electrostatics - Bothra Classes

1

Electric Charge :

Charge is the property associated with matter due to which it produces and experiences electrical

and magnetic effects. The excess or deficiency of electrons in a body gives the concept of charge.

1. Types of charge:

Positive charge: It is the deficiency of electrons as compared to proton.

Negative charge: It is the excess of electrons as compared to proton.

SI unit of charge: ampere � second i.e. Coulomb, Dimension : [A T]

Practical units of charge are ampere � hour (= 3600 C) and faraday (= 96500 C)

Millikan calculated quanta of charge by ‘Highest common factor’ (H.C.F) method and it is equal

to charge of electron.

1C = 3 � 109 stat coulomb, 1 Faraday = 96500 C.

2. Specific Properties of Charge :

Charge is a scalar quantity: It represents excess or deficiency of electrons,

Charge is transferable : If a charged body is put in contact with another body, then charge can be

transferred to another body.

Charge is always associated with mass: Charge cannot exist without mass though mass can

exist without charge. When body is given positive charge, its mass decreases and when body is

given negative charge its mass increases.

Charge is quantized : The quantization of electric charge is the property by virtue of which all

free charges are integral multiple of a basic unit of charge represented by e. Thus charge q of a

body is always given by q = ne, n = positive or negative integer

Note : Charge on a proton = (–) charge on an electron = 1.6 � 10–19C

Charge is conserved : In an isolated system, total charge does not change with time, though

individual charge may change i.e. charge can neither be created nor destroyed. Conservation of

charge is also found to hold good in all types of reactions either chemical (atomic) or nuclear.

No exceptions to the rule have ever been found.

Charge is invariant: Charge is independent of frame of reference, i.e. charge on a body does

not change whatever be its speed.

Accelerated charge radiates energy:

Attraction – Repulsion : Similar charges repel each other while opposite charges attract.

3. Methods of Charging

Friction : If we rub one body with other body, electrons are transferred from one body to the

other. Transfer of electrons takes places from lower work function body to higher work function

body.

ELECTROSTATICS 1

2

Electrostatic induction :

If a charged body is brought near a metallic neutral body, the charged body will attract opposite

charge and repel similar charge present in the neutral body As a result of this one side of the

neutral body becomes negative while the other positive, this process is called ‘electrostatic

induction’. Charging a body by induction (in four successive steps)

Some important facts associated with induction-

• Inducing body neither gains nor loses charge

• The nature of induced charge is always opposite to that of inducing charge

• Induction takes place only in bodies (either conducting or non conducting) and not in particles.

Conduction:

The process of transfer of charge by contact of two bodies is known as conduction. If a charged

body is put in contact with uncharged body, the uncharged body becomes charged due to transfer

of electrons from one body to the other.

• The charged body loses some of its charge (which is equal to the charge gained by the uncharged

body)

• The charge gained by the uncharged body is always lesser than initial charge present on the

charged body.

• Flow of charge depends upon the potential difference of both bodies.

[No potential difference � No conduction].

• Positive charge flows from higher potential to lower potential,while negative charge flows from

lower to higher potential.

Coulomb’s Law

The electrostatic force of interaction between two static point electric charges is directly

proportional to the product of the charges, inversely proportional to the square of the distance

between them and acts along the straight line joining the two charges.

If two points charges q1 and q

2 separated by a distance r. Let F be the electrostatic force between

these two charges.

According to Coulomb’s law,

1 2F q q� and 2

1F

r�

1 2

2e

kq qF

r� where

29

2

0

19 10

4

Nmk

C��

� � �� , k = coulomb’s constant or electrostatic force constant

Coulomb’s law in vector form

12F ��

force on q1 due to 1 2

2 212ˆ

kq qq r

r�

1 2

21 122ˆ

kq qF r

r�

�

(here 12r is unit vector from q

1 to q

2)

3

Coulomb’s law in terms of position vector

1 2

12 1 23

1 2

( )| |

kq qF r r

r r�

� � ��

Principle of superposition :

The force is a two body interaction, i.e., electrical force between two point charges is independent

of presence or absence of other charges and so the principle of superposition is valid, i.e., force on

a charged particle due to number of point charges is the resultant of forces due to individual point

charges, i.e., 1 12 13

........F F F� � ��

When a number of charges are interacting, the total force on a given charge is vector sum of the

forces exerted on it by all other charges individually.

Some important points regarding Coulomb’s law and Electric force :

The law expresses the force between two point charges at rest. The law is based on physical

observations and is not logically derivable from any other concept. Experiments till today reveal

its universal law.

The law is analogous to Newton’s law of gravitation 1 2

2

m mG

r� with the difference that:

(a) Electric force between charged particles is much stronger than gravitational force, i.e., FE > > F

G.

This is why when both FE and F

G are present, we neglect F

G.

(b) Electric force can be attractive or repulsive while gravitational force is always attractive.

(c) Electric force depends on the nature of medium between the charges while gravitational force

does not.

(d) The force is an action-reaction pair, i.e., the force which one charge exerts on the other is equal

and opposite to the force which the other charge exerts on the first.

The force is conservative, i.e., work done in moving a point charge once round a closed path

under the action of Coulomb’s force is zero. The net Coulomb’s force on two charged particles

in free space and in a medium filled up to infinity are 1 2

2

0

1

4

q qF

r�� . So

0'

FK

F

��

� � , where K is

Dielectric constant of a medium and defined as numerically equal to the ratio of the force on

two point charges in free space to that in the medium filled upto infinity.

Net electric force on both particles change in the presence of dielectric but force due to one

charge particle on another charge particle does not depend on the medium between them. Electric

force between two charges does not depend on neighbouring charges.

Equilibrium of Charged Particles:

In equilibrium net electric force on every charged particle is zero. The equilibrium of a charged

particle, under the action of coulombic forces alone can never be stable.

Equilibrium of three point charges

Two charges must be of like nature as 1 2

2 20

( )q

KQ q KQ qF

x r x� � �

Third charge should be of unlike nature as 1

1 2 1

2 20

Q

KQ Q KQ qF

r x� � � , therefore

1

1 2

Qx r

Q Q�

� and

4

1 2

2

1 2

Q Qq r

Q Q

�

�

Equilibrium of symmetric geometrical point charged system :

Value of Q at centre for the system to be in state of equilibrium

(i) For equilateral triangle 3

qQ

� (ii) For square

(2 2 1)

4

qQ

��

Equilibrium of suspended point charge system :

For equilibrium position cosT mg� � and 2 2

2 2sin tan e

e

FkQ kQT F

mgx x mg� ��

If � is small then tan�1

2 2 2 33

2

0

2sin

2 2 2

x x kQ kQ Qx x

mg mgx mg� �

��

� � � � � � � � � � �

� �

� �

If whole set up is taken into an artificial satellite ( 0)eff

g � then2

24

e

kqT F� �

�

Electric Field

In order to explain ‘action at a distance’, i.e., ‘force without contact’ between charges it is assumed

that a charge or charge distribution produces a field in space surrounding it. So the region

surrounding a charge (or charge distribution) in which its electrical effects are perceptible is

called the electric field of the given charge.

