CHAPTER
1
ELECTROSTATICS
SUB- TOPICS

Introduction

Electric Charge

Property of charge

Electric field

Lines of force

Electric Potential Energy

Electric Potential

Potential due to a point

Gauss’s Law

INTRODUCTION
 The branch of physics concerning with the study of electric charge at rest is known as
electrostatics. Electrostatics deals with the study of forces, fields and potentials
arising from static charges.

ELECTRIC CHARGE
 It is a basic property associated with the elementary particles like electrons, protons
etc. which is responsible for electric force.
 Charges are of two types, viz. positive charge and negative charge.
 SI unit of charge is coulomb abbreviated as C.
 Electronic charge e = 1.60218  10-19 C.
 Charge of electron = e
 Charge of proton = +e
PHYSICS X CLASS

PROPERTY OF CHARGE
Quantization of charge
If protons and electrons are the only charge carries in the universe, all observable charges
must be integral multiples of e. If object contains n1 protons and n2 electrons, the net
charge on the object is
n1 (e) + n2(e) = (n1 – n2)e
Note: Latest theoretical models on constituents of protons and neutrons require
1
2
new particles called quarks which carry    e or    e charge. But the same models
3
3
also predict that free quarks do not exist. Therefore, these do not violate the statement of
the law of quantization of charge.
Illustration : 1
Find the number of basic charge in 1 C.
Solution
If 1 C contains n units of basic charge e, then
1C
n
= 6.25  1018
1.6  1019 C

Conservation of charge:
It is possible to create or destroy charged particles but it is not possible to create or
destroy net charge. In beta decay process, a neutron converts itself into a proton and a
0fresh electron is created. The charge however, remains zero before and after the event.

The magnitude of charge is not affected by its motion i.e. charge is invariant.

Like charges repels each other while unlike charges attract each other. Repulsion is a sure
test of electrification. A charged body may attract a neutral body or an oppositely charged
body but it always repels a similarly charged body.

A charge at rest produces only electric field around itself.

A charge having unaccelerated motion produces electric as well as magnetic field around
itself.
A charge having accelerated motion emits electromagnetism radiation also in addition to
producing electric and magnetic fields.
A body can be charged by means of friction, conduction and induction.
Charging a body implies transfer of charge (electrons) from one body to another.
Positively charged body means loss of electrons, i.e. deficiency of electrons. Negatively
charged body means excess of electrons.
Creation of charge by friction




 When amber is rubbed with wool, it becomes electrically charged.
 An automobile becomes charged when it travels through the air.
 A paper sheet becomes charged when it passes through a printing machine.
 A gramophone record becomes charged when cleaned with a dry cloth.
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
Charging a body by means of induction.
This method is preferable since the same number of bodies without loss of charge. When
electrified rods are brought near light objects (even when light object is not a conductor),
the rod induce opposite charges on the near surface of the objects and similar charges
move to the farther side of the object.
A
A
++
++



+
+
+
Neutral
Net charge is zero
(a)
(b)
Illustration : 2
To charge a neutral sphere positively without touching it.
+
+
+
+
(a)
(d)


+
+
+
+
(b)
+
+
+
+







(c)
+
+
+
+
(e)
Take uncharged metallic sphere on an insulating stand. Figure (a).
Bring a negatively charged rod close to the metallic sphere. Figure (b).
Connect the sphere to the ground by conducting wire. The electrons to the ground
while positive charges at near end will remain held there due to attractive forces of the
rod. Figure (c).
Disconnect the sphere from the ground. Figure (d).
Remove the electrified rod. The positive charge will spread uniformly over the sphere.
Illustration : 3
Three metallic sphere A, B and C having charges +2C, OC, 2C respectively are brought
in contact and then removed. It is found that final charge distribution on sphere A and B
are 7 C and 4C respectively. What is the charge on sphere C.
Solution
Net charge of isolated system (A + B + C) must be conserved, hence
+2C + 0 + (2C) = +7C + (4C) + qC
Hence, qC = 3C
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
COULOMB’S LAW
 When the linear size of charged bodies are much smaller than the distance separately
them, then the charged bodies are treated as point charges. Force between two point
charges q1, q2 separated by a distance r in vacuum, the magnitude of force between
them is given by
K | q1q 2 |
1
F
where k 
= 8.987  109 Nm2/C2  9  109 Nm2/C2
2
r
4 o
o is called the permittivity of free space and
o = 8.854  10-12 C2N1m-2
Direction of force is decided by sign of q1q2.
q1
q2
F12 
k | q1q 2 |
r2
F21 
r
k | q1q 2 |
r2
q1q 2  0
q1
q2
F12
F21
r
F12  F21 
q1q 2  0
k | q1q 2 |
r2
Illustration : 4
Two particles A and B having charges 8  10- C and 2  10-6 C respectively are held
fixed with a separation of 20 cm. Where should a third charged particle be placed so that
it does not experience a net electric force.
Solution
As the net electric force on C should be equal to zero, the force due to A and B must be
opposite in direction. Hence, the particle must be placed on the line AB. As A and B have
charges of opposite signs, C cannot be between A and B. Also, A has larger magnitude of
charge than B. Hence, C should be placed closer to B than A.
A
B
C
20 cm
The force due to A =
Q is the charge on C.
The force due to B =
x
k  8 106 C  Q
 20 cm  x 
2
k  2 106 C  Q
x2
To have zero resultant
Force due to A = Force due to B
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 20cm  x 
2

