2it + 3tj . What is the (c) – f (d) 2f
electric field at the point of position vector rv0 = 8it - 5tj
23. The given figure gives electric lines of force due to
(a) 1200 V/m (b) 0.04 V/m two charges q1 and q2. What are the signs of the two
(c) 900 V/m (d) 4500 V/m charges?
Electric lines of force, Gauss’s Law
19. A square of side ‘a’ is lying in xy plane such that two q1 q2
of its sides are lying on the axis. If an electric field
Ev = E0 xkt is applied on the square. The flux passing
through the square is: (a) Both are negative
E0 a3 (b) Both are positive
(a) E0 a3 (b) 2 (c) q1 is positive but q2 is negative
E0 a3 E0 a2 (d) q1 is negative but q2 is positive
(c) 3 (d) 2
24. Three positive charges of equal value q are placed at
20. Figure (a) shows an imaginary cube of edge length L. the vertices of an equilateral triangle. The resulting
A uniformly charged rod of length 2L moves towards lines of force should be sketched as in:
left at a small but constant speed v. At t = 0, the left
end of the rod just touches the centre of the face of the
cube opposite to it. Which of the graphs shown in fig. (a) (b)
(b) represents the flux of the electric field through the
cube as the rod goes through it?
Flux d
b a
A. v
B. c
2L
(c) (d)
L time
(a) a (b) b
(c) c (d) d
25. The volume charge density as a function of distance X
21. Figure shows two large cylindrical from one face inside a unit cube is varying as shown in
shells having uniform linear charge – + the figure. Then the total flux (in S.I. units) through the
densities + λ and – λ. Radius of r cube if (ρ0 = 8.85 × 10−12 C/m3) is:
inner cylinder is ‘a’ and that of
density
outer cylinder is ‘b’. A charged a
v

particle of mass m, charge q b 0

revolves in a circle of radius r.
Then, its speed ‘v’ is: (Neglect gravity and assume the
radii of both the cylinders to be very small in
comparison to their length.) (in m)
 Electrostatics
(a) 1/4 (b) 1/2 31. For an infinite line of charge having charge density λ
(c) 3/4 (d) 1 lying along x-axis, the work required in moving charge
q from C to A along arc CA is:
26. A charge q is placed at the centre of the cubical vessel
B
(with one face open) as shown in figure. The flux of the
C

electric field through the surface of the vessel is a

A
d • d a

++++++++++++ X
qm qm
d
(a) zero (b) q/ε0 (a) rf log e 2 (b) 4rf log e 2
q 0 0
(c) 4f (d) 5q/6ε0 qm qm 1
0 (c) 4rf log e 2 (d) 2rf log e 2
0 0

Electric potential, potential energy 32. Two isolated metallic solid spheres of radii R and 2R
27. Two equal positive charges are kept at points A and B. are charged such that both of these have same charge
The electric potential, while moving from A to B along density σ. The spheres are located far away from each
straight line : other, and connected by a thin conducting wire. Then
(a) continuously increases the new charge density on the bigger sphere is.
(b) remains constant 5v 5v
(a) 6 (b) 3
(c) decreases then increases
7v 7v
(d) increases then decreases (c) 6 (d) 3
28. A charge + q is fixed at each of the points x = x0,
33. An infinite long plate has surface charge density σ. As
x = 3x0, x = 5x0, ...... upto infinity on the x-axis and shown in the fig. a point charge q is moved from A to
a charge – q is fixed at each of the points x = 2x0, B. Net work done by electric field is:
x = 4x0, x = 6x0, ..... upto infinity. Here x0 is a positive
.......

constant. Take the electric potential at a point due to a

Q
charge Q at a distance r from it to be 4re r . Then
0
the potential at the origin due to the above system of B(x1,0) A(x2,0)
charges is:
q
(a) 0 (b) 8re x ,n2
.......

0 0

(a) 2e ^ x1 - x2h (b) 2e ^ x2 - x1h

q ,n2 vq vq
(c) ∞ (d) 4re x
0 0 0 0

(c) e ^ x2 - x1h (d) e ^ x1 + x2h

vq vq
29. A non-conducting ring of radius 0.5 m carries a total
charge of 1.11 × 10−10 C distributed non-uniformly 0 0

on its circumference producing an electric field Ev 34. Figure shows an electric line of force which curves
every where in space . The value of the line integral along a circular arc. The magnitude of electric field
l=0
intensity is same at all points on this curve and is equal
# - Ev $ dl (l = 0 being centre of the ring) in volts is: to E. If the potential at A is V, then the potential at B is:
l=3

(Approximately) E
A
(a) + 2 (b) − 1
B
R
(c) − 2 (d) zero 
30. A charged particle ‘q’ is shot from a large distance i
with speed v towards a fixed charged particle Q. It (a) V – ERθ (b) V – 2ER sin 2
approaches Q upto a closest distance r and then returns. (c) V + ERθ (d) V + 2ER sin 2
i
If q were given a speed ‘2v’, the closest distance of
approach would be: 35. A charged particle of charge ‘Q’ is held fixed and
q Q
another charged particle of mass ‘m’ and charge ‘q’
v r (of the same sign) is released from a distance ‘r’. The
(a) r (b) 2r impulse of the force exerted by the external agent on
r r the fixed charge by the time distance between ‘Q’ and
(c) 2 (d) 4
‘q’ becomes 2r is:
Physics
(a) Qq/4re 0 mr (b) Qqm/4rf 0 r Dipoles
(c) 0 (d) cannot be determined 41. The force between two short electric dipoles separated
by a distance r is directly proportional to:
36. Three charges +q, – q, and + 2q are placed at the vertices
of a right angled triangle (isosceles triangle) as shown. (a) r2 (b) r4
The net electrostatic energy of the configuration is: (c) r– 2 (d) r– 4
+2q
42. Two short electric dipoles are placed as shown (r is the
distance between their centres). The energy of electric
interaction between these dipoles will be:
a P1

+q a -q

(a) - a ^ 2 + 1 h a ^ 2 + 1h
Kq 2
Kq2
(b) P2
C
(c) - a ^ 2 - 1 h
Kq2
(d) None of these
(C is centre of dipole of moment P2)
37. A uniform electric field of 400 V/m exists in space as 2k P1 P2 cos i - 2k P1 P2 cos i
(a) (b)
shown in graph. Two points A and B are also shown r3 r3
with their co-ordinates. The potential difference - 2k P1 P2 sin i - 4k P1 P2 cos i
(c) (d)
VB – VA in volts is: r3 r3
y 43. Three dipoles each of dipole
P
E=400 V/m moment of magnitude p are placed
tangentially on a circle of radius R
B (0,3cm)
in its plane positioned at equal 1200 1200
x
angle from each other as shown in
A (-4cm, 0)
16° the figure. The magnitude of P 1200 P
electric field intensity at the centre
of the circle will be:
4kp 2kp
(a) – 12 V (b) 15 V (a) (b)
R 3
R3
(c) 12 V (d) 8 V kp
(c) (d) 0
38. A uniform electric field pointing in positive x-direction R3
exists in a region. Let A be the origin, B be the point on 44. Three electric point charges q, q and – 2q are placed at
the x-axis at x = + 1 cm and C be the point on the y-axis the three corners of an equilateral triangle of side l. The
at y = + 1 cm. Then the potentials at the points A, B and magnitude of electric dipole moment of the system of
C satisfy: three particles is:
(a) VA < VB (b) VA > VB (a) ql (b) 2ql
(c) VA < VC (d) VA > VC (c) 3 ql (d) 4ql
39. Two equal point charges are fixed at x = – a and 45. In which of the following cases, force acting on the
x = + a on the x-axis. Another point charge Q is placed dipole could be zero
at the origin. The change in the electrical potential
 
energy of Q when it is displaced by a small distance x 
along the x-axis, is approximately proportional to: +q 
(a) –q

(b)
(a) x (b) x2 
+q
 –q
(c) x3 (d) 1/x 

40. Potential difference between centre and the surface of


sphere of radius R and uniform volume charge density 
r within it will be: –q
(c)
+q

tR2 tR2
(a) 6e (b) 4e
0 0 

tR2
(c) 0 (d) 2e
0 (d) None
 Electrostatics
46. An electric dipole consists of charges ± 2.0 × 108 C (a) zero (b) 25 V/m
separated by a distance of 2.0 × 10–3 m. It is placed (c) 50 V/m (d) 100 V/m
near a long line charge of linear charge density
4.0 × 104 C/m as shown in figure, such that the negative 51. Two small conductors A and B are Q
given charges q1 and q2 respectively.

/
/\
/\
charge is at a distance of 2.0 cm from the line charge.