Electric field at a point is characterized either by a vector function of position E�

called

‘electric field intensity’ or by a scalar function of position V called ‘electric potential’. The

electric field in a certain space is also visualized graphically in terms of ‘lines of force.’ So

electric field intensity, potential and lines of force are different ways of describing the same

field.

Intensity of electric field due to point charge :

Electric field intensity is defined as force per unit infinitesimal test charge.

02 30

0

ˆlimq

F kq kqE r r

q r r��

��

Note : Test charge (q0) is a fictitious charge that exerts no force on nearby charges but experiences

forces due to them.

Properties of electric field intensity :

• It is a vector quantity. Its direction is the same as the force experienced by positive charge.

• Electric field due to positive charge is always away from it while due to negative charge always

towards it.

• Its unit is Newton/coulomb and its dimensional formula is [MLT–3A–1]

• Force on a point charge is in the same direction of electric field on positive charge and in

opposite direction on a negative charge. F qE��

• It obeys the superposition principle that is the field intensity at a point due to a charge

distribution is vector sum of the field intensities due to individual charges.

Electric field intensities due to various charge distributions:

Due to discrete distribution of charge: Field produced by a charge distribution for discrete

distribution:

5

By principle of superposition intensity of electric field due to thi charge 13

1

ip

kqE r

r�

� �

� Net electric field due to whole distribution of charge 13

1 1

p

kqE r

r� �

� �

Continuous distribution of charge

Treating a small element as particle 3

0

1

4

dqE r

r��

Due to linear charge distribution 2

dE k

r

��

�

�

[ � = charge per unit length]

Due to surface charge distribution 2

s

dsE k

r

�� [� = charge per unit area]

Due to volume charge distribution 2

v

dvE k

r

�� [ � = charge per unit volume]

Electric field strength at a general point due to a uniformly charged rod:

As shown in figure, if P is any general point in the

surrounding of rod, to find electric field strength at P, we

consider an element on rod of length dx at a distance x

from point O as shown in figure. Now if dE be the electric

field at P due to the element, then

2 2( )

kdqdE

x r�

� ,

Here Q

dq dxL

�

Electric field strength in x-direction due to dq at P is

2 2 2 2

sinsin sin

( ) ( )x

kdq kQdE dE dx

x r L x r

��

� ��

Here we have tanx r �� and 2secdx r d� ��

Thus 2

2 2

secsin

secx

kQ r ddE

L r

� ��

�� = sin

kQd

Lr� �

Net electric field strength due to dq at point P in x-direction is

Similarly, electric field strength at point P due to dq in y-direction is

2 2cos cos

( )y

kQdxdE dE

L r x� ��

�

Again we have tanx r �� and 2secdx r d� �� . Thus we have

2

2 2

seccos cos

secy

kQ r kQdE d d

L Lrr

��

��

Net electric field strength at P due to dq in y-direction is

1

1

2

2

1 2cos [ sin ] [sin sin ]

y y

kQ kQ kQE dE d

Lr Lr Lr

��

�

� � � � ��

�

� � � � � ��

Thus electric field at a general point in the surrounding of a uniformly charged rod which

subtends angles 1

� and 2

� at the two comers of rod can be given as in x-direction :

6

2 1cos cosx

kQE

Lr� �� and in y-direction 1 2[sin sin ]y

kQE

Lr� ��

Electric field due to a uniformly charged ring

Case -I: At its centre Here by symmetry we can say that electric field strength at centre due to every

small segment on ring is cancelled by the electric field at centre due to the segment exactly

opposite to it. The electric field strength at centre due to segment AB is cancelled by that due to

segment CD. This net electric field strength at the centre of a uniformly charged ring is E0 = 0.

Case II: At a point on the axis of Ring

Here we’ll find the electric field strength at point P due to the ring which is situated at a distance

x from the ring centre. For this we consider a small section of length d� on ring as shown. The

charge on this elemental section is 2

Qdq d

R�� [Q = total charge of ring]

Due to element d� , electric field strength dE at point P can begiven as 2 2( )

KdqdE

R x�

�

The component of this field strength dE sin� which is normal to the axis of ring will be cancelled

out due to the ring section opposite to d� .The component of electric field strength along the axis

of ring dE cos � due to all the sections will be added up. Hence total electric field strength at point

P due to the ring is2 2 2

2 2 2 2 3/ 2 2 2 3/22 20 0 0

cos( ) 2 ( ) 2 ( )

R R R

P

kdq x kQx kQxE dE d d

R x R R x R R xR x

� � �

��

� � � � ��

� � � ��

2 2 3/2 2 2 3/ 2[2 ]

2 ( ) ( )

kQx kQxR

R R x R x�

��

� �

Electric field strength due to a charged circular arc at its centre :

Figure shows a circular arc of radius R which subtend an angle � at its centre. To find electric

field strength at C, we consider a polar segment on arc of angular width d� at an angle � from

the angular bisector XY as shown.The length of elemental segment is Rd� , the charge on this

element Q

dq d��

� ��

�

7

Due to this dq, electric fild at centre of arc C is given as 2

kdqdE

R�

Now electric field component due to this segment sindE � which is perpendicular to the

angular bisector gets cancelled out in integration and net electric field at centre will be along

angular bisector which can be calculated by integrating cosdE � within limits from 2

� to

2

�.

Hence net electric field strength at centre C is / 2

/2

cosC

E dE

�

�

�

� � .