2
x2
Giving x = 20 cm
Illustration : 5
Three equal charges, each having a magnitude 2  10-6 C, are placed at the three corners
of a triangle of side 3 cm, 4 cm and 5 cm. Find the force on the charge at the right-angled
corner.
Solution
C
3 cm
FAB

A
B
4 cm
FAC
F
Solution is shown in the figure.
FAB 
FAC 
k  2 106 C 
 4 cm 
2
2

9 109   2 106 
9 109   2 106 
 3cm 
2
Thus net force on A is F =
16 104
2
 22.5 N
2
 40 N
 FAB    FAC 
2
2
 45.9 N
This resultant makes an angle  with BA where tan  = 40/22.5 = 16/9
Illustration : 6
A charge Q is to be divided on two objects. What should be the values of the charges on
the objects so that the force between the objects can be maximum.
Solution
Suppose one object receives a charge q and the other Q – q. The net force between the
objects is
kq  Q  q 
F
d2
Where d is the separation between them, for F to be maximum,
dF
k
 0  2  Q  2q  , q = Q/2
dq
d
Thus, the charge should be divided equally on the two objects.
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
ELECTRIC FIELD
 Electric field due to point charge
The electric field due to a charge Q (source charge) at a point in space is defined as
the force that a unit positive charge would experience if placed at that point. Thus,
F
E  , where q is test charge.
q
F
Q
k | Q |q
r2
q
q
Q
r
F
r
(a) If Q is negative
kQq
r2
(b) If Q is positive
Hence electric field at a distance of r from source charge Q is given by
E
F kQ

q r2
Illustration : 7
Find the electric field at point C
5  109 C
A
10 cm
5  109 C
C
B
10 cm
Solution
Field at C due to charge particle at A is towards right and
k  5 109 
EA 
2
 20 cm 
Field at C due to charge particle at B is towards right and
k  5 109 
EB 
2
10 cm 
Hence net field at C is towards right
5
5
 k  5 109 
 9 109   5 109 
8
N / C  5.62 103 N / C
And Enet = E B  8
2
2
10
10 cm 
Illustration : 8
Three charges, each equal to q, are placed at the three corners of a square of side a. Find the
electric field at the forth corner.
Solution
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EC
y
EB
A
45o
D
q
B
x
EA
q
q
C
kq
EA  2  EC , EB 
a

kq
2a

2
E
kq
2a 2
kq ˆ
kq
i ; E C  2 ˆj
2
a
a
kq 
kq ˆ ˆ
EB 
cos 45o ˆi  sin 45o ˆj 
ij
2 
2 2 a2
2a
EA 

 

 kq
kq  ˆ  kq
kq  ˆ
i  2 
Hence E NET  E A  E B  E C   2 
j
2 
2 2a   a
2 2 a2 
a
 2 2  1  kq
o
| E NET | 
 2 , making angle 45 with AD.
2
a


Illustration : 9
Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The
distance of each corner from the centre is a. Find the electric field at the centre of the pentagon.
Solution
A
B
q
q
E
O
q C
q
D
If we put a charge q at the corner E also, the field at O will be zero by symmetry. Thus,
the field at the centre due to the charges at A, B, C and D is equal and opposite to the field
due to the charge q at E alone.
kq
The field at O due to the charge q at E is 2 along EO.
a
kq
Thus the field at O due to the given system of charges is 2 along OE.
a
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
LINES OF ELECTRIC FORCE
 Lines of force are drawn in such a way that the tangent to a line of force gives the
direction of the resultant electric field.
 The electric field is proportional to the lines per unit area if the liens originate
isotropically from the charge.
 More the crowding more will be the magnitude of field.
 Field lines start from positive charges and end at negative charges.
 In a charge-free region, electric field lines can be taken to continuous curves without
any breaks.
 To have unique direction of resultant electric field, two files lines can never cross
each other.
 Electrostatic field lines do not form any closed loops.
 They are conservative in nature.
 Field lines are always perpendicular to surface of the conductor.
(a) Field lines due to point charge
–
q
+
q
(b) Field lines due to two positive point charges.
+q E= 0 +q
(c) Field lines due to two positive and negative point charge.
+
q
–q
Illustration : 10
Consider the situation shown in figure. What are the signs of Q1 and Q2? If the lines are
drawn in proportion to the charge, what is the ratio Q1/Q2?
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Q
2
Q
1
Solution
Q1 is positive, Q2 is negative.
Q1
6
1
 