/\
/\
/\
Now they are placed inside a hollow A

/\
\
Find the force acting on the dipole /\
metallic conductor (c) carrying a
\
/
q1 \/\/ /\ C
\
/\/\/\
\

charge Q. If all the three conductors B

A, B and C are connected by
2.0 cm q2

– + conducting wires as shown, the

charges on A, B and C will be respectively:
(a) 0.6 N towards the line charge q1 + q2 q1 + q2
(b) 0.1 N away from the line charge
(a) 2 , 2 ,Q
(c) 0.9 N to wards the line charge Q + q1 + q3 Q + q1 + q2 Q + q1 + q2
(b) 3 , 3 , 3
(d) 1.5 N away from the line charge
q1 + q2 + Q q1 + q2 + Q
47. A system of two electric dipoles, each of dipole moment (c) 2 , 2 ,0
having magnitude P are arranged in the configuration (d) 0, 0, Q + q1 + q2
shown in figure. The electrostatic interaction energy of
52. An ellipsoidal cavity is carved
this system of two dipoles is:
y within a perfect conductor. A A
positive charge q is placed at B
the center of the cavity . The
q

x points A and B are on the cavity

(–a, 0) O (a, 0) surface as shown in the figure.
Then:
(a) Electric field near A in the cavity = electric field
near B in the cavity
5 kp 7kp2
2
(a) 4 3 (b) (b) Charge density at A = Charge density at B
a a3
(c) Potential at A = Potential at B
kp2
(c) (d) None of these (d) Total electric field flux through the surface of the
4a3
cavity is q/ε0.
Self energy, Energy density E
53. A positive point charge Q is kept (as
48. Two uniformly charged non-conducting hemispherical shown in the figure) inside a neutral
shells each having uniform charge density σ and radius
conducting shell whose centre is at C. An C
R form a complete sphere (not stuck together) and
external uniform electric field E is applied. Q
surround a concentric spherical conducting shell of
radius R/2. If hemispherical parts are in equilibrium Then:
then minimum surface charge density of inner (a) Force on Q due to E is zero
conducting shell is: (b) Net force on Q is zero
(a) – 2σ (b) − σ/2 (c) Net force acting on Q and conducting shell
(c) − σ (d) 2σ considered as a system is zero
49. Assume a piece of flat cloud has area 25 × 106 m2 and (d) Net force acting on the shell due to E is zero.
electric potential 105 V. If the height of cloud is 0.75 54. A charge Q is kept at the centre of a conducting sphere
km, then energy stored by electric field between earth of inner radius R1 and outer radius R2. A point charge
and cloud will be: (potential of Earth = 0) q is kept at a distance r (> R2) from the centre. If q
(a) 250 J (b) 750 J experiences an electrostatic force 10 N then assuming
(c) 1225 J (d) 1475 J that no other charges are present, electrostatic force
Conductors experienced by Q will be:
50. A charge q is uniformly distributed over a large plastic (a) – 10 N (b) 0
plate. The electric field at a point P close to the centre (c) 20 N (d) none of these
and just above the surface of the plate is 50 V/m. If the 55. A dipole having dipole moment p is placed in front
plastic plate is replaced by a copper plate of the same of a solid uncharged conducting sphere as shown in
geometrical dimensions and carrying the same uniform the diagram. The net potential at point A lying on the
charge q, the electric field at the point P will become: surface of the sphere is:
Physics
A
kq2
R
r (a) 0 (b) 2b

kq2 kb2 kq2 kq2
p
(c) 2b - 2a (d) 2a - 2b
59. A thin spherical conducting shell of radius R has a
kp cos z kp cos2 z charge q. Another charge Q is placed at the centre of
(a) (b)
r2 r2 the shell. The electrostatic potential at a point P at a
2kp cos2 z
(c) zero (d) distance R/2 from the centre of the shell is:
r2
2Q 2Q 2q
56. A thin, metallic spherical shell Q (a) 4rf R (b) 4rf R - 4rf R
0 0 0
contains a charge Q on it. A
2Q q ^q + Q h
point charge – q is placed at the (c) 4rf R + 4rf R (d)
0 0 4rf 0 R
centre of the shell and another –q q
charge q′ is placed outside it as 60. The figure shows a charge q placed inside a cavity in an
shown in fig. The force on the uncharged conductor. Now if an external electric field
charge at the centre is is switched on
(a) zero (b) towards left
(c) towards right (d) upward c
57. Consider the situation of the previous problem. The q
force on the central charge due to the shell is
(a) upward (b) towards left
(c) towards right (d) zero (a) only induced charge on outer surface will
58. A point charge q is brought from redistribute
infinity (slowly so that heat developed b (b) only induced charge on inner surface will
in the shell is negligible) and is placed q a redistribute
at the centre of a conducting neutral (c) both induced charge on outer and inner surface
spherical shell of inner radius a and will redistribute
outer radius b, then work done by (d) force on charge q placed inside the cavity will
external agent is: change

EXERCISE - 3

Numerical Type 3. A non-conducting disc of mass = 2 kg, total charge

= +1C uniformly distributed, is placed on a rough
1. Linear charge densities of the two rods are given as horizontal non-conducting surface with its cross-
- m0 x + m0 y section in vertical plane as shown. A uniform horizontal
m1 = and m2 = . If the dipole moment
l 2
l2 electric field E is now switched on. Find the maximum
, then find ]m + ng .
m m0 l magnitude of electric field E in (N/C) so that the disc
of the system of rods is n
y rolls purely. (g = 10 m s-2)
E
(0, l) +1C

2

1  =0.2
x 4. A square loop of side l having y
(0, 0) (l, 0)
uniform linear charge density
2. A particle of charge q and mass m moves rectilinearly l is placed in an x-y plane as B C
under the action of an electric field E = A - Bx, where shown in the figure. There is a l
A D
B is positive constant and x is a distance from the point non-uniform electric field x
where the particle was initially at rest. The distance v a t l l
E = (x + l) i , where a and l
aA l
travelled by the particle till it comes to rest is .
bB are constants. The resultant electric force on the loop
Find (a + b). is having the value 2nall. Find the value of n.
 Electrostatics
5. A problem of practical interest is 10. A neutral spherical conductor having cavity A and
to make a beam of electrons B of radius ‘1cm’ and ‘2cm’ respectively. Radius of
turn through 90°. This can be spherical conductor is 10 cm. If cavity ‘A’ contains
done with the parallel plates charge 2mC and ‘B’ contains charge of -1mC. Electric
shown in the figure. An electron potential at the centre of conductor is k × 104 volt. Then
what is the value of k.
with kinetic energy 8.0 x 10-17
J enters through a small hole in Electron
the bottom plate. The strength +2C
1cm
of electric field that is needed if the electron is to 5cm
emerge from an exit hole 1.0 cm away from the 10cm
entrance hole, traveling at right angle to its original -1C 5cm
direction is.... × 105 N/C. 2cm

6. A non-conducting infinite

rod is placed along the 
c-axis, the upper half of the 11. Two metal balls of radii R1 and R2 such that R2 = 2R1 is
rod (lying along z $ 0) is connected to a long thin conducting wire and has total
charged positively with a 0 charge Q = 21mC. Then the ball of radius R1 is placed
uniform linear charge R inside a grounded metal sphere of radius R = 3R1 as
density +l. while the lower shown. What charge (in mC) will flow through the wire
2R connecting R1 and R2 in this process ?
half (z < 0) is charged
negatively with uniform 
R
linear charge density -l. R2
The origin is located at the junction of the positive and
negative halves of the rod. A uniformly charged annular R1
disc (surface charge density: s0) of inner radius R and
outer radius 2R is placed in the x-y plane with its centre
at the origin. The force on the rod due to the disc is
yv 0 mR 12. An uniformly charged non
8f 0 . Find the value of y. conducting soild sphere has a radius
7. An electric field Ev = E0 it + E0 tj exists in space. The R. Total energy of system (inside + R 4R

a outside) is U1. If this sphere is now

flux through a triangular loop with vertices at d , 0, 0 n, covered by a conducting spherical
2
a a a E0 a 2
d , 0, a n and d , a, n is . Find b. layer of thickness 4R, the total energy of system
2 2 2 b U
reduces to U2. Find value of U1 .
8. An arrangement of source charges produces electric 2

potential V = 5000 x2 along the x-axis, where V is in 13. Two charges 3 C and q are placed at (2, 0) and
volt and x is in metre. If a charge particle of mass 1 g (0, 2). The direction of intensity of electric field at (2,
and charge 1 nC is present in this field and its turning 2) makes an angle 30º with y-axis then q is _____
points are at ! 8.0 cm, then what is the particle’s
maximum speed (in mm/s). qC
9. Two insulating plates, shown in the figure, are the
uniformly charged in such a way that the potential C
difference between them is V2 – V1 = 20 V (i.e., plate
2 is at a higher potential). The plates are separated by 14. Find th e potential difference VAB (inVolt) between A (2 m,
distance d = 0.1 m and can be treated as infinitely large. 1m, 0m) and B(0m, 2m, 4m) in an electric field,
An electron is released from rest on the inner surface of Ev = _ xit - 2yjt + zkt i V/m .
plate 1. Its speed when it hits plate 2 is 2.65 × 10n m/s.
Find n. (q = 1.6 × 10-19 C, m0 = 9.11 10-31 kg) 15. The electric flux for Gaussian surface q4

that enclose the charge particles in

Y q2
free region is in
q1
0.1 m 3 2 –1
__ × 10 Nm C (given q1 = – 14
X
nC, q2 = 78.85 nC, q3 = – 56 nC and q3