Electric field strength due to a uniformly surface charged disc :

If there is a disc of radius R, charged on its surface with surface charge density � , we wish to

find electric field strength due to this disc at a distance x from the centre of disc on its axis at

point P shown in figure.To find electric field at point P due to this disc, we consider an

elemental ring of radius y and width dy in the disc as shown in figure. The charge on this

elemental ring 2dq ydy� �� [Area of elemental ring 2ds y dy�� ]

Now we know that electric field strength due to a ring of radius R, charge Q. at a distance x

from its centre on its axis can be given as 2 2 3/2

( )

kQxE

x R�

�

Here due to the elemental ring electric field strength dE at point P can be given as

2 2 3/2 2 2 3/ 2

2

( ) ( )

kdqx k y dy xdE

x R x y

� ��

� �

Net electric field at point P due to this disc is given by integrating above expression from 0 to R

as

8

2 2 3/2 2 2 3/2 2 2 2 200 0

0

2 2 1 12 1

2( ) ( )

RR R

k xy dy y dy xE dE k x k x

xx y x y x y x R

� � ��

�

� � � ��

� � � � ��

If R >> x; 0

2Pve E

��

� �

Electric Lines Of Force

Electric lines of electrostatic field have following properties

• Imaginary

• Can never cross each other

• Can never be closed loops

• The number of lines originating or terminating on a charge is proportional to the magnitude of

charge. In rationalised

MKS system 01/ � electric lines are associated with unit charge, so if a body encloses a charge

q, total lines of force associated with it (called flux) will be 0

/q � .

• Lines of force ends or starts normally at the surface of a conductor.

• If there is no electric field there will be no lines of force.

• Lines of force per unit area normal to the area at a point represents magnitude of intensity,

crowded lines represent

Strong field while distant lines weak field.

• Tangent to the line of force at a point in an electric field gives the direction of intensity.

Electric Flux

The word “flux” comes from a Latin word meaning “to flow” and you can consider the flux of a

vector field to be a measure of the flow through an imaginary fixed element of surface in the field.

Electric flux is defined as EE dA� � !��

This surface integral indicates that the surface in question is to be divided into infinitesimal

elements of area dA�

and the scalar quantity E dA!��

is to be evaluated for each element and

summed over the entire surface.

Important points about electric flux

(i) It is a scalar quantity

(ii) Units (V-m) and N-m2/C

Dimensions : [ML3T

–3A

–1]

(iii) The value of � does not depend upon the distribution of charges and the distance between

them inside the closed surface.

Electric Flux through a Circular Disc :

Figure shows a point charge q placed at a distance � from a disc of radius R. Here we wish to

find the electric flux through the disc surface due to the point charge q. We know a point

charge q originates electric flux in radially outward direction. The flux originated in cone

shown in figure passes through the disc surface.

9

To calculate this flux, we consider an elemental ring on disc surface of radius x and width dx as

shown, Area of this ring (strip) is 2dS x dx�� . The electric field due to q at this elemental ring

is given as 2 2( )

kqE

x�

� �If d� Is the flux passing through this elemental ring, then

2 2 3/ 2 2 2 3/ 2

0 00 0

1 cos2 2 2( ) ( )

R R

o

q x dx q x dx qd

x x� � �

� � ��

� ��

� �,

where 2 2

cosR

� ��

�

�

Electric flux through the lateral surface of a cylinder due to a point charge :

Figure shows a cylindrical surface of length L and radius R. On its axis at its centre a point

charge q is placed, Here we wish to find the flux coming out from the lateral surface of this

cylinder due to the point charge q. For this we consider an elemental strip of width dx on the

surface of cylinder as shown. The area of this strip is 2dS R dx�� .

The electric field due to the point charge on the strip can be given as 2 2( )

kqE

x R�

� .

If d� is the electric flux through the strip, then

2

2 2 2 2 3/22 2cos 2 2

( ) ( )

kq R dxd EdS Rdx KqR

x R x Rx R� � � ��

� ��

Total flux through the lateral surface of cylinder

/ 22

2 2 3/ 2 2 20 /2

2 ( ) 4

L

oL

qR dx qd

x R R� �

� �

�

� � ��

� ��

�

This situation can also be easily handled by using the concepts of Gauss’s law.

Gauss’s Law

It relates the total flux of an electric field through a closed surface to the net charge enclosed by

that surface and according to it, the total flux linked with a closed surface is 01/ � times the

charge enclosed by the closed surface i.e., 0

S

qE ds

�! ��

.

10

Regarding Gauss’s law it is worth noting that

" Flux through Gaussian surface is independent of its shape.

" Flux through Gaussian surface depends only on charges present inside Gaussian surface.

" Flux through Gaussian surface is independent of position of charges inside Gaussian surface.

" Electric field intensity at the Gaussian surface is due to all the charges present (inside as well as

outside)

" In a close surface incoming flux is taken negative while outgoing flux is taken positive.

" In a Gaussian surface 0� � does not imply E = 0 but E = 0 at all the points of the surface

implies 0� � .

" Gauss’s law and Coulomb’s law are equivalent, i.e., if we assume Coulomb’s law we can prove

Gauss’s law and vice-versa. To prove Gauss’s law from Coulomb’s law consider a hypothetical

spherical surface [called Gaussian-surface] of radius r with point charge q at its centre as

shown in figure. By Coulomb’s law intensity at a point P on the surface will be, 3

0

1

4E r

r��

.

And hence electric flux linked with area 3

0

1

4

qds E ds r ds

r�� !

�� , here direction of r

� and ds��

are

same, i.e., 2

2 2

0 0

1 1(4 )

4 4S So

q q qE ds ds r

r r�

��

��

� � ,,,,Which is the required result. Though here in

proving it we have assumed the surface to be spherical, it is true for any arbitrary surface

provided the surface is closed,

(a) If a enclosed surface (not enclosing any charge) is placed in an electric field (either uniform or

non- uniform) total flux linked with it will be zero.

(b) If an enclosed surface encloses a charge q, then total flux

linked with the body will be 0

S

qE ds

�! ��

�

11

From this expression it is clear that the flux linked with an enclosed surface is independent of the

shape and size of the body and position of charge inside it. [figure]

Note : So in case of enclosed symmetrical surface with charge at its centre, flux linked with each

half will be 0

1( )

2 2E

q�

��

� � ��

and the symmetrical enclosed surface has n identical faces with point

charge at its centre, flux linked with each face will be 0

E q

n n

��

� ��

� �

" Gauss’s law is a powerful tool for calculating electric field intensity in case of symmetrical

charge distribution by choosing a Gaussian – surface in such a way that E�

is either parallel or

perpendicular to its various faces. As an example, consider the case of a plane sheet of charge

having charge density � . To calculate E at a point P close to it consider a Gaussian surface in

the form of a ‘pill box’ of cross-section S as shown in figure.

The charge enclosed by the Gaussian-surface = S� and the flux linked with the pill box

= ES + 0 + ES = 2ES (as E is parallel to curved surface and perpendicular to plane faces)

So from Gauss’s law, 0

1( )

Eq�

�� ,

0 0

12 ( )

2ES S E

��

� ��

" If 0E ��

, 0E ds� � ! ��

� , so q = 0 but if q = 0, 0E ds! ��

� So E�

may or may not be zero.