Q2
12
2

ELECTRIC POTENTIAL ENERGY
 Charge in electric potential energy of the system is negative of the work done by the
electric force as configuration of the system changes.
 Potential energy between two point charges.
Consider a system of two charges q1 and q2. Suppose, the charge q1 is fixed at a
point A and the charge q2 is taken from a point B to a point C along the line ABC.
Let AB = r1 and AC = r2.
r
A
dr
B
Q2
C
Consider a small displacement of the charge q2 in which its distance from q1
changes from r to r + dr.
The electric force on the charge q2 is F =
kq1q 2
toward AB.
r2
The work done by this force in the small displacement dr is
dw =
kq1q 2
dr
r2
Total work done as the charge moves from B to C is
1 1
kq1q 2
dr  kq1q 2   
2
r
 r1 r2 
r1
r2
w
The change in potential energy
U(r2) – U(r1) is, therefore
1 1
U(r2) – U(r1) = W  kq1q 2   
 r1 r2 
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We choose the potential energy of the two charge system to be zero when they
have infinite separation, i.e. U = 0.
The potential energy when the separation is r is U(r) = U(r) – U() =
kq1q 2
r
 It may be noted that potential energy depends essentially on the separation between
the charges and is independent of spatial location.
 If there are three charges q1, q2 and q3 there are three pairs. The potential energy of
the system is equal to the sum of the potential energies of the three pairs. Similarly
for an N-particle system.
Illustration : 11
Three particles, each having a charge of 10 c, are placed at the vertices of an equilateral
triangle of side 10 cm. Find the work done by a person in pulling them apart to infinite
separations.
Solution
The potential energy of the system in the initial condition is
U
3K 10 C 10 C 
 27 J
10 cm 
When the charges are infinitely separated, the potential energy is reduced to zero. If we
assume that the charge do not get kinetic energy in the process, the total mechanical
energy of the system decreases by 27 J. Thus, the work done by the person on the system
is 27 J.

Work done by external agent without changing kinetic energy
= Ufinal configuration  Uinitial configuration

ELECTRIC POTENTIAL
The potential at a point A is equal to the change in electric potential energy per unit test
charge when it is moved from the reference point to the point A.
VA = VA – VP = (UA – UP)/q
P is reference point.
As potential energy is a scalar quantity, potential is also a scalar quantity. Thus, if V 1 is
the potential at a point due to a charge q1 and V2 is the potential at the same point due to a
charge q2, the potential due to both charges is V2 + V1.

Electric potential due to a point source charge q at a distance r from it is given by
V

10
1 q
4o k r
The change in potential per unit distance is called potential gradient. It may be expressed
as dV/dr.
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
The electric field at a point is related to the negative potential gradient as follows
E
dV
dr

From the above relation the unit of electric field may be expressed as volt per metre
(V/m)

Electric potential energy (U) of a point charge is defined as the work done in bringing
the point charge from infinity to any point in the electric field.

The work done in bringing a number of charges from infinity to their respective positions
in an electric field is termed as electric potential energy of the system.

The expression for the potential energy of two charge system is given by
U12 
1 q1q 2
4o k r
Here, r = separation between two charges.

Equipotential surface: The surface on which the electric potential is same is called
equipotential surface .

Shapes of equipotential surfaces due to various charge distributions are as follows:
Charge Distribution
Figure
Shape of
equipotential surface
(i) Point charge
Spherical
Equipotential
surface
q
(ii) Spherical shell
of charge (q)
Equipotential
surface
Spherical
+ + +q
+
+
+ +
+
spherical
Shell
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(iii) Nonconducting
sphere of charge
(q)
Spherical
Equipotential
surface
++
++q+++
sphere of
charge q
(iv) Infinite linear
charge
distribution
Cylindrical
Equipotential
surface
Infinite line of
charge q
(v) Electric dipole
q
Plane
Plane
equipotential
surface
+q
Electric dipole
(vi) Infinite sheet of
charge
Plane
Infinite sheet
of charge
q +
+ +
++
+
+
+ +
Equipotential
surface

The lines of electric force are perpendicular to the equipotential surfaces.

The electric field is conservative. That is the work done in moving a test charge does not
depend on the path followed. On the other hand it depends on the initial and final
positions of the test charge.

The work done in moving a test charge from one point on the equipotential surface to
another on the same surface is zero.