1 2 q4 = 12.23 nC)
Physics
16. A dipole consists of two particles one with charge 18. An infinite, uniformly charged sheet with 
+ 1 mC and mass 1 kg and the other with charge surface charge densities s cuts through a
– 1 mC and mass 3 kg separated by a distance of 3 m. spherical Gaussian surface of radius R at a R
x
For small oscillations about its equilibrium position, distance x from its center, as shown in the
the angular frequency, when placed in a uniform figure. The electric flux F through the
kr ^ R2 - x2h v
electric field of 30 V/m is _____ 10–1 rad/s. Gaussian surface is f0 find
the value of k ?
17. Two infinite line charges, each having a uniform charge
density l, pass through the midpoints of two pairs of 19. A dipole of dipole moment Pv = 2it - 3tj + 4kt is placed
opposite faces of a cube of edge L as shown in figure. at point A(2, – 3, 1). The electric potential due to this
The modulus of the total electric flux due to both the dipole at the point B (4, – 1, 0) is equal to (All the
line charges through the face is ABCD is lL/(k∈0) find parameters specified here are in S.I. units) – ______ ×
the value of k? 109 volts.
H 20. An electric dipole is kept on the axis of a uniformly
G
D
charged ring at distance R/ 2 from the centre of the
C ring. The direction of the dipole moment is along the
E axis. The dipole moment is P, charge of the ring is Q
F and radius of the ring is R. The force on the dipole is
A B nearly ____.

EXERCISE - 4

Single Option Correct 1. | Q1 | > | Q2 |

1. Figure shows a system of three 2. Q1 is positive in nature
concentric metal shells A, B and C 3. A and B are equilibrium points.
with radii a, 2a and 3a respectively.
Shell B is earthed and shell C is 3a a 4. C is a point of unstable equilibrium for positive
given a charge Q. Now if shell C is charge for displacement along the line joining Q1
A
2a B
connected to shell A, then the final C and Q2. Then which of the above statements are
charge on the shell B is equal to: correct.
4Q 8Q
(a) – 13 (b) – 11 (a) 1 and 3 (b) 1, 2 and 3
5Q 3Q (c) 1, 2 and 4 (d) 1, 2, 3 and 4
(c) – 3 (d) – 7
4. An oil drop of mass M is found floating freely between
2. AB and CD are two large non parallel metallic plates the plates of a parallel plate condenser (which consists
connected to a battery as shown. If σA and σB be charge of two large metallic plates having equal and opposite
densities at points A and B respectively then:
A
charges), the plates being horizontal and the lower
B
plate carrying a charge + Q. The area of each plate is A
and the distance of separation between them is D. The
charge on the oil drop must be (g is the acceleration
V
D
C due to gravity)
(a) σA > σB (b) σA < σB Aq ! 0 MgA
(a) QM (b) Q
(c) σA > σB (d) σA < σB
3. The curve represents the distribution of potential (V) - gAQ - MgAQ
(c) D (d)
along the straight line joining the two charges Q1 and e0
Q2 (separated by a distance r): 5. Two smooth spherical non conducting shells each of
V
radius R having uniformly distributed charge Q and
− Q on their surfaces are released on a smooth non-
conducting surface when the distance between their
centres is 10 R. The mass of A is m and that of B is 2 m.
x
Q1 A Q2 B C

The speed of A just before A and B collide is: [Neglect

gravitational interaction]
r
 Electrostatics
Q –Q 10. Each of the two long parallel threads carries a uniform
m 2m charge l per unit length. The threads are separated by a
distance l. Find the maximum magnitude of the electric
A B

/////////////////////////////////////////////////////////////// field strength in the symmetry plane of this system

10R located between the threads.
2kQ 2
4kQ2 (a) Emax = l/2pe0l (b) Emax = l/3pe0l
(a) 15mR (b) 15mR (c) Emax = 2l/pe0l (d) Emax = l/pe0l
8kQ2 16kQ2 11. The charge per unit length of the four
(c) 15mR (d) 15mR quadrant of the ring is 2l, – 2l, l and – l
6. On a semicircular ring of radius = 4R, charge respectively. The electric field at the
+ 3q is distributed in such a way that on one quarter centre is
+ q is uniformly distributed and on another quarter m m
+ 2q is uniformly distributed. Along its axis a smooth (a) – 2rf R it (b) 2rf R tj
0 0
nonconducting and uncharged pipe of length 6 R is fixed 2m
axially as shown. A small ball of mass m and charge + q (c) 4rf R it (d) None
0
is thrown from one end of pipe with speed u. The speed
‘u’ so that ball will come out from the other end of pipe is: 12. If the electric potential of the inner metal
sphere is 10 volt and that of the outer shell a
q
+ ++ is 5 volt, then the potential at the centre b

+
+ +
+ will be:
+q u + (a) 10 volt (b) 5 volt
+
+ 4R
(c) 15 volt (d) 0
13. A neutral conducting spherical shell is kept near a
2q
6R
charge q as shown. The potential at point P due to the
3q2 7q 2
induced charges is c K = 4rf m
(a) u > (b) u > 1
40re 0 Rm 40re 0 Rm 0

3q2 9q 2
(c) u ≥ 40re 0 Rm (d) u > 40re 0 Rm r c
q
7. A particle of mass m and charge Q is placed in an p
x
electric field E which varies with time t as E = E0 sin wt.
It will undergo simple harmonic motion of amplitude kq - kq
QE02 QE0 (a) r (b) r
(a) (b) kq kq kq kq
m~2 m~2 (c) r - x (d) x - r
QE0 QE
(c) (d) m~0 14. Find the potential function V (x, y) of an electrostatic
m~ 2

field Ev = 2axy it + a _ x2 - 3y2 itj where a is a constant.

8. Charge density of the given surface is
s. Then electric field strength at the
R ay3 ay2
(a) V0 + ax2 y - 3 (b) V0 - axy2 - 3
centre (of quarter sphere) is C
ay3
v v (c) V0 + axy2 + 3 (d) V0 - ax2 y + ay3
(a) (b)
4 2 f0 2 2 f0 15. Find the potential j of an uncharged conducting sphere
(c)
v v
(d) 2f outside of which a point charge q is located at a distance
2 f0 0 l from the sphere’s centre.
q q
9. A system consists of a thin charged wire ring of radius (a) { = 4rf l (b) { = rf l
0 0
R and a very long uniformly charged thread oriented q q
along the axis of the ring,with one of its ends coinciding (c) { = 3rf l (d) { = 5rf l
0 0
with the centre of the ring. The total charge of the ring 16. A conducting sphere of radius a has charge Q on it. It is
is equal to q. The charge of the thread (per unit length) enclosed by a neutral conducting concentric spherical
is equal to l. Find the interaction force between the shell having inner radius 2a and outer radius 3a. Find
ring and the thread. self energy of outer shell.
mq 5 kQ
2
11 kQ
2
(a) Zero (b) 4re R (a) 12 a (b) 12 a
0
mq mq kQ2
(c) 2re R (d) re R (c) 2a (d) none
0 0
Physics
17. In uniform electric filed E = 10 N/C as shown in 22. A ring of radius R carries a non-
figure, find (a) VA – VB (b) VB – VC uniform charge of linear density l =
l0 cos q (see in the figure)Magnitude
of the net dipole moment of the ring
is:
(a) pR2l0 (b) 2pR2l0
rR2
(a) 15 volt, 20 volt (b) 10 volt, 30 volt (c) 2 m0 (d) 4pR2l0
(c) – 10 volt, 20 volt (d) 15 volt, 25 volt 23. A charge ‘q’ is placed in a mug as shown in the
18. A charged particle having some mass is resting figure. The radius of the circular face is ‘r’ and
in equilibrium at a height H above the centre of a distance between the centre of the circular face q
uniformly charged non‑conducting horizontal ring of and charge is ‘r’. Flux through the circular face
radius R. The force of gravity acts downwards. The is
equilibrium of the particle will be stable
(a) f ^1 - 2 h
q q
(b) 2f d1 + n
1
R 2
(a) for all values of H (b) only if H > 0 0
2
q q
(d) 2f d1 - n
1
R R (c) f
(c) only if H < (d) only if H = 0 2
2 2 0