If a dipole is enclosed by a closed surface then, q = 0, so 0E ds! ��

� , but 0E #�

.

Note: If instead of plane sheet of charge, we have a charged conductor, then as shown in figure (B)

0in

E � .

So E

� = ES and hence in this case 0

E��

� . This result can be verified from the fact that intensity

at the surface of a charged spherical conductor of radius R is, 2

0

1

4

qE

R�� with 24q R� �� .So for

a point close to the surface of conductor, 2

2

00

1(4 )

4E R

R

��

��

12

Electric field due to solid conducting or hollow sphere:

For outside point (r>R) :

Using Gauss’s theorem 0

qE ds

�$

! ��

� � At every point on the Gaussian

surface ||E ds��

; cos 0E ds E ds E ds! � % ��

0

qE ds

�$

! ��

� [E is constant over the Gaussian surface]

For surface point r = R : 2

04S

qE

R��

For Inside point (r < R) ; Because charge inside the conducting sphere or hollow is zero,

(i.e. 0q$ � )

So 0

0 0in

qE ds E

�$

! � � � ��

�

Electric field due to solid non conducting sphere :

Outside (r > R)

2

2

0 0

44

P

q q qE ds E r E

r�

� � ��$

! � � � � � ��

�

At surface (r = R) 2

04S

qE

r�� , Put r = R

Inside (r < R) : From Gauss’s theorem 0

qE ds

�$

! ��

� Where q$ charge contained within Gaussian

surface of radius r

2

2

0 0

(4 )4

q qE r E

r�

� � �$ $

� � � ………… (i)

As the sphere is uniformly charged, the volume charge density (charge/volume) � is constant

throughout the sphere 34

3

q

R

��

� � charge enclosed in gaussian surface

34

3q r� �� $ � � �

�

33

3 3

4

3(4 / 3)

q qrr q

R R�

�� $ ��

� � ,

13

put this value in equation (i) 3

0

1

4in

qrE

R��

Electric field due to an infinite line distribution of charge:

Let a wire of infinite length is uniformly charged having a constant linear charge density � . P is

the point where electric field is to be calculated. Let us draw a coaxial Gaussian cylindrical

surface of length � .

From Gauss’s theorem

1 2 3

1 2 3

0s s s

qE dS E dS E dS

�! � ! � ! ��

1E dS&��

so 1 0E dS! ��

and 2E dS&��

so 2 0E dS! ��

0

2q

E r��

� ��3

||E dS� ��

�

Charge enclosed in the Gaussian surface

q �� So 0 0

22

E r Er

� ��

� ��

�� or

2kE

r

�� where

0

1

4k

��

Dielectric In Electric Field

Let 0

E�

be the applied field, Due to polarization, electric field is P

E�

. The resultant field is E�

.

For homogeneous and isotropic dielectric, the direction of P

E�

is opposite to the direction of 0

E�

.

So, Resultant field is E = E0 – E

P

Electric Potential

Electric potential is a scalar property of every point in the region of electric field. At a point in

electric field potential is defined as the interaction energy of a unit positive charge. If at a point

in electric field a charge q0 has potential energy U, then electric potential at that point can be

given as 0

UV

q� joule/coulomb

Potential energy of a charge in electric field is defined as work done in bringing the charge

from infinity to the given point in electric field. Similarly we can define electric potential as

“work done in bringing a unit positive charge from infinity to the given point against the

electric forces. So we can say that

r

V E dr'

� !��

Electric Potential due to a point charge in its surrounding

The potential at a point P at a distance r from the charge q 0

P

UV

q� .

Where U is the potential energy of charge q0 at point p, 0

kqqU

r� .

14

Thus potential at point P is P

kqV

r�

Electric Potential due to a charge Rod :

Figure shows a rod of length L, uniformly charged with a charge Q. Due to this we’ll find electric

potential at a point P at a distance r from one end of the rod as shown in figure.

For this we consider an element of width dx at a distance x from the point P.Charge on this

element is

The potential dV due to this element at point P can be given by using the result of a point charge

as kdq kQ

dV dxx Lx

� �

Net electric potential at point P : ln

r L

r

kQ kQ r LV dV dx

Lx L r

� ��

Electric potential due to a charged ring :

Case-I : At Its centre

To find potential at the centre C of the ring, we first find potential dV at centre due to an elemental

charge dq on ring which is given as kdq

dVR

�

Total potential at C is kdq kQ

V dVR R

� � ��

As all dq’s of the ring are situated at same distance R from the ring centre C, simply the potential

due to all dq’s is added as being a scalar quantity, we can directly say that the total electric

potential at ring centre is kQ

R. Here we can also state that even if charge Q is non-uniformly

distributed on ring, the electric potential C will remain same.

Case-II : At a point on axis of ring

We find the electric potential at a point P on the axis of ring as shown, we can directly state the

result as here also all points of ring are at same distance 2 2x R� the point P, thus the potential at

15

P can be given as 2 2P

kQV

R x�

�

Electric potential due to a uniformly charged disc :

Figure shows a uniformly disc of radius R with surface charge density� coul/ m2. To find

electric potential at point P we consider an elemental ring of radius y and width dy, charge on

this elemental ring is 2dq y dy� �� . Due to this ring, the electric potential at point P can be

given as 2 2 2 2

2kdq k y dydV

x y x y

� �! !� �

� �

Net electric potential at Point P due to whole disc can be given as

2 2 2 2

0 2 2 00 0 0

2 2 2

RR y dyV dV x y x R x

x y

� � ��

� � � �� ! � � � � � � � ��

� �

Potential Difference Between Two points in electric field:

Potential difference between two points in electric field can be defined as work done in displacing

a unit positive charge from one point to another against the electric forces.

If a unit +ve charge is displaced from a point A to B as shown work required can be given asB

B AA

V V E dx � !��

If a charge q is shifted from point A to B, work done against electric forces can be given as W = q

(VB – V

A)

If in a situation work done by electric forces is asked, we use W = q (VA – V

B)

If VB < V

A, then charges must have tendency to move toward B (low potential point) it implies

that electric forces carry the charge from high potential to low potential points. Hence we can say

that in the direction of electric field always electric potential decreases.

Electric potential due to hollow or conducting sphere

At outside sphere

According to definition of electric potential, electric potential at point P

2

04

r rq

V E dr drr��' '

� ! � � ��

2

04

out

qE

r��

��

2

0 0 0

1 1

4 4 4

rrq q q

V drr rr�� ''

� ��

At surface : 2

04

R Rq

V E dr drr��' '

� ! � � ��

2

04

out

qE

r��

��

16

2 2

0 0 0

1 1

4 4 4

RRq q q

V drRr r�� ''

� � � ��

Inside the surface � Inside the surface 0dV

Edr

� � � V = constant so 0

4

qV

R��

Electric potential due to solid non conducting sphere

At outside sphere : Same as conducting sphere.