Electric potential of the earth is arbitrarily assumed to be zero.

The surface of a conductor is an equipotential surface because electric field inside a
conductor is zero which implied that potential remains constant.
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
If a cavity exists inside a conductor, then field strength inside the cavity is zero
irrespective of the fact that field exists outside the conductor. This results in electrostatic
shielding.

Electrostatic shielding is the effect due to which two divisions (inner and outer) of a
closed conducting shell are independent of each other in respect to electric fields.

GAUSS’S LAW

ELECTRIC FLUX
The total number of lines of force that pass through a closed surface in an electric field is
called electric flux. It is denoted by .

Electric flux coming out of a small surface element dS is given by
d  E.dS  EdScos
n̂
E
dS
dScos

O

Value of total electric flux emerging out of a closed surface is given by
   d   E. dS
S
S

dS

E
O
(i) If E is parallel to the closed surface dS, then  = 90o. That is, E and dS are mutually
perpendicular.
Hence d  E.dS  EdScos90o  0 (because cos 90o = 0)
 =0
(ii) If E is in the direction of area vector, that is E and dS are in the same direction, then  =
0o.
Hence,   E.dS  EdScos0  EdS
(because cos 0 = 1)
 max = ES
This is the maximum value of electric flux.
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(iii) If E is in a direction opposite to that of the unit normal vector, then  = 180o.
Hence, d = E.dS  EdScos180 o  EdS (because cos 180o = 1)
 min = ES
This is the minimum value of electric flux.
(iv) If closed surface is situated in uniform electric field, then  = 0.
GAUSS’S LAW
The statement of the Gauss’s law may be written as follows:
The flux of the net electric field through a closed surface equals the net charge enclosed
by the surface divided by o. In symbols,
qin
 E.dS  
o
where qin is the ent charge enclosed by the surface through which the flux is calculated.

Electric flux is a scalar quantity.

Units of electric flux are: (i) Nm2C-1 (ii) Vm (iii) JmC-1

Dimensional formula for electric flux is [ML3T-3I-1].

Electric flux entering a closed surface is taken as negative and that emanating out of the
closed surface is taken as positive.

The net electric flux coming out of a closed surface depends upon the nature as well as
the quantity of the charges enclosed by that surface. It also depends on the medium
present.

The value of electric flux is independent of the distribution of charges and the separation
between them inside the closed surface .
Illustration : 12
A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is
constructed with its center at the periphery of the ring. Find the flux of the electric field
through the surface of the sphere.
A
Ring
O
O1
Sphere
B
Solution
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From the geometry of the figure, OA = OO1 and O1A = O1O. Thus OAO1 is an equilateral
triangle. Hence AOO1 = 60o or AOB = 120o.
There are AO1B of the ring subtends an angle 120o at the center O. Thus, one third of the
ring is inside the sphere.
The charge enclosed by the sphere = Q/3. From Gauss’s law, the flux of the electric field
through the surface of sphere is Q/3o.
GAUSSIAN SURFACE


The imaginary surface enclosing the position or distribution of charges is defined as
Gaussian surface.
(i) Gaussian surface is spherical for a point charge, conducting and non-conducting spheres.
Gaussian surface
O q
(point charge)
+q
O
Gaussian surface
Conducting surface
Gaussian surface
+q
O
Uniformly charged
non-conducting
sphere
(ii) The Gaussian surface is cylindrical for infinite sheet of charge, infinite line of charge,
charged cylindrical conductor etc.
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Gaussian surface
+
+
+
+
+
+
Infinite line
of charge
+
+
+
+
+
+
+
+
+
Infinite sheet of
charge
+
+
+
Cylindrical
Gaussian surfaace

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
KEY POINTS

Electrostatic force between two point charges separated by distance r is F 
where k 

1
= 8.987  109 Nm2/C2  9  109 Nm2/C2
4 o
Electric field at a distance of r from source charge Q is given by
E

F kQ

q r2
1 1
U(r2) – U(r1) = W  kq1q 2   
 r1 r2 
The expression for the potential energy of two charge system is given by
U12 

1 q
4o r
The electric field at a point is related to the negative potential gradient as follows
E

1 q1q 2
4o r
Electric potential due to a point source charge q at a distance r from it is given by
V

K | q1q 2 |
r2
dV
dr
Value of total electric flux emerging out of a closed surface is given by
   d   E. dS =
S
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q in
o
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
ASSIGNMENT – I
1*.
Which of the following is not a static electric field?
(A)
(B)
(C)
(D)
2.
A charge Q is distributed with the ratio x : y between two point objects so that they
experience maximum electrostatic force at a distance R of separation. Then x : y =
(A) 1 : 2
(B) 1 : 1
(C) 2 : 1
(D) 2 : 1
3*.
A positive charge Q is placed at the centre of
a square of side l. Then the net force acting on
the central charge Q due to the other four
charges Q, 2Q ,3Q and 4Q placed at the
vertices of the square is
Q2
3Q 2
(A)
(B)
î
ĵ
2ol 2
2ol 2
y
4Q
3Q
Q +
x
Q
2Q
3Q 2
2 Q2