19. A solid metal sphere of radius R has a 24. A charged small metal sphere A hangs from a
charge + 2Q. A hollow spherical shell +2Q silk thread S, which makes an angle q with a
of radius 3R placed concentric with R large charged nonconducting sheet having
the first sphere has net charge – Q. The 3R uniform surface charge density s is

potential difference between the –Q proportional to
S
spheres is (a) sin q (b) cos q
Q Q A
(a) rf R (b) 3rf R (c) tan q (d) cot q
0 0
Q Q
(c) 2rf R (d) 7rf R 25. A charge is placed at centre of circular face of cylinder
0 0
of radius ‘r’ and length ‘r’ flux through remaining
20. Consider the dipole pv kept in a space of electric field
curved surface is {other than opposite circular face}
a shown. The dipole will move
q q q
(a) 2f - 2f <1 - F (b) <1 - F
1 1
0 0 2 2f 0 2



q q 1 q
(c) 2f - f < F (d) f <1 - F


1
p
0 0 2 0 2

 26. Three identical metal plates with large Q –2Q
surface areas are kept parallel to each


other as shown in figure. The leftmost

(a) upwards (b) downwards plate is given a charge Q, the
(c) towards right (d) towards left rightmost a charge – 2Q and the
21. An infinite nonconducting sheet at middle one remains neutral. Find the
x = 0 carries a uniform surface charge appearing on the outer surface

charge density s. A thin rod of O  of the rightmost plate.
x
length 2L has a uniform linear – C Q Q
(a) - 2 (b) + 2
charge density l on one half and – l 
Q Q
on the other half. The rod is hinged (c) 2f (d) f
0 0
at its mid point C at x = 2L and lies in the XOZ plane
making an angle a with the positive X-axis as shown in 27. Plates A and B constitute an isolated, charge parallel-
figure. The torque xv experienced by the rod is plate capacitor. The inner surfaces (I and IV) of A and
4vmL2 t B have charges + Q and – Q respectively. A third plate
ke 0 j sin a . The value of k is C with charge + Q is now introduced midway between
(a) 1 (b) 2 A and B. Which of the following statements is not
(c) 3 (d) 8 correct?
 Electrostatics
(d) The torque on the dipole due to the field must be
zero
33. Figure shows a charge Q
placed at the centre of open
face of a cylinder as shown
in figure. A second charge
(a) The surfaces I and II will have equal and opposite q is placed at one of the
charges. positions A, B, C and D,
(b) The surfaces III and IV will have equal and out of which positions A and D are lying on a straight
opposite charges. line parallel to open face of cylinder. In which
(c) The charge on surface III will be greater than Q. position(s) of this second charge, the flux of the electric
field through the cylinder remains unchanged ?
(d) The potential difference between A and C will be (a) A (b) B
equal to the potential difference between C and B. (c) C (d) D
Multiple Options Correct 34. A and B are two conducting
B

28. Two equal negative charges – q each are fixed at the concentric spherical shells. A is ++ +
++
given a charge Q while B is + A +
+
points (0, a) and (0, – a) on the y-axis .A positive + +
uncharged. If now B is earthed as +++ +
charge Q is released from rest at the point (2a, 0) on shown in figure. Then:
the x-axis. The charge Q will:
(a) The charge appearing on inner surface of B is
(a) Execute simple harmonic motion about the origin – Q.
(b) At origin velocity of particle is maximum. (b) The field inside and outside A is zero.
(c) Move to infinity (c) The field between A and B is not zero.
(d) Execute oscillatory but not simple harmonic (d) The charge appearing on outer surface of B is zero.
motion.
35. An oil drop has a charge – 9.6 × 10–19 C and mass
29. A non-conducting solid sphere of radius R is uniformly 1.6 × 10–15 gm. When allowed to fall, due to air
charged. The magnitude of the electric field due to the resistance force it attains a constant velocity. Then
sphere at a distance r from its centre. if a uniform electric field is to be applied vertically
(a) increases as r increases, for r < R to make the oil drop ascend up with the same
(b) decreases as r increases, for 0 < r < ∞ constant speed, which of the following are correct.
(c) decreases as r increases, for R < r < ∞. (g = 10 ms–2) (Assume that the magnitude of resistance
(d) is discontinuous at r = R force is same in both the cases)
30. Which of the following quantites depends on the choice (a) The electric field is directed upward
of zero potential or zero potential energy? (b) The electric field is directed downward
1
(a) Potential at a particular point (c) The intensity of electric field is 3 # 102 N C–1
(b) Change in potential energy of a two-charge system 1
(d) The intensity of electric field is 6 # 105 N C–1
(c) Potential energy of a two - charge system
(d) Potential difference between two points 36. At distance of 5 cm and 10 cm outwards from the
31. The electric field intensity at a point in space is equal in surface of a uniformly charged solid sphere, the
magnitude to: potentials are 100 V and 75 V respectively. Then:
(a) Magnitude of the potential gradient there (a) Potential at its surface is 150 V.
(b) The electric charge there (b) The charge on the sphere is (5/3) × 10−9 C.
(c) The magnitude of the electric force, a unit charge (c) The electric field on the surface is 1500 V/m.
would experience there (d) The electric potential at its centre is 225 V.
(d) The force, an electron would experience there 37. An electric dipole is kept in the electric field produced
32. An electric dipole is placed (not at infinity) in an by a point charge.
electric field generated by a point charge. (a) dipole will experience a force.
(a) The net electric force on the dipole may be zero (b) dipole will experience a torque.
(b) The net electric force on the dipole will not be zero (c) it is possible to find a path (not closed in the field
(c) The torque on the dipole due to the field may be on which work required to move the dipole is zero.
zero (d) dipole can be in stable equilibrium.
Physics
38. Select the correct alternative: (a) It is possible to consider a spherical surface of
(a) The charge gained by the uncharged body from a radius a and whose centre lies within the square
charged body due to conduction is equal to half of shown, through which total flux is + ve
the total charge initially present. (b) It is possible to consider a spherical surface of
(b) The magnitude of charge increases with the radius a and whose centre lies within the square
increase in velocity of charge shown through which total flux is – ve
(c) Charge cannot exist without matter although (c) It is possible to consider a spherical surface of
matter can exist without net charge radius a and whose centre lies within the square
(d) Between two non-magnetic substances repulsion shown through which total flux is zero
is the true test of electrification (electrification
(d) There are two points within the square at which
means body has net charge)
electric field is zero.
39. The electric potential decreases uniformly from
43. A positively charged thin metalic ring of radius R is
180 V to 20 V as one moves on the X-axis from
fixed in the xy-plane with its centre at the origin O. A
x = – 2 cm to x = + 2 cm. The electric field at the origin:
negatively charged particle P is released from rest at
(a) must be equal to 40 V/cm.
the point (0, 0, z0) where z0 > 0 . Then the motion of P
(b) may be equal to 40 V/cm.
is:
(c) may be greater than 40 V/cm.
(d) may be less than 40 V/cm. (a) Periodic, for all values of z0 satisfying 0 < z0 < ∞
(b) Simple harmonic, for all values of z0 satisfying
40. The electric field produced by a positively charged
0 < z0 ≤ R
particle, placed in an xy-plane is 7.2 (4i + 3j) N/C
at the point (3 cm, 3 cm) and 100 it N/C at the point (c) Approximately simple harmonic, provided z0 << R
(2 cm, 0). (d) Such that P crosses O and continues to move along
(a) The x-coordinate of the charged particle is the negative z-axis towards z = − ∞.
– 2 cm. 44. Select the correct statement:
(b) The charged particle is placed on the x-axis. (Only force on a particle is due to electric field)
(c) The charge of the particle is 10 × 10–12 C. (a) A charged particle always moves along the electric
line of force.
(d) The electric potential at the origin due to the charge
is 9V. (b) A charged particle may move along the line of force
41. A uniform electric field of strength E exists in a region. (c) A charge particle never moves along the line of
An electron (charge – e, mass m) enters a point A with force
velocity V tj . It moves through the electric field and (d) A charged particle moves along the line of force
exits at point B. Then: only if released from rest.
Vy 45. A negative point charge placed at the point A is