At Surface : Same as conducting sphere.

Inside the sphere

r

V E dr'

� !��

1 2

R r

R

V E dr E dr'

� ��

� ��

2 3

R r

R

kq kqrV dr dr

r R'

� ��

2

3

1

2

rR

R

kq rV kq

r R'

� ��

2 2

3 3

1

2 2

r RV kq

R R R

� ��

� �2 2

33

2

kqV R r

R� ��

Equipotential Surfaces

For a given charge distribution, locus of all points having same potential is called ‘equipotential

surface’.

Equipotential surfaces can never cross each other (otherwise potential at a point will have two

values which is absurd)

Equipotential surfaces are always perpendicular to direction of electric field.

If a charge is moved from one point to the other over an equipotential surface then work done

( ) 0AB AB B A

W U q V V� � � B AV V��

Shapes of equipotential surfaces

The intensity of electric field along an equipotential surface is always zero.

Electric Potential Gradient

The maximum rate of change of potential at right angles to an equipotential surface in an

electric field is defined as potential gradient. E V grad V� ( � � �

Note : Potential is a scalar quantity but the gradient of potential is a vector quantity

In Cartesian co-ordinates ˆ ˆV V VV i j k

x y z

� �) ) )( � � �� ) ) )� �

�

Electrostatic Potential Energy

Potential energy of a system of particles is defined only in conservative fields. As electric field is

also conservative, we define potential energy in it. Potential energy of a system of particles we

define as the work done in assembling the system in a given configuration against the interaction

forces of particles. Electrostatic potential energy is defined in two ways,

17

(i) Interaction energy of charged particles of a system

(ii) Self energy of a charged object

Electrostatic Interaction Energy :

Electrostatic interaction energy of a system of charged particles is defined as the external work

required to assemble the particles from infinity to the given configuration. When some charged

particles are at infinite separation, their potential energy is taken zero as no interaction is there

between them. When these charges are brought close to a given configuration, external work is

required if the force between these particles is repulsive and energy is supplied to the system,

hence final potential energy of system will be positive. If the force between the particle is attractive,

work will be done by the system and final potential energy of system will be negative.

Interaction Energy of a system of two charged particles :

Figure shows two + ve charges q1 and q

2 separated by a distance r. The electrostatic interaction

energy of this system can be given as work done in bringing q2 from infinity to the given separation

from q1.

It can be calculated as 1 2

2

r rkq q

W F dx dxx' '

� " � � ��

[–ve sign shows that x is decreasing]

W = 1 2kq q

r = U [ interaction energy]

If the two charges here are of opposite sign, the potential energy will be negative as 1 2kq qU

r�

Interaction Energy for a system of charged particles :

When more than two charged particles are there in a system, the interaction energy can be given

by sum of interaction energies of all the pairs of particles. For example if a system of three

particles having charges q1, q

2 and q

3 is given as shown in figure.The total interaction energy of

this system can be given as 1 3 2 31 2

3 2 1

kq q kq qkq qU

r r r� � �

Self Energy of charged object :

Electrostatic self energy of a charged object is defined as the external work required to assemble

the charge on the object bringing it from infinity.

Self energy of uniformly charged hollow sphere

Let at any time charge on sphere be ‘q’work done in bringing

a small charge dq from infinity and adding it to the sphere will be

kqdW dq

R

� ��

So total work required will be, 2

02

Qkq kQ

W dqR R

� ��

So self energy of uniformly charged hollow sphere is

18

Self energy of uniformly charged solid sphere

Let at any time charge accumulated on sphere be ‘q’ upto

radius ‘r’work done in bringing a small charge dq from

infinity and adding it to the sphere will be,

324

( 4 )3

kq k rdW dq r dr

r r

��

� ��

,

where 34

3

Q

r�

��

� ��

(volume charge density)

So total work required will be,

3

0

4

3

Rk r

Wr

��

� ��

� � 2

( 4 )r dr� �2

3

5

kQ

R�

So self energy of uniformly charged solid sphere is2

3

5

kQ

R

Electric Dipole

A system of two equal and opposite charges separated by a certain distance is called electric

dipole, shown in figure. Every dipole has a characteristic property called dipole moment. It is

defined as the product of magnitude of either charge and the separation between the charges,

given as

In some molecules, the centres of positive and negative charges do not coincide. This results in

the formation of electric dipole. Atom is non-polar because in it the centres of positive and negative

charges coincide. Polarity can be induced in an atom by the application of electric field. Hence it

can be called as induced dipole.

Dipole Moment: Dipole moment p = q d

(i) Vector quantity , directed from negative to positive charge

(ii) Dimension : [LTA], Units ; coulomb � metre (or C-m)

(iii) Practical unit is “debye” = Two equal and opposite point charges each having charge 10–10

Frankline (= e) and separation of o

1A then the value of dipole moment ( p�

) is 1 debye.

1 Debye = 10 1010 10 Fr m � �20 30

910 3.3 10

3 10

C mC m

��

��

Dipole Placed In Uniform Electric Field :

Figure shows a dipole of dipole moment p�

placed at an angle � to the direction of electric

field. Here the charges of dipole experience force of magnitude qE in opposite direction as

shown. [ ( ) ] 0netF qE q E� � ��

19

Thus we can state that when a dipole is placed in a uniform electric field, net force on the

dipole is zero. But as equal and opposite forces act with a separation in their line of action, they

produce a couple which tend to align the dipole along the direction of electric field. The

magnitude of torque due to this couple can be given as

* = Force � separation between lines of actions of forces

= sin sinqE d pE� ��

r F d qE qd E p E* � � � � � � � ��

Work done In Rotation of a Dipole in Electric field :

When a dipole is placed in an electric field at an angle � , the torque on it due to electric field is

sinpE* �� . Work done in rotating an electric dipole from 1

� to 2

� [uniform field]

dW d* �� so W dW d* �� and 1 2 2 1 1 2(cos cos )W U U pE� � � � � ��

Eg. 0 180[1 ( 1)] 2W pE pE� � �

0 90(1 0)W pE pE� � �

If a dipole is rotated from field direction ( 0 )� � % to � then (1 cos )W pE ��

Electrostatic potential energy :

Electrostatic potential energy of a dipole placed in a uniform field is defined as work done in

rotating a dipole from a direction perpendicular to the field to the given direction i.e.,

90

90

sin cosW pE d pE p E

�

� � � �%�

%

� � � "��

.