ĵ
(D)
î
2
2
 o l
2ol
If a charge q is placed at the centre of the line joining two equal charges +Q and +Q, the
system of the three charges will be in equilibrium, if q is
Q
Q
(A) 
(B) 
2
4
Q
(C) 4Q
(D) 
2 2
(C)
4.
5*.
Q
The ratio of q and Q so as to make the system in
equilibrium is
(A) 1 : 3
(B) 1:  3
(C) 1: 3
(D) 3 :1
q
Q
18
Q
IIT Foundation Programme
Electrostatics
6.
A conducting sphere has a charge Q. It touches three identical spheres one after the other.
The final charge of the sphere is
(A) Q/3
(B) 2Q/3
(C) Q/6
(D) Q/8
7.
Two point charges +8q and 2q are located at x = 0 and x = L respectively. The location
of a point on the x-axis at which the net electric field due to these two point charges is
zero is
(A) 2L
(B) L/4
(C) 8L
(D) 4L
8*.
Four charges equal to Q and placed at the four corners of a square and a charge q is at its
centre. If the system is in equilibrium, the value of q is
Q
Q
1 2 2
(A)  1  2 2
(B)
4
4
Q
Q
(C)  1  2 2
(D)
1 2 2
2
2








9.
A charge Q is placed at each of opposite corners of a square. A charge q is placed at each
of the other two corners. If the net electrical force on Q is zero, then Q/q equals
1
(A) 
(B) 2 / 2
2
(C) 1
(D) 1
10*.
Six charges of equal magnitude, 3 positive and 3
negative are to be placed on PQRSTU corners of a
regular hexagon, such that field at the centre is double
that of what it would have been if only one +ve charge is
placed at R.
(A) +,+,+,,,
(B) ,+,+,+,,
(C) ,+,+,,+,
(D) +,,+,,+,
P
U
R
O
T
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Q
S
19
PHYSICS X CLASS
ASSIGNMENT – II
1*.
The field pattern is shown due to two
point charges 1 and 2. Then q1/q2 =
1
(A) 
(B) 2
2
1
(C)
(D) none of these
2
1
a
2
2.
An external agent does a work of 100 J in bringing a test charge q = 10 -2 C slowly from
infinity to a point P. If the charges producing field and potential are kept fixed at their
respective position, the electric potential at P is
(A) 0.5  104 volt
(B) 104 volt
4
(C) 10 volt
(D)1 volt
3.
A fixed point charge q does a negative work when a negative charge q ' approaches to q.
Then, the potential due to q at any point is
(A) +ve
(B) –ve
(C) zero
(D) dependent on q '
4.
When we take an electron slowly away from the fixed nucleus, the external work done is
(A) +ve
(B) –ve
(C) zero
(D) sometime –ve
5*.
Potential of a charge q1is +10 V at the point P. When we bring another charge q2 around
point P, the potential at P becomes 4 V. Then q2 is
(A) +ve
(B) –ve
(C) 2q2/5
(D) +5q2/2
6.
An electron is taken from 500 V to +1500 V. The work done by the electric field is
(A) 1.6  10-16 J
(B) 3.2  10-16 J
-16
(C) 1.6  10 J
(D) 3.2  10-16 J
7*.
The charge Q is shifted slowly from C to D, vertices of
square of side l. The work done by the external agent is
KQ2
KQ2
(A) 
(B) 
3 2
3 2 2
l
l
KQ2
KQ2
(C)  2  2
(D) 2  2
l
l







Q
B

A
Q
20
Q
C
D
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Electrostatics
8*.
A charge Q is shifted from the origin to a
point P(a, a). the work done by the
electrical force is
KQq 
1 
(A)
1


a 
5
KQq
(B)