Vx a a

+2q A +2q

(a) in stable equilibrium along x-axis

2amV2 t (b) in unstable equilibrium along y-axis
(a) Ev =- i (c) in stable equilibrium along y-axis
ed2
(b) Rate of work done by the electric field at B is (d) in unstable equilibrium along x-axis
4ma2 V3
. 46. Four charges of 1 mC, 2 mC, 3 mC, and – 6 mC are
d3
(c) Rate of work by the electric field at A is zero. placed one at each corner of the square of side 1 m. The
2aV square lies in the x-y plane with its centre at the origin.
(d) Velocity at B is d it + Vjt . (a) The electric potential is zero at the origin.
42. A system of two dipoles placed in the way as shown in (b) The electric potential is zero everywhere along the
figure: x-axis only of the sides of the square are parallel to
+q a –q
x and y axis.
(c) The electric potential is zero everywhere along the
a z-axis for any orientation of the square in the x-y
plane.
(d) The electric potential is not zero along the z-axis
–q +q except at the origin.
 Electrostatics
47. Potential at a point A is 3 volt and at a point B is 7 volt, PASSAGE-2 (Question 53 to 55)
an electron is moving towards A from B.
If a charge q is moving towards the the centre of an earthed
(a) It must have some K.E. at B to reach A conducting sphere of radius b with uniform velocity v.
(b) It need not have any K.E. at B to reach A Conducting sphere is earthed with an ammeter and resistance
(c) to reach A it must have more than or equal to 4 eV in series (as shown)
K. E. at B.
(d) when it will reach A, it will have K.E. more then or
at least equal to 4 eV if it was released from rest at B.
48. An electric dipole is placed at the centre of a sphere. r
Mark the correct answer
(a) the flux of the electric field through the sphere is zero 53. Net charge on surface of conductor at the instant when
(b) the electric field is zero at every point of the sphere. particle is at x distance from centre.
(c) the electric potential is zero everywhere on the sphere. (a) + q (b) – q
(d) the electric potential is zero on a circle on the - qx b
(c) b (d) - q x
surface.
49. Which of the following statements are correct? 54. Reading of ammeter at this instant
xv v
(a) Electric field calculated by Gauss law is the field (a) q 2 (b) q b
b
due to only those charges which are enclosed qbv
inside the Gaussian surface. (c) (d) Zero
x2
(b) Gauss law is applicable only when there is a 55. Potential of the sphere will be
symmetrical distribution of charge.
(a) Zero at any instant (b) Increasing
(c) Electric flux through a closed surface will depend
(c) Decreasing (d) Infinite
only on charges enclosed within that surface
(d) None of these PASSAGE-3 (Question 56 to 58)
Comprehension Type A small conducting spherical shell with inner
radius a and outer radius b is concentric
PASSAGE-1 (Question 50 to 52)
with a larger conducting spherical shell
A solid metal sphere of radius R has a charge with inner radius c and outer radius d.
+2Q.
+2Q
R The inner shell has total charge + 2q, and
A hollow spherical conducting shell of the outer shell has charge + 4q.
3R
radius 3R placed concentric with the first
56. Find the electric field in terms of q and the distance
sphere has net charge – Q. –Q
r from the common centre of the two shells for
50. Find the electric field between the spheres at a (i) r < a (ii) a < r < b
distance r from the centre of the inner sphere. kq - kq
(a) 0, 0 (b) ,
[R < r < 3R]. r2 r2
Q Q 2kq - 2kq kq
(a) (b) (c) , 2 (d) 0, 2
2rf 0 r 2
rf 0 r2 r2 r r
Q Q
(c) (d) 57. Find the electric field in terms of q and the distance r
f 0 r2 5rf 0 r2 from the common centre of the two shells for
51. What would be the final distribution of charges if the (i) b < r < c (ii) c < r < d
spheres are joined by a conducting wire. 2q 2kq
(a) zero on outer and Q on inner sphere (a) ,0 (b) ,0
r2 r2
(b) + Q on inner and – Q on outer sphere
kq kq kq 2kq
(c) zero on inner and Q on outer sphere (c) 2 , 2 (d) 2 , 2
(d) – Q on inner and + 2Q on outer sphere r r r r
58. Find the electric field in terms of q and the distance r
52. If the inner sphere is earthed, what is the charge
from the common centre of the two shells for r > d.
on it.
2Q Q kq 2kq
(a) 3 (b) 2 (a) (b)
r2 r2
5Q Q 6kq 3kq
(c) 2 (d) 3 (c) (d)
r2 r2
Physics
PASSAGE-4 (Question 59 to 61) PASSAGE-6 (Question 65 to 67)
+
Two insulated light rods of length l and 2l An empty thick conducting shell of inner + b
+
are placed in xy plane such their mid point
+q
radius a and outer radius b is shown in a
–q
figure. If it is observed that the inner face of

+
is origin and they are free to rotate in xy
the shell carries a uniform charge density – + +
plane about z-axis. Two + q charges are –q
s and the surface carries a uniform charge
+
fixed at two ends of bigger rod and two – +q
density ‘s’.
q charges are fixed at two ends of smaller 65. If a point charge qA is placed at the center of the shell,
rod. then choose the correct statement(s)
(a) The charge must be positive
59. What is electric dipolemoment of system ?
(b) The charge must be negative
3ql 5ql (c) The magnitude of charge must be 4psa2
(a) 2 (b) 2 (d) The magnitude of charge must be 4ps(b2 – a2)
(c) ql (d) Zero
66. If another point charge qB is also placed at a distance c
60. Electric field at point (a, 0, 0) is E. Now if a >> l then (> b) from the center of shell, then choose the correct
1 1 statements
(a) E \ (b) E \ vq A b2
a2 a3 (a) force experienced by charge A is
f 0 c2
1 1
(c) E \ (d) E \ 5 (b) force experienced by charge A is zero
a4 a
vqB b
61. Work done to rotate smaller rod through 180° about (c) The force experienced by charge B is
f 0 c2
z-axis
kq2 kq A qB
(a) 0 (b) l (d) The force experienced by charge B is
c2
67. Choose the correct statement related to the potential of
2kq2 - kq2
(c) (d) the shell in absence of qB
l l
(a) Potential of the outer surface is more than that of
PASSAGE-5 (Question 62 to 64) the inner surface because it is positively charged
Four initially uncharged thin, large, plane (b) Potential of the outer surface is more than that of
identical metallic plates A, B, C and D are +q the inner surface because it carries more charge
(c) Both the surfaces have equal potential
arranged parallel to each other as shown.
–q
vb
Now plates A, B, C and D are given (d) The potential of the outer surface is f
–q 0
charges Q, 2Q, 3Q and 4Q respectively. +q Matching Column Type
Plates A and D are connected by a metallic 68. Column Ι gives certain situations involving two thin
wire while plates B and C are connected conducting shells connected by a conducting wire via
by other metallic wire then after Electrostatic equilibrium is a key K. In all situations, one sphere has net charge + q
reached. and other sphere has no net charge. After the key K is
pressed, column ΙΙ gives some resulting effects. Match
62. Charge on left surface of plate B will be the figures in Column Ι with the statements in Column ΙΙ.
3Q
(a) zero (b) 2 Column I Column II
- 9Q 5Q
(c) (d) (a) initially no (p) Charge flows
2 2
+q
net charge
through
63. Charge on plate D after earthing it will be K
connecting
5Q wire
(a) zero (b) 2 shell I
- 5Q shell II
(c) 2 (d) 6Q
(b) +q initially no (q) Potential
64. Total charge on plates after earthing plates A and B will be K
net charge
energy of
system of
(a) zero (b) 6Q
spheres
- 7Q shell II decreases.
(c) 2 (d) – 2Q shell I
 Electrostatics
y
(c) initially no (r) No heat is
net charge
produced.
+q
3 p ĵ
K
x
p î
shell I R

shell II

(d) +q (s) The shell Ι has Column I Column II

no charge after (a) The coordinate(s) of point
initially no
equilibrium is on circle where potential is (p) c R, 3R m
net charge 2 2
K reached. maximum:
(b) The coordinate(s) of point on
circle where potential is zero: (q) c- R , - 3R m
shell I 2 2
(c) The coordinate(s) of point
on circle where magnitude (r) c- 3 R , R m
shell II
2 2
of electric field intensity is
69. Column Ι gives a situation in which two dipoles of 1 4p
dipole moment pit and 3 pjt are placed at origin. 4rf 0 R3
A circle of radius R with centre at origin is drawn as (d) The coordinate(s) of point
on circle where magnitude (s) c 3R, - R m
shown in figure. Column ΙΙ gives coordinates of certain 2 2
of electric field intensity is
positions on the circle. Match the statements in Column 1 2p
Ι with the statements in Column ΙΙ. 4rf 0 R3
70.
Column I Column II
(a) V Potential along x-axis (p) Field at origin is zero

(b) 10V 20V 30 Equipotential lines (q) Zero field at some point (s) other than origin
y

0V

(c) y Equipotential lines (r) Field at + x immediately after origin is in the –ve x-direction
10V
10V
10V

10V x

(d) y (s) Field at + x immediately after origin is in the + ve

x-direction.
Equipotential lines
15V
20V
30V

40V x
Physics
Previous Year (JEE Main)

Numerical Type 7. Two infinite planes each with uniform surface charge
density + v are kept in such a way that the angle
1. For a charged spherical ball, electrostatic potential between them is 30°. The electric field in the region
inside the ball varies with r as V = 2ar2 + b. Here a shown between them is given by : [2020]
and b are constant and r is the distance from the center. Y
The volume charge density inside the ball is - maf . +
The value of m is ______ [2023]
2. 27 identical drops are charged at 22 V each. They
combine to form a bigger drop. The potential of the 30o
+ X
bigger drop will be _____ V . [2022]
(a) 2e 7(1 + 3 ) yt - 2 ] (b) 2e 0 ;c1 - 2 m yt - 2 E
v xt v 3 xt
3. Two small spheres each of mass 10m are suspended
0
from a point by threads each 0.5 m long. They are
(c) 2e ;(1 + 3 ) yt + 2 D (d) e ;c1 +
v xt v 3 m t xt E
equally charged and repel each other to a distance of y+ 2
a 0 0 2
0.20 m. The charge on each sphere is 21 # 10 -8 C.
8. Three charged particles A, B and C with charges –4q,
The value of a will be ____. [Given g = 10 ms -2 ] 2q and –2q are present on the circumference of a circle