E�

is a conservative field so whatever work is done in rotating a dipole from 1

� to 2

� is just

equal to change in electrostatic potential energy 1 2 2 1 1 2

(cos cos )W U U pE� � � � � ��

Work done in rotating an electric dipole in an electric field :

Suppose at any instant, the dipole makes an angle� with the electric field. The torque acting on

dipole.

( 2 sin ) sinqEd q E pE* � �� ! ��

The work done in rotating dipole from 1

� to 2

�

2 2

1 1

sinW d pE d

� �

� �

* � � ��

1 2(cos cos )W pE � ��

2 1( cos )U U U pE � � �

Force on an electric dipole in Non-uniform electric field E :

If in a non-uniform electric field dipole is placed at a point where electric field is E, the

interaction energy of dipole at this point U p E� !��

. Now the force on dipole due to electric

field U

Fr

+�

+

20

If dipole is placed in the direction of electric field then dE

F pdx

�

Electric Potential due to DipoleAt axial point

Electric potential due to +q charge 1( )

kqV

r�

�

Electric potential due to –q charge 2( )

kqV

r

�

� �

Net electric potential 1 2 2 2 2 2

2

( ) ( ) ( )

kq kq kq kpV V V

r r r r

��

�

� � � �

If r > > > 2

kpV

r� ��

At equatorial point

Electric potential of P due to +q charge 1

kqV

x�

Electric potential of P due to –q charge 2

kqV

x�

Net potential 1 20

kq kqV V V

x x� � � � 0V� �

At general point 2 3

0 0

cos

4 4

p p rV

r r

��

!� �

� �

Electric field due to an electric dipole :

Figure shows an electric dipole placed on x-axis at origin. Here we wish to find the electric

field and potential at a point O having coordinates ( , )r � . Due to the positive charge of dipole

electric field at O is in radially outward direction and due to the negative charge it is radially

inward.

3

2 cosr

V kpE

r r

�)� �

)and 3

1 sinV kpE

r r�

��

)� �

)

Thus net electric field at point O,

2 2 2

31 3cosnet r

kpE E E

r� ��

If the direction of Enet

is at an angle � from radial direction, then

1 1tan tan

2r

E

E

��

Thus the inclination of net electric field at point 0 is ( )� ��

At a point on the axis of a dipole :

Electric field due to +q charge 1 2( )

kqE

r�

�

Electric field due to –q charge 2 2( )

kqE

r�

� �

21

Net electric field 1 2 2 2 2 2 2

4

( ) ( ) ( )

kq kq kq rE E E

r r r

��

�

�

� � � [ 2p q� �� = Dipole moment]

If r >>> � then 3

2kpE

r�

At a point on equitorial line of dipole

Electric field due to +q charge 1 2

kqE

x� ;

Electric field due to –q charge 2 2

kqE

x�

Vertical component of E1 and E

2 will cancel each other and horizontal components will be added

So net electric field at P

1 2cos cosE E E� ��

1 2[ ]E E��

1 2

22 cos cos

kqE E

x� �� cos

x� ��

� and 2 2x r� � �

3 2 2 3/ 2 2 2 3/ 2

2 2

( ) ( )

kq kq kpE

x r r� � �

� �

� �

� �

If r ,,, � , then 3

kpE

r� or 3

kpE

r

��

�

Electrostatic Pressure :

Force due to electrostatic pressure is directed normally outwards to the surface.Force on small

element ds of charged conductordF = (Charge on ds) � Electric field = 2

0 0

( )2 2

ds ds� �

��

�

Inside 1 2 1 2

0E E E E � � � Just outside 1 2 2 2

0

22

E E E E E��

� � � � � (E1 is field due to point chargege

on the surface and E2 is field due to rest of the sphere).

The electric force acting per unit area of charged surface is defined as electrostatic pressure.

2

02

electrostatics

dFP

dS

��

� �

Equilibrium of liquid charged surfaces (Soap bubble)

Pressure (forces) act on a charged soap bubble , due to

(i) Surface tension PT (inward)

(ii) Air outside the bubble Po (inward)

(iii) Electrostatic pressure Pe (outward)

(iv) Air inside the bubble P1

(outward)

in state of equilibrium inward pressure = outward pressure PT + P

o = P

1 + P

e

Excess pressure of air inside the bubble (Pex

) = P1 – P

o = P

T – P

e

22

Conductor And Its Properties [For Electrostatic Condition]

(i) Conductors are materials which contains large number of free electrons which can move freely

inside the conductor.

(ii) In electrostatics, conductors are always equipotential surfaces,

(iii) Charge always resides on surface of conductor.

(iv) If there is a cavity inside the conductor having no charge then charge will always reside only

on outer surface of conductor.

(v) Electric field is always perpendicular to conducting surface.

(vi) Electric field Intensity near the conducting surface is given by formula 0

ˆE n��

��

0 0

ˆ ˆ;A B

A BE n E n� ��

� ��

and 0

ˆC

CE n��

��

(vii) When a conductor is grounded Its potential becomes zero.

(viii) When two conductors are connected there will be charge flow till their potential becomes

equal.

(ix) Electric pressure at the surface of a conductor is given by formula2

02

P��

� , where� is the

local surface charge density.

Some other important results for a closed conductor

(i) If a charge q is kept in the cavity then –q will be induced on the inner surface and +q will

be induced on the outer surface of the conductor (it can be proved using gauss theorem)

(ii) If a charge q is kept inside the cavity of a conductor and conductor is given a charge Q then –q

charge will be induced on inner surface and total charge on the outer surface will be q + Q. (it can

be proved using gauss theorem)

(iii) Resultant field, due to q (which is inside the cavity) and induced charge on S1 at any point outside

S1 (like B,C) is zero. Resultant field due to q + Q on S

2 and any other charge outside S

2,,,, at any

point inside of surface S2 (like A, B) is zero

23

(iv) Resultant field in a charge free cavity in a closed conductor is zero. There can be charges

outside the conductor and on the surface also. Then also this result is true. No charge will be

induced on the inner most surface of the conductor.

(v) Charge distribution for different types of cavities in conductors

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

Note : In all cases charge on inner surface S1 = –q and on outer surface S

2 = 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 S

2may change if some charges(s) is/are kept

outside the conductor.