5 1
P(a, a)
(a, 0)

q
(a, 0)
+
+q
O
x
a
KQq 
1 
(C) 
1


a 
5
KQq
(D)
5 1
a


9.
Bringing an electron nearer to the nucleus, potential energy of the system (atom) :
(A) increases
(B) decreases
(C) remains constant
(D) may change
10.
A proton and an -particle are released from rest. After moving through same potential
difference, the ratio of their speeds is
(A) 1 : 2
(B) 2 :1
(D) 1: 2
(C) 2 : 1
11*.
An electric charge 10-3 C is placed at the origin (0, 0) of X-Y coordinate system. Two
points A and B are situated at
2, 2 and (2, 0) respectively. The potential difference

between the points A and B will be
(A) 9 V
(C) 2 V

(B) zero
(D) 4.5 V
12.
On moving a charge of 20 C by 2 cm, 2 J of work is done. Then the potential difference
between the points is
(A) 0.1 V
(B) 8 V
(C) 2 V
(D) 0.5 V
13.
Three charges Q, +q and +q are placed at the
vertices of a right-angled isosceles triangle as
shown in figure. The net electrostatic energy of
the configuration is zero, if Q is equal to
q
2q
(A) 
(B) 
1 2
2 2
(C) 2q
(D) +q
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Q
a
+q
+q
a
21
PHYSICS X CLASS
ASSIGNMENT – III
1*.
2.
A point charge q is placed at the origin. The net
flux of electric field (E) passing through the
curved surface S is
q
(A)
o
q
(B)
8 o
(C) zero
q
(D)
4 o
y
S
+
q
x
z
The ratio of E passing through the surface S1 and S2 is
(A) 1 : 1
(B) 2 : 1
(C) 3 : 1
(D) 1 : 3
S2
S1
Q
+
4Q/3
3.
4.
A Gaussian surface S enclosed two charges q1 = q,
q2 = q and q3 = q. The net flux passing through S is
q
(A) 0
(B)
o
q
q'
(C) 
(D)
o
o
22
q1
q3
q2
S
If the electric flux entering and leaving an enclosed surface respectively is 1 and 2,the
electric charge inside the surface will be
(A)  2  1  o
(B)  2  1  / o
(C)  2  1  / o
5.
P
(D)  2  1  o
A Gaussian surface in the figure is shown by dotted line.
The electric field on the surface will be
(A) due to q1 and q2 only
(B) due to q2 only
(C) zero
(D) due to all
q1
q2
q1
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6.
A point charge +q is placed at the mid-point of a cube of side L. The electric flux
emerging from the cube is
(A) q/o
(B) q/(6L2o)
(C) 6qL2/o
(D) zero
7*.
Figure shows four charges q1, q2, q3 and q4 fixed in
space. Then the total flux of electric field through a
closed surface S, due to all charges q1, q2, q3 and q4 is
(A) not equal to the total flux through S due to charges
q3 and q4
(B) equal to the total flux through S due to charges q3
and q4
(C) zero if q1 + q2 = q3 + q4
(D) twice the total flux through S due to charges q3 and
q4 if q1 + q2 = q3 + q4.
q2
q1
q3
q4
8.
Choose the correct statement
(A) Gauss’ law is valid only for charges placed in vacuum
(B) Gauss’ law is valid only for symmetrical charge distribution
(C) The flux of electric field through a closed surface due to all the charges (inside or
outside) is equal to the flux due to the charges enclosed by the surface
(D) None of the above is a correct statement
9.
A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked
with the surface?
q
2q
(A)
(B)
4 o
3 o
q
6q
(C)
(D)
6 o
o
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PHYSICS X CLASS
COMPETITIVE CORNER
Straight Objective Type
This section contains multiple choice questions. Each question has 4 choices (A), (B), (C), (D), out of which ONLY
ONE is correct. Choose the correct option.
1.
The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the
other by 0.01 m, then the force between them will become
(A) 7.20 N
(B) 11.25 N
(C) 22.50 N
(D) 45 N
2.
Two point charges +3 C and +8 C repel each other with a force of 40 N. If a charger of
5 C is added to each of them, then the force between them will become
(A) 10 N
(B) +10 N
(C) +20 N
(D) 20 N
3.
A point charge Q is placed at the mid-point of a line joining two charges, 4q and q. If the
net force on charge q is zero, then Q must be equal to
(A) q
(B) q
(C) 2q
(D) 4q
4.
Two point charges +e and +2e are at 16 cm away from each other. Where should another
charger q be placed between them so that the system remains in equilibrium?
(A) 24 cm from +e
(B) 12 cm from +e
(C) 80 cm from +e
(D) 8 cm from +e
5.
A charge Q is placed at the corner of a cube. The electric flux through all the six faces of
a cube is
(A) Q/o
(B) Q/6o
(C) Q/8o
(D) Q/3o
6.
A charge q is placed at the center of the open end of cylindrical vessel. The flux of the
electric field through surface of the vessel is
q
(A) zero
(C)
24
q
2o
q
o
2q
(D)
o
(B)
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Electrostatics
7.
Flux coming out from a positive charge placed in air is
(A) o
(B) o1
(C)  4o 
1
(D) 4 o
8.
If the electric flux entering and leaving an enclosed surface respectively is 1 and 2, the
electric charge inside the surface will be
(A)  1  2  o
(B)  2  1  o
(C)  1  2  / o
(D)  2  1  / o
9.
The work done by an agency to carry a 10C charge from infinity to a point in
electrostatic field is 50 J. The potential at that point is
(A) 0.2 V
(B) 5 V
(C) 5 V
(D) 500 V
10.
The electrostatic potential energy of a charge of 5 C at a point in the electrostatic field is
50 J. The potential at that point is
(A) 0.1 V
(B) 5 V
(C) 10 V
(D) 250 V
11.
E = dV/dr. Here negative sign signified that
(A) E is opposite to V
(B) E is negative
(C) E increases when V decreases
(D) E is directed in the direction of decreasing V.
12.
Two small spheres each carrying a charge q are placed 1 m apart. The electric force
between them is F. If one sphere is taken around the other. The work done is
(A) F
(B) 2F
(C) F/2
(D) zero
Multiple Correct Answer Type
This section contains multiple choice questions. Each question has 4 choices (A), (B), (C), (D), out of