[2021] of radius d. The charged particles A, C and centre O of
Single Option Correct the circle formed an equilateral triangle as shown in
figure. Electric field at O along x-direction is: [2020]
4. A point charge 2 # 10 -2 C is moved from P to S in y
a uniform electric field of 30 NC-1 directed along 2q -4q
positive x-axis. If coordinates of P and S are (1,2,0) m B d A
o
150 d
and (0,0,0) m respectively, the work done by electric
field will be [2023] 30o
O 30o X
(a) - 1200 mJ (b) 600 mJ d
(c) - 600 mJ (d) 1200 mJ C
5
-2q
5. A vertical electric field of magnitude 4.9 # 10 N/C
3q 3q
just prevents a water droplet of a mass 0.1 g from (a) (b)
falling. The value of charge on the droplet will be: rf o d 2
4rf o d2
(Given g = 9.8 m/s 2 ) [2022] 3 3q 2 3q
(c) (d)
(a) 1.6 # 10 -9 C (b) 2.0 # 10 -9 C 4rf o d 2
rf o d2
(c) 3.2 # 10 -9 C (d) 0.5 # 10 -9 C 9. Three charges +Q, q, +Q are placed respectively, at
distance 0, d/2 and d from the origin, on the x-axis.
6. A charge 'q' is placed at one corner of a cube as shown
If the net force experienced by +Q, placed at x = 0, is
in figure. The flux of electrostatic field E through the zero, then value of q is [2019]
shaded area is: [2021] (a) +Q/4 (b) –Q/2
Z
(c) +Q/2 (d) –Q/4
10. Two point charges q1( 10 μC) and q2(–25 mC) are
placed on the x-axis at x = 1 m and x = 4 m respectively.
The electric field (in V/m) at a point y = 3 m on1 y-axis is
Y :take 4rf = 9 # 109 Nm2 C -2D
1
[2019]
q 0

X (a) (63 it − 27 tj ) ×102 (b) (81 it - 81 tj ) ×102

q q (c) ( − 63 it + 27 tj ) ×102 (d) ( − 81 it + 81 tj ) ×102
(a) 4f (b) 24f
0 0
11. A charge Q is distributed over three concentric
q q
(c) 48f (d) 8f spherical shells of radii a, b, c (a < b < c) such that
0 0
their surface charge densities are equal to one another.
Note: Shaded region in the original JEE-Main - 2021 The total potential at a point at distance r from their
Question has been changed. common centre, where r < a, would be [2019]
 Electrostatics
Q Q ab + bc + ca 17. A long cylindrical shell carries positive surface charge
4rf 0 ]a + b + cg
(a) (b) 12rf abc σ in the upper half and negative surface charge – σ
0

Q ]a + b + cg Q ]a + b + c2g
2 2 in the lower half. The electric field lines around the
4rf 0 ]a + b + c g 4rf 0 ]a3 + b3 + c3g
(c) 2 2 2 (d) cylinder will look like figure given in: (figures are
schematic and not drawn to scale) [2015]
12. Charge is distributed within a sphere of radius R with
A -2r/a (a) (b)
a volume charge density ρ(r) = e , where A and
r2
a are constants. If Q is the total charge of this charge
distribution, the radius R is [2019]
1
(a) a log f Q p (b) a log c1 - Q m
(c) (d)
1 - 2raA 2raA
1
(c) 2 log c1 - m (d) a log f Q p
a Q
2raA 2 1 - 2raA 18. Assume that an electric field E = 30x2 it exists in
13. Three concentric metal shells A, B and C of respective space. Then the potential difference VA – VO, where VO
radii a, b and c (a < b < c) have surface charge densities is the potential at the origin and VA the potential at x =
+s, –s and +s respectively. The potential of shell B is: 2 m is: [2014]
(a) 120 V/m (b) –120 V/m
(a) ! : b - c + aD (b) ! : a - b + cD [2018]
σ 2 2
σ 2 2

0 c 0 a (c) – 80 V/m (d) 80 V/m

(c) ! 0 ; E
σ a2 - b2 σ ; b2 - c2 E
19. Two charges, each equal to q, are kept at x = – a and x
b + c (d) ! 0 b +a = a on the x-axis. A particle of mass m and charge q0 =
q
14. An electric dipole has a fixed dipole moment pv , which 2 is placed at the origin. If charge q0 is given a small
makes angle θ with respect to x-axis. When subjected displacement (y << a) along the y-axis, the net force
to an electric field Ev1 = Eit , it experiences a torque acting on the particle is proportional to: [2013]
Tv1 = xkt . When subjected to another electric field (a) y (b) – y
Ev2 = 3 E1 tj , it experiences a torque Tv2 =- Tv1 The 1 1
(c) y (d) - y
angle θ is: [2017]
(a) 30° (b) 45° 20. A charge Q is uniformly distributed over a long rod
(c) 60° (d) 90° AB of length L as shown in the figure. The electric
potential at the point O lying at distance L from the end
15. The region between two concentric A is: [2013]
spheres of radii ‘a’ and ‘b’, O A B
respectively (see figure), has volume a
L L
A Q 3Q
charge density ρ = r where A is a Q (a) 8re L (b) 4re L
b 0 0
constant and r is the distance from
Q Q ln 2
the centre. At the centre of the (c) 4re L ln 2 (d) 4re L
spheres is a point charge Q. The value of A such that 0 0

the electric field in the region between the spheres will be 21. In a uniformly charged sphere of total charge Q and
constant, is: [2016] radius R, the electric field E is plotted as function
Q 2Q of distance from the centre. The graph which would
(a) (b)
2r ^b2 - a2h r ^a 2 - b 2 h correspond to the above will be : [2012]
2Q Q
E E
(c) (d)
ra2 2ra2
16. A uniformly charged solid sphere of radius R has (a) (b)
potential V0 (measured with respect to ∞) on its
surface. For this sphere the equipotential surfaces with
R R
3V 5V 3V V
r r
potentials 2 0 , 4 0 , 4 0 and 40 have radius R1, R2, E E

R3 and R4 respectively. Then [2015]

(c) (d)
(a) R1 = 0 and R2 < (R4 – R3) (b) 2R < R4
(c) R1 = 0 and R2 > (R4 – R3)
(d) R1 ≠ 0 and (R2 – R1) > (R4 – R3) R r R r
Physics
22. This question has statement-1 and statement-2. Of the Statement-2: The electric field at a distance r(r < R)
four choices given after the statements, choose the one tr
that best describes the two statements. [2012] from the centre of the sphere is 3f
0

An insulating solid sphere of radius R has a uniformly (a) Statement 1 is true, Statement 2 is true, Statement
positive charge density ρ. As a result of this uniform
charge distribution there is a finite value of electric 2 is the correct explanation of Statement 1.
potential at the centre of the sphere, at the surface of (b) Statement-1 is true, Statement-2 is true;
the sphere and also at a point out side the sphere. The
Statement-2 is not the correct explanation of
electric potential at infinity is zero.
statement-1.
Statement-1: When a charge ‘q’ is taken from the
centre to the surface of the sphere its potential energy (c) Statement 1 is true Statement 2 is false.
qt
changes by 3f . (d) Statement 1 is false Statement 2 is true.
0

Previous Year (JEE Advanced)

Single Option Correct 3

(a) If rB = 2 , then the electric field is zero
1. An electric dipole is formed by two charges +q and everywhere outside B.
–q located in xy-plane at (0, 2) mm and (0, –2) mm,
3
respectively, as shown in the figure. The electric potential (b) If rB = 2 , then the electric potential just outside B
at point P(100, 100) mm due to the dipole is V0. The k
charges +q and –q are then moved to the points (–1, 2) is e .
0
mm and (1, –2) mm, respectively. What is the value of (c) If rB = 2, then the total charge of the configuration
electric potential at P due to the new dipole? [2023]
is 15rk.
y
5
P (100, 100) mm (d) If rB = 2 , then the magnitude of the electric field
13rk
just outside B is e .
0

3. A thin spherical insulating shell of radius R carries a

q (0, 2 mm) uniformly distributed charge such that the potential
(1, 2) mm at its surface is V0. A hole with a small area a4pR2
( a << 1) is made on the shell without affecting the rest
O x of the shell. Which one of the following statements is
correct? [2019]
q (1, 2) mm (a) The magnitude of electric field at a point, located
(0, 2) mm on a line passing through the hole and shell’s
V V
(a) 40 (b) 20 center, on a distance 2R from the center of the
V0 3V aV
(c) (d) 4 0 spherical shell will be reduced by 2R0
2
(b) The magnitude of electric field at the center of the
2. In the figure, the inner (shaded) region A represents a aV
shell is reduced by 2R0
sphere of radius rA = 1, within which the electrostatic
charge density varies with the radial distance r from the (c) The ratio of the potential at the center of the shell
center as t A = kr, where k is positive. In the spherical R
to that of the point at 2 from center towards the
shell B of outer radius rB, the electrostatic charge
2k 1-a
density varies as t B = r . Assume that dimensions hole will be 1 - 2a
are taken care of. All physical quantities are in their SI (d) The potential at the center of the shell is reduced
units. [2022] by 2aV0
4. The electric field E is measured at a point P ^0, 0, d h
rA
generated due to various charge distributions and
the dependence of E on d is found to be different for
A
different charge distributions. List-I contains different
B rB
relations between E and d. List-II describes different
Which of the following statement(s) is(are) correct? electric charge distributions, along with their locations.
 Electrostatics
Match the functions in List-I with the related charge 6. Consider a uniform spherical charge distribution of
distributions in List-II. [2018] radius R1 centred at the origin O. In this distribution, a
List-I List-II spherical cavity of radius R2, centred at P with distance
OP = a = R1 – R2 (see figure) is made. If the electric
P.
E is indepen-
1.
A point charge Q at the field inside the cavity at position rv is Ev ]rvg, then the
dent of d origin correct statement(s) is (are) [2015]
A small dipole with point
chrarges Q at ^0, 0, ,h and
R2