Sharing of Charges :

Two conducting hollow spherical shells of radii R1 and R

2 having charges Q

1 and Q

2 respectively

and separated by large distance, are Joined by a conducting wire. Let final charges on spheres are

q1 and q

2 respectively,

24

Potential on both spherical shell become equal after Joining, therefore1 2 1 1

1 2 2 2

;Kq Kq q R

R R q R� � ......(i)

and 1 2 1 2q q Q Q� � � ......(ii)

f rom (i) and (i i ) 1 2 1

1

1 2

( )Q Q Rq

R R

��

� ; 1 2 2

2

1 2

( )Q Q Rq

R R

��

�

ratio of charges

2

1 1 1 1 1

2

2 2 22 2

4;

4

q R R R

q R RR

� ��

� �

ratio of surface charge densities 1 1

2 2

R

R

��

�

Ratio of final charges 1 1

2 2

q R

q R�

Ratio of final surface charge densities. 1 2

2 1

R

R

��

�

Example 1. If a charged body is placed near a neutral conductor, will it attract the conductor or repel

it?

Solution :

If a charged body (+ve) is placed left side near a neutral conductor, (–ve) charge will

induce at left surface and (+ve) charge will induce at right surface. Due to positively

charged body –ve induced charge will feel attraction and the +ve induced charge will feel

repulsion. But as the –ve induced charge is nearer, so the attractive force will be greater

than the repulsive force. So the net force on the conductor due to positively charged body

will be attractive. Similarly, we can prove for negatively charged body also.

From the above example we can conclude that. “A charged body can attract a neutral

body.”

If there is attraction between two bodies then one of them may be neutral. But if there is

repulsion between two bodies, both must be charged (similarly charged).So “repulsion is

the sure test of electrification”.

Example 2. A positively charged body ‘A’ attracts a body ‘B’ then charge on body ‘B’ may be:

(A) positive (B) negative (C) zero (D) can’t say

Answer : B, C

Example 3. Charge conservation is always valid. Is it also true for mass?

Solution : No, mass conservation is not always. In some nuclear reactions, some mass is lost and it

is converted into energy.

25

Example 4. What are the differences between charging by induction and charging by conduction ?

Solution : Major differences between two methods of charging are as follows :

(i) In induction, two bodies are close to each other but do not touch each other while in

conduction they touch each other. (Or they are connected by a metallic wire)

(ii) In induction, total charge of a body remains unchanged while in conduction it changes.

(iii) In induction, induced charge is always opposite in nature to that of source charge

while in conduction charge on two bodies finally is of same nature.

Example 5. Find out the electrostatic force between two point charges placed in air (each of +1 C) if

they are separated by 1m.

Solution : Fe =

1 2

2

kq q

r=

9

2

9 10 1 1

1

� � �= 9×109 N

From the above result, we can say that 1 C charge is too large to realize. In nature, charge

is usually of the order of -C

Example 6. A particle of mass m carrying charge q1 is revolving around a fixed charge –q

2 in a circular

path of radius r. Calculate the period of revolution and its speed also.

Solution :0

1

4��1 2

2

q q

r = mr.2 =

2

2

4 mr

T

�’

T2 =

2 2

0

1 2

(4 ) (4 )r mr

q q

�� or T = 4�r

0

1 2

mr

q q

��

and also we can say that

1 2

2

04

q q

r�� =

2mv

r� V =

1 2

04

q q

mr��

Example 7. A point charge qA = + 100 µc is placed at point A (1, 0, 2) m and another point charge

qB= +200µc is placed at point B (4, 4, 2) m. Find :

(i) Magnitude of electrostatic interaction force acting between them

(ii) Find AF�

(force on A due to B) and BF�

(force on B due to A) in vector form

Solution : (i)

Value of F : 2

A Bkq q

Fr

��

=

9 6 6

22 2 2

(9 10 ) (100 10 ) (200 10 )

(4 1) (4 0) (2 2)

� � �

� � = 7.2 N

(ii) Force on ,BF�

= 3| |

A Bkq qr

r

��

26

=

9 6 6

32 2 2

(9 10 )(100 10 )(200 10 ) ˆˆ ˆ(4 1) (4 0) (2 2)

(4 1) (4 0) (2 2)

i j k � � � � � � � � �

� � = 7.23 4ˆ ˆ5 5

i j� ��

N

SimilarlyAF�

= 7.2N3 4ˆ ˆ5 5

i j� � � ��

N

Action(AF�

)and Reaction (BF�

) are equal but in opposite direction.

Example 8. Three equal point charges of charge +q each are moving along a circle of radius R and a point

charge –2q is also placed at the centre of circle (as shown in figure). If charges are revolving

with constant and same speed in the circle then calculate speed of charges

Solution : F2 – 2F

1 cos 30º =

2mv

R� 2

( ) (2 )K q q

R–

2

2

2( )

( 3 )

Kq

Rcos 30 =

2mv

R

�/2 1

23

kqv

Rm

� ��

� �

Example 9. Five point charges, each of value q are placed on five vertices of a regular hexagon of side

L. What is the magnitude of the force on a point charge of value –q coulomb placed at the

centre of the hexagon?

Solution : Method-I : If there had been a sixth charge +q at the remaining vertex of hexagon,

force due to all the six charges on –q at O would have been zero (as the forces due to

individual charges will balance each other), i.e., 0RF ��

27

Now if f�

is the force due to sixth charge and F�

due to remaining five charges.

From F�

+ f�

= 0 i.e. F� = – f

�

or, |F| = | f | =0

1

4�� 2

q q

L

� =

2

2

0

1

4

q

L��

NetF�

= ODF�

=

2

2

1

4

q

L�� 0 along OD

Method-II :In the diagram, we can see that force due to charge A and D are opposite to

each other

OFF�

+ OCF�

= 0�

....(i)

Similarly OBF�

+OEF�

= 0�

....(ii)

SoOFF�

+ OBF�

+OCF�

+ODF�

+ OEF�

= NetF

�

Using (i) and (ii)NetF� =

ODF�

=

2

2

1

4

q

L�� 0 along OD.

Example 9 A thin straight rod of length l carrying a uniformly distributed charge q is located in vacuum.

Find the magnitude of the electric force on a point charge ‘Q’ kept as shown in the figure.

Solution : As the charge on the rod is not point charge, therefore, first we have to find force on

charge Q due to charge over a very small part on the length of the rod. This part, called

element of length dy can be considered as point charge.

Charge on element, dq = �dy =q

dy�

Electric force on ‘Q’ due to element

28

= 2

. .K dq Q

y = 2

. . .

.

K Q q dy

y �

All forces are along the same direction,

� F = dF� . This sum can be calculated using integration,

therefore, F =

a+l

2

y=a

KQqdy

y��

=KqQ

�=

a

a

1

y

��

� �

�

= KQ.q 1 1

a a

� � � �� =

KQq

( )a a � �

Note : (1)The total charge of the rod cannot be considered to be placed at the centre

of the rod as we do in mechanics for mass in many problems.