which ONE or MORE is correct. Choose the correct options.
13.
Charges Q1 and Q2 lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and  be the flux of E over S.
(A) If Q1 changes, both E and  will change
(B) If Q2 changes, E will change but  will not change
(C) If Q1 = 0 and Q2  0 then E  0 but  = 0
(D) If Q1  0 and Q2 = 0 then E = 0 but   0
14.
P is a point on an equipotential surface S. The field at P is E
(A) E must be perpendicular to S in all cases
(B) E will be perpendicular to S only if S is a plane surface
(C) E cannot have a component along a tangent to S
(D) E may have a nonzero component along a tangent to S if S is a curved surface
IIT Foundation Programme
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PHYSICS X CLASS
15.
A point charge is brought in an electric field. The electric field at a nearby point
(A) will increase if the charge is positive
(B) will decrease if the charge is negative
(C) may increase if the charge is positive
(D) may decrease if the charge is negative
16.
Which of the following quantities do not depend on the choice of zero potential or zero potential
energy?
(A) potential at a point
(B) potential difference between two points
(C) potential energy of a two-charge system
(D) change in potential energy of a two-charge system
17.
S1 and S2 are two equipotential surfaces on which the potentials are not equal
(A) S1 and S2 cannot intersect
(B) both S1 and S2 cannot be plane surfaces
(C) In the region between S1 and S2, the field is maximum where they are closest to each other
(D) A line of force from S1 to S2 must be perpendicular to both
18.
In a uniform electric field, equipotential surfaces must
(A) be plane surfaces
(B) be normal to the direction of field
(C) be spaced such that surfaces having equal differences in potential are separated by equal
distances
(D) have decreasing potentials in the direction of the field
Linked Comprehension Type
This section contains paragraphs. Based upon each paragraph multiple choice questions have to be answered. Each
question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Choose the correct option.
Comprehension – 1
The earth has a net electric charge that causes a field at points near its surface. The charge on the earth is
supposed to be a result of an atmospheric battery created between ionosphere and the earth. The electric
field near the earth’s surface believed to be 150 N/C and directed towards the center of the earth. A man
suggested that this electric field may be used in flying.
19.
What magnitude and sign of charge would a 60 kg human have to acquire to overcome his or her
weight?
(A) 6 C
(B) 4 C
(C) 6 C
(D) 4C
20.
What would be the force of repulsion between two people with the said charge when they are 100
m apart?
(A) 1.44  105 C
(B) 14.4  105 C
(C) 1.44  106 C
(D) 1.44  107 C
Comprehension – 2
The imaging drum of a photocopier is positively charged to attract negatively charged particles of toner.
Near the surface of drum, its electric field has magnitude 1.4  105 NC-1. A toner particle is to be attached
to the drum with a force that is 10 times the weight of the particle. Assume toner particles are made of
carbon 12
6 C.
26
IIT Foundation Programme
Electrostatics
21.
Find the charge to mass ratio of the charged toner particle
(A) 7.0  10-4 Ckg-1
(B) 7.0  10-3 Ckg-1
-5
-1
(C) 7.0  10 Ckg
(D) 7.0  10-4 Cg-1
22.
Find the number of carbon atoms that for each excess electron on a toner particle
(A) 1.15  108
(B) 1.15  107
9
(C) 1.15  10
(D) 1.15  1010
Assertion – Reason Type question
This section contains certain number of questions. Each question contains Statement – 1 (Assertion) and
Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is
correct Choose the correct option.
(A) Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1.
(B) Statement–1 is True, Statement–2 is True; Statement–2 is not a correct explanation for Statement–
1.
(C) Statement–1 is True, Statement–2 is False.
(D) Statement–1 is False, Statement–2 is True.
23. Statement 1: If a proton and an electron are placed in the same uniform electric field, they
experience different acceleration.
Statement-2: Electric force on a test charge is independent of the mass of the test charge.
24. Statement-1: The electric flux through the Gaussian’s surface enclosing charges – q, 2q and –q is
zero.
Statement-2: Total charge enclosed by the Guassian surface is 4q.
Matrix Match Type
This section contains Matrix-Match Type questions. Each question contains statements given in two
columns which have to be matched. Statements (A, B, C, D) in Column–I have to be matched with
statements (p, q, r, s) in Column–II. The answers to these questions have to be appropriately bubbled as
illustrated in the following example.
25.
In column I there is a system of two charge particles separated by distance 2r. In column
II, E and V are the magnitude of field and potential at x due to system of charges
respectively. Match the column I with column II.
Column – I
r
r
(A)
Q
(B)
Q
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Q
x
r
Column – II
(P)
E=0
(Q)
E=
r
x
Q
kQ
4r 2
27
PHYSICS X CLASS
x
(C)
Q
60o
60o
r
(R)
V=0
(S)
V=
2kQ
r
(T)
V=
kQ
r
Q
r
x
(D)
Q
60o
r
60o
Q
r
Integer Answer Type
26.
A pendulum bob of mass 80 mg and carrying charge of 2×10–8 C is at rest at a certain
angle () with the vertical in a horizontal uniform electric field of 40,000 Vm–1. If
tan   x . Find x
27.
Three particles each having charge 10 C are placed at the vertices of an equilateral
triangle of side length 30 cm. Find work done by the field (in J) when the separation
between them become infinite.