Q. E ? 1/d 2.
P
- Q at ^0, 0, ,h a R1
[take 2, << d ] O

an infinite line charge

coincident with the line
R. E ? 1/d 2 3.
x-axis, with uniform linear
charge density m (a) E is uniform, its magnitude is independent of R2

but its direction depends on r
Two infinite wires carrying
uniform linear charge (b) E is uniform, its magnitude depends on R2 and its

density parallel to the direction depends on r
x-axis. The one along (c) E is uniform, its magnitude is independent of a but
4. ^ y = 0, z = , h has a charge

S. E ? 1/d3 its direction depends on a
density + λ and the one (d) E is uniform and both its magnitude and direction
along ^ y = 0 , z =- ,g has

depend on a
a charge density - λ. Take 7. Charges Q, 2Q and 4Q are uniformly distributed in
2, << d three dielectric solid spheres 1, 2 and 3 of radii R/2, R
Infinite plane charge and 2R respectively, as shown in figure. If magnitudes
coincident with the xy- of the electric fields at point P at a distance R from
5. the centre of spheres 1, 2 and 3 are E1 E2 and E3
plane with uniform surface
charge density. respectively, then [2014]
(a) P " 5; Q " 3, 4; R " 1; S " 5; P P
P " 5; Q " 3; R " 1; S " 2;
P
(b) R
R
P " 5; Q " 3; R " 1, 2; S " 4;
R
(c) Q
2Q 4Q

(d) P " 4; Q " 2, 3; R " 1; S " 5; R/2

5. The figures below depict two situations in which two 2R
infinitely long static line charges of constant positive Sphere 1 Sphere 2 Sphere 3
line charge density λ are kept parallel to each other. In (a) E1 > E2 > E3 (b) E3 > E1 > E2
their resulting electric field, point charges q and – q are
kept in equilibrium between them. The point charges (c) E2 > E1 > E3 (d) E3 > E2 > E1
are confined to move in the x direction only. If they 8. Let E1(r), E2(r) and E3(r) be the respective electric
are given a small displacement about their equilibrium fields at a distance r from a point charge Q, an infinitely
positions, then the correct statement(s) is (are) [2015] long wire with constant linear charge density λ, and an
infinite plane with uniform surface charge density σ. If
    E1(r0) = E2(r0) = E3(r0) at a given distance r0, then
x x m
+q –q (a) Q = 4spr02 (b) r0 = 2rv [2014]
(c) E1(r0/2) = 2E2(r0/2) (d) E2(r0/2) = 4E3(r0/2)
(a) Both charges execute simple harmonic motion. 9. Four charge Q1,Q2,Q3 and Q4, of same magnitude
(b) Both charges will continue moving in the direction are fixed along the x-axis at x = – 2a, – a, + a and +
of their displacement.
2a, respectively. A positive charge q is placed on the
(c) Charge + q executes simple harmonic motion
while charge – q continues moving in the direction positive y axis at a distance b > 0. Four options of the
of its displacement. signs of these charges are given in List-I. The direction
(d) Charges – q executes simple harmonic motion of the forces on the charge q is given in List-II Match
while charge + q continues moving in the direction List-1 with List-II and select the correct answer using
of its displacement. the code given below the lists. [2014]
Physics
q(0, b) 7q
(c) When x = 2q, the potential at O is
4 3 re 0 a
3q
(d) When x = –3q, the potential at O is .
Q1 Q2 Q3 Q4 4 3 re 0 a
12. A disk of radius R with uniform positive charge density
(–2a, 0) (–a, 0) (+a, 0) (+2a, 0)

List-I List-II v is placed on the xy plane with its center at the origin.
The Coulomb potential along the z-axis is
P. Q1, Q2, Q3, Q4, all positive 1. +x
V (z) = 2e ^ R2 + z2 - z h
v

Q. Q1, Q2 positive Q3, Q4 negative 2. – x 0

A particle of positive charge q is placed initially at
R. Q1, Q4 positive Q2, Q3 negative 3. + y rest at a point on the z axis with z = z0 and z0 > 0. In
S. Q1, Q3 positive Q2, Q4 negative 4. – y addition to the Coulomb force, the particle experiences
2ce

Code: a vertical force Fv =- ckt with c > 0. Let b = qv 0 .
(a) P-3, Q-1, R-4, S-2 (b) P-4, Q-2, R-3, S-1 Which of the following statement(s) is(are) correct?
[2022]
(c) P-3, Q-1, R-2, S-4 (d) P-4, Q-2, R-1, S-3 1 25
(a) For b = 4 and z0 = 7 R, the particle reaches
10. Consider a thin spherical shell of radius R with its
centre at the origin, carrying uniform positive surface the origin.
charge density. The variation of the magnitude of the 1 3
(b) For b = 4 and z0 = 7 R, the particle reaches the
electric field E ] r g and the electric potential V(r) origin.
with the distance r from the centre, is best represented 1 R
by which graph? [2012] (c) For b = 4 and z0 = , the particle returns
3
E(r) V(r) E(r) V(r) back to z = z0 .
(d) For b > 1 and z0 > 0, the particle always reaches
(a) (b) the origin.
13. A uniform electric field, E =- 400 3 yt NC -1
0 R r 0 r is applied in a region. A charged particle of mass m
R
E(r) V(r) carrying positive charge q is projected in this region
E(r) V(r)
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
(c) (d)
entry point into the field as shown schematically in the
q
0 r
figure. Take m = 1010 Ckg -1 . Then- [2020]
0 r R
R

Multiple Options Correct E

11. Six charges are placed around a q
regular hexagon of side length a
as shown in the figure. Five of q 90° q
them have charge q, and the u
remaining one has charge x. The O  T
perpendicular from each charge a
q q
to the nearest hexagon side 5m
passes through the center O of
the hexagon and is bisected by x (a) the particle will hit T if projected at an angle 45º
the side. from the horizontal
Which of the following statement(s) is(are) correct in
(b) the particle will hit T if projected either at an angle
SI units [2022]
30º or 60º from the horizontal
(a) When x = q, the magnitude of the electric field at 5
O is zero. (c) time taken by the particle to hit T could be 6
(b) When x = – q, the magnitude of the electric field at ns as well as 52 ns
q
O is 5
6re 0 a2 (d) time taken by the particle to hit T is 3 ns
 Electrostatics
14. Two identical non-conducting solid spheres of same in the figure. Ther permittivity of free space is f 0 . Which
mass and charge are suspended in air from a common of the following statements is (are) true? [2018]
point by two non-conducting, massless strings of same
length. At equilibrium, the angle between the strings is (a) The electric flux through the shell is 3 Rλ/ε0
a . The spheres are now immersed in a dielectric liquid (b) The z-component of the electric field is zero at all
of density 800 kg m−3 and dielectric constant 21. If the the points on the surface of the shell
angle between the strings remains the same after the (c) The electric flux through the shell is 2 Rλ/ε0
immersion, then [2020]
(a) electric force between the spheres remains (d) The electric field is normal to the surface of the
unchanged shell at all points
(b) electric force between the spheres reduces 18. A point charge + Q is placed just outside an imaginary
(c) mass density of the spheres is 840 kg m−3 hemispherical surface of radius R as shown in the
(d) the tension in the strings holding the spheres figure. Which of the following statements is/are
remains unchanged correct? [2017]
+Q
15. 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 R
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 (a) The circumference of the flat surface is an
the following option(s) is/are correct? equipotential surface.
[e0 is permittivity of free space] [2019]
(b) The electric flux passing through the curved
(a) If h < 8R/5 and r = 3R/5 then f = 0 Q
surface of the hemisphere is - 2f d1 -
1
(b) If h > 2R and r > R then f = Q/e0 n
0 2
(c) If h > 2R and r = 4R/5 then f = Q/5e0
(d) If h > 2R and r = 3R/5 then f = Q/5e0 (c) Total flux through the curved and the flat surfaces
Q
16. An electric dipole with dipole moment 0 ^it + tj h is
P is f
0
2 (d) The component of the electric field normal to the
held fixed at the origin O in the presence of an uniform flat surface is constant over the surface.
electric field of magnitude E0. If the potential is
constant on a circle of radius R centered at the origin as 19. Two non-conducting solid spheres of radii R and 2R,
shown in figure, then the correct statement(s) is/are: having uniform volume charge densities ρ1 and ρ2
respectively, touch each other. The net electric field at
(∈0 is permittivity of free space. R>> dipole size) [2019] a distance 2R from the centre of the smaller sphere,
along the line joining the centres of the spheres, is zero.
t
The ratio t1 can be; [2013]
2
32
(a) – 4 (b) - 25
32
(c) 25 (d) 4