28
IIT Foundation Programme
Electrostatics

KEY & HINTS
ELECTROSTATICS
ASSIGNMENT – I
1.
2.
3.
4.
5.
(C)
(B)
(D)
(B)
(B)
6.
7.
8.
9.
10.
(D)
(A)
(D)
(B)
(C)
ASSIGNMENT – II
1.
2.
3.
4.
5.
6.
7.
(A)
(C)
(B)
(A)
(B)
(B)
(D)
8.
9.
10.
11.
12.
13.
(A)
(B)
(B)
(B)
(A)
(B)
ASSIGNMENT – III
1.
2.
3.
4.
5.
(B)
(C)
(A)
(C)
(D)
6.
7.
8.
9.
(A)
(B)
(C)
(C)
COMPETITIVE CORNER
1.
2.
(B)
5  (0.06)2 = F  (0.04)2
Hence F = 11.25 N
(A)
In the second case, the charges will be 2 C and +3 C.
Here 40  3  8 , F  2  3
Which gives F  
3.
6
 40  10 N
24
(A)
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29
PHYSICS X CLASS
Net force on q is
Q
4q
r
F
4.
q
r
1 4q  q
1 Qq

0
2
4o  2r 
4o r 2
Hence q2 + Qq = 0
This gives Q = q
(D)
q
+e
x
+2e
16 – x
For the system to be in equilibrium, we have,
force on charge q due to +e = Force on charge q due to +2e
That is,
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
30
1 eq
1
2eq

0
2
4o x
4o 16  x 2
 2x2 = (16 – x)2
 x2 + 32x – 256 = 0
On solving, we get x = 8 cm
(C)
Conceptual
(C)
Conceptual
(B)
Conceptual
(B)
Conceptual
(C)
Potential = potential energy / test charge.
(C)
Potential = potential energy / test charge.
(D)
(D)
Electric field is conservative. So, work done in taking a charge on the closed path is zero.
(A), (B), (C)
Conceptual
(A), (C)
Conceptual
(C), (D)
Conceptual
(B), (D)
Conceptual
(A), (C), (D)
Conceptual
(A), (B),(C), (D)
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Electrostatics
19.
Conceptual
(D)
60  10 = q E
or q =
20.
(D)
F
21.
60 10
 4C
150
4  4  9  109
100 
2
= 144  105 N
(A)
qE = 10 mg
q log 9.8  10
= 7.0  10-4 Ckg-1


m E 1.4 105
22.
(D)
10mg 12  103  9.8 10

E
1.4  105
84  107
Number of electrons n 
1.6  1019
12
84  10
n
1.6
6.023  1023  1.6
Number of carbon atoms =
= 1.15  1010 per excess e
84  1012
q
23.
24.
25.
(B)
Conceptual
(C)
Conceptual
A – P, S, B – R, C –T, D – Q, R
Conceptual
26.
27.
(1)
Let T be the tension in the thread of the pendulum. Various forces acting on the pendulum
are (i) weight (mg) of the bob of the pendulum. (ii) Tension (T) in the thread. (iii) Force
(F) due to electric field (F = qE).
T cos   mg & T sin   qE  tan   qE / mg
(9)
The Ui = 3k  (10C)2 / 30 cm = 9 J
Uf = 0
Work done by field = Ui – Uf = 9 J


 
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31