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

carrying uniform volume charge densities + ρ and – ρ,
(a) Total electric field at point A is E A = 2 E 0 _ Si + Tj i
respectively, are placed such that they partially overlap,
(b) Total electric field at point B is E B = 0 as shown in the figure. At all points in the overlapping
region: [2013]
(c) R = b 4r !0 E l
P 1/3

0 0
(d) The magnitude of total electric field on any two  –
points of the circle will be same.
17. An infinitely long thin nonconducting R2
wire is parallel to the z-axis and carries R1

a uniform line charge density λ. It

pierces a thin non-conducting spherical (a) the electrostatic field is zero
shell of radius R in such a way that the (b) the electrostatic potential is constant
arc PQ subtends an angle 120c at the (c) the electrostatic field is constant in magnitude
centre O of the spherical shell, as shown (d) the electrostatic field has same direction
Physics
21. Six point charges are kept at the vertices of a regular y
hexagon of side L and centre O, as shown in the figure.
1 q
Given that K = 4rf 2 , which of the following
0 L
statement (s) is (are) correct ? [2012] q
p
F L E x
+q –q (0, 0)
P l

S T 25. Two large circular discs separated by a distance

A D
+2q
O
–2q of 0.01 m are connected to a battery via a switch as
shown in the figure. Charged oil drops of density 900
R kg m−3 are released through a tiny hole at the center
B
+q –q
C of the top disc. Once some oil drops achieve terminal
(a) the electric field at O is 6K along OD velocity, the switch is closed to apply a voltage of 200
V across the discs. As a result, an oil drop of radius 8
(b) The potential at O is zero
# 10−7 m stops moving vertically and floats between
(c) The potential at all points on the line PR is same
the discs. The number of electrons present in this oil
(d) The potential at all points on the line ST is same.
drop is ________. (neglect the buoyancy force, take
Comprehension Numerical Type acceleration due to gravity = 10 ms−2 and charge on an
Q electron (e) = 1.6 # 10–19 C) [2020]
Two point charges –Q and + are placed in the xy-
3 Switch
plane at the origin (0, 0) and a point (2, 0), respectively,
as shown in the figure. This results in an equipotential 0.01 m
200 V
circle of radius R and potential V = 0 in the xy-plane
with its center at (b, 0). All lengths are measured in
meters. 26. A point charge q of mass m is suspended vertically by
y
a string of length l. A point dipole of dipole moment
p is now brought towards q from infinity so that the
Q charge moves away. The final equilibrium position of
x the system including the direction of the dipole, the
Q angles and distances is shown in the figure below. If
+
3
the work done in bringing the dipole to this position
22. The value of R is ___ meter [2021] is N×(mgh), where g is the acceleration due to gravity,
then the value of N is _________ . (Note that for three
23. The value of b is ___ meter [2021]
coplanar forces keeping a point mass in equilibrium,
Numerical Type F
is the same for all forces, where F is any one
sin i
24. One end of a spring of negligible unstretched length
of the forces and i is the angle between the other two
and spring constant k is fixed at the origin (0,0). A
point particle of mass m carrying a positive charge q forces) [2020]
is attached at its other end. The entire system is kept
on a smooth horizontal surface. When a point dipole 
l
pv pointing towards the charge q is fixed at the origin,
the spring gets stretched to a length , and attains a
new equilibrium position (see figure below). If the q
l 
point mass is now displaced slightly by D, << , 2 l sin 2
from its equilibrium position and released, it is found
h
1 k
to oscillate at frequency m . The value of d is
d p
______. [2020]

 Electrostatics
27. An infinitely long uniform line charge distribution of Integer Type
charge per unit length λ lies parallel to the y-axis in the
28. A charge q is surrounded by a closed surface consisting
3
y-z plane at z = 2 a (see figure). If the magnitude of an inverted cone of height h and base radius R, and
of the flux of the electric field through the rectangular a hemisphere of radius R as shown in the figure. The
surface ABCD lying in the x-y plane with its centre at nq
electric flux through the conical surface is 6e (in SI
mL 0
the origin is nf (ε0 permittivity of free space), then units). The value of n is _______. [2022]
0
the value of n is [2015]
z

L
D
C 3
a
2
a
y
O
A B
x

Answer Key
Exercise 1
1. (d) 2. (a) 3. (a) 4. (c) 5. (a) 6. (b) 7. (c) 8. (b) 9. (a) 10. (b)
11. (d) 12. (b) 13. (c) 14. (c) 15. (a) 16. (a) 17. (c) 18. (a) 19. (b) 20. (a)
21. (a) 22. (a) 23. (b) 24. (c) 25. (d) 26. (c) 27. (b) 28. (a) 29. (b) 30. (c)
31. (d) 32. (c) 33. (d) 34. (a) 35. (a) 36. (d) 37. (a) 38. (b) 39. (d) 40. (b)
41. (b) 42. (d) 43. (b) 44. (d) 45. (b) 46. (c) 47. (c) 48. (c) 49. (c) 50. (d)
51. (c) 52. (d) 53. (d) 54. (d) 55. (d) 56. (b) 57. (c) 58. (d) 59. (c) 60. (c)
61. (a) 62. (c) 63. (a) 64. (a) 65. (c)

Exercise 2
1. (c) 2. (d) 3. (b) 4. (d) 5. (d) 6. (b) 7. (b) 8. (d) 9. (c) 10. (c)
11. (a) 12. (a) 13. (c) 14. (a) 15. (c) 16. (a) 17. (b) 18. (d) 19. (b) 20. (d)
21. (a) 22. (c) 23. (a) 24. (b) 25. (c) 26. (d) 27. (c) 28. (d) 29. (a) 30. (d)
31. (a) 32. (a) 33. (a) 34. (a) 35. (b) 36. (c) 37. (c) 38. (b) 39. (b) 40. (a)
41. (d) 42. (b) 43. (b) 44. (c) 45. (c) 46. (a) 47. (c) 48. (a) 49. (d) 50. (c)
51. (d) 52. (c) 53. (d) 54. (b) 55. (b) 56. (a) 57. (b) 58. (c) 59. (c) 60. (a)

Exercise 3
1. (5) 2. (3) 3. (12) 4. (5) 5. (1) 6. (8) 7. (2) 8. (8) 9. (6) 10. (9)
11. (2) 12. (3) 13. (1) 14. (3) 15. (1) 16. (1) 17. (2) 18. (1) 19. (2) 20. (0)

Exercise 4
1. (b) 2. (b) 3. (c) 4. (b) 5. (c) 6. (a) 7. (b) 8. (a) 9. (b) 10. (d)
11. (a) 12. (a) 13. (c) 14. (d) 15. (a) 16. (a) 17. (c) 18. (b) 19. (b) 20. (d)
21. (d) 22. (a) 23. (d) 24. (c) 25. (a) 26. (a) 27. (d) 28. (b, d) 29. (a, c)
30. (a, c) 31. (a, c) 32. (b, c) 33. (a, d) 34. (a, c, d) 35. (b, c) 36. (a, b, c, d) 37. (a, c)
Physics
38. (c, d) 39. (b, c) 40. (b, c, d) 41. (a, b, c, d) 42. (a, b, c) 43. (a, c) 44. (b)
45. (c, d) 46. (a, c) 47. (a, c) 48. (a, d) 49. (c) 50. (a)
51. (c) 52. (d) 53. (c) 54. (c) 55. (a) 56. (a) 57. (b)
58. (c) 59. (d) 60. (c) 61. (a) 62. (d) 63. (c)
64. (a) 65. (a, c) 66. (b) 67. (c, d)
68. a→(p, q), b→(p, q), c→(p, q, s), d→(r, s) 69. a→(p, q), b→(r, s), c→(p, q), d→(r, s)
70. a→(p, q, r), b→(r), c→(p, q), d→(s)

Previous Year (JEE Main)

1. (12) 2. (198) 3. (20) 4. (c) 5. (b) 6. (b) 7. (b) 8. (a) 9. (d) 10. (a)
11. (c) 12. (d) 13. (c) 14. (c) 15. (d) 16. (a) 17. (c) 18. (c) 19. (a) 20. (d)
21. (c) 22. (c)

Previous Year (JEE Advanced)

1. (b) 2. (b) 3. (c) 4. (b) 5. (c) 6. (d) 7. (c) 8. (c) 9. (a) 10. (d)
11. (a,b, c) 12. (a,c,d) 13. (b, c) 14. (a, c) 15. (a, b, d)
16. (b, c) 17. (a,b) 18. (a, b) 19. (b, d) 20. (c, d)
21. (a, c) 22. (R = 01.73) 23. (b = 03.00) 24. (3.14) 25. (6)
26. (2) 27. (6) 28. (3)