Note : (2) If a >>l then,F = 2

KQq

a

i.e.Behavior of the rod is just like a point charge.

Example 10. Two equal positive point charges ‘Q’ are fixed at points B(a, 0) and A(–a, 0). Another test

charge q0 is also placed at O(0, 0). Show that the equilibrium at ‘O’ is

(i) Stable for displacement along X-axis.

(ii) Unstable for displacement along Y-axis.

Solution : (i)

Initially AOF�

+ BOF�

= 0 � | |AOF�

= | |BOF�

= 0

2

KQq

a

When charge is slightly shifted towards + x axis by a small distance +x, then.

| |AOF�

< | |BOF�

Therefore, the particle will move towards origin (its original position). Hence, the

equilibrium is stable.

29

(ii) When charge is shifted along y axis:

After resolving components, net force will be along y axis So, the particle will not

return to its original position & it is unstable equilibrium. Finally, the charge will

move to infinity.

Example 11. A particle of mass m and charge q is located midway between two fixed charged

particles each having a charge q and a distance 2� apart. Prove that the motion of the

particle will be SHM if it is displaced slightly along the line connecting them and released.

Also find its time period.

Solution : Let the charge q at the mid-point is displaced slightly to the left. The force on the displaced

charge q due to charge q at A,

F1=

0

1

4��

2

2( )

q

x��

The force on the displaced charge q due to charge at B,

F2 =

0

1

4��

2

2( )

q

x �

Net restoring force on the displaced charge q.

F = F2 – F

1 or F =

0

1

4��

2

2( )

q

x �–

0

1

4��

2

2( )

q

x��

or F =

2

04

q

�� 2 2

1 1

( x) ( x)

� � �

�� =

2

04

q

�� 2 2 2

4

( )

x

x �

�

Since �>> x, � F =

2

4

0

q x

��

� or F =

2

3

0

q x

��

Hence we see that F � x and it is opposite to the direction of displacement. Therefore, the

motion is SHM.

30

T = 2m

k� , (here k =

2

3

0

q

� � )T=

3

0

22

m

q

��

�

Example 12. Find out mass of the charge Q, so that it remains in equilibrium for the given

configuration.

Solution :

� 4 Fcos� = mg

� 4×3/2

22

2

KQq

h� �

��

� h = mg � m = 3/2

22

4

2

KQqh

g h� �

��

�

Example 13. Electrostatic force experienced by –3-C charge placed at point ‘P’ due to a system ‘S’

of fixed point charges as shown in figure is ˆ ˆ(21 9 )F i j� ��

µN.

(i) Find out electric field intensity at point P due to S.

(ii) If now, 2-C charge is placed and –3 -C is removed at point P then force experienced by it

will be.

Solution : (i) F qE��

� ˆ ˆ(21 9 )i j µN� = –3µC ( )E�

�/ E�

= – 7 i – 3 jN

C

(ii) Since the source charges are not disturbed the electric field intensity at ‘P’ will remain

same.

F�

2-C = +2( E

�) = 2(–7 i – 3 j )= (–14 i – 6 j ) -N

Example 14. Calculate the electric field intensity which would be just sufficient to balance the weight

of a particle of charge –10 -C and mass 10 mg. (Take g = 10 ms2)

Solution : As force on a charge q in an electric field E�

is F�

q = q E�

So, according to given problem:

[W = weight of particle]

31

| | | |qF W��

i.e., |q|E = mg

i.e., E =| |

mg

q= 10 N/C., in downward direction.

Example 15. Find out electric field intensity at point A (0, 1m, 2m) due to a point charge –20-C

situated at point B( 2 m, 0, 1m).

Solution : E = 3| |

KQr

r

�� = 2

ˆ| |

KQr

r� �/ r

� = P.V. of A – P.V. of B (P.V. = Position vector)

= ˆˆ ˆ– 2i j k� � | r� | = 2 2 2( 2) (1) (1)� � = 2

E =

9 69 10 ( 20 10 )

8

� � � ˆˆ ˆ– 2i j k� � = – 22.5 × 103ˆˆ ˆ– 2i j k� � N/C.

Example 16. Two point charges 2-c and – 2-c are placed at points A and B as shown in figure. Find out

electric field intensity at points C and D. [All the distances are measured in meter].

Solution : Electric field at point C (EA, E

B are magnitudes only and arrows represent directions).

Electric field due to positive charge is away from it while due to negative charge, it is

towards the charge. It is clear that EB> E

A.

� ENet

= (EB – E

A) towards negative X-axis

= 2 2

(2 ) (2 )

( 2) (3 2)

K c K c- - towards negative X-axis = 8000 ˆ(– )i N/C

Electric field at point D :

Since magnitude of charges are same and also

AD = BD

So, EA = E

B

Vertical components of AE�

and BE�

cancel each other while horizontal components are in

the same direction.

So, Enet

= 2EA cos� = 2

2. (2 )

2

K c- cos45°

32

=

610

2

K � =

9000 ˆ2

i N/C.

Example 17. Six equal point charges are placed at the corners of a regular hexagon of side ‘a’. Calculate

electric field intensity at the centre of hexagon?

Answer : Zero (By symmetry)

Similarly electric field due to a uniformly charged ring at the centre of ring :

Example 18. Find out electric field intensity at the centre of a uniformly charged semicircular ring of

radius R and linear charge density �.

Solution : � = linear charge density.

The arc is the collection of large no. of point charges. Consider a part of ring as an element

of length Rd� which subtends an angle d� at centre of ring and it lies between � and � + d�

dE��

= dEx i

+ dEy j ; E

x = 0xdE �� (due to symmetry)

& Ey = ydE� =

0

sindE

�

�� =K

R

�

0

sin .d

�

� �� = 2K

R

�

33

Example 19. Positive charge Q is distributed uniformly over a circular ring of radius a. A point particle

having a mass m and a negative charge –q, is placed on its axis at a distance y from the

centre. Find the force on the particle. Assuming y << a, find the time period of oscillation

of the particle if it is released from there. (Neglect gravity)

Solution : When the negative charge is shifted at a distance x from the centre of the ring along its

axis then force acting on the point charge due to the ring :

FE = qE (towards centre) = q 2 2 3/2( )

KQy

a y

� ��

If a >> y thena2 + y2~ a2

� FE =

0

1

4�� 3

Qqy

a (Towards centre)

Since, restoring force FE� y, therefore motion of charge the particle will be

S.H.M.

Time period of SHM

T = 2�m

k= 2�

3

04

m

Qq

a��

=

1/23 3

016 ma

Qq

� ��

Example 20. Calculate electric field at a point on axis, which at a distance x from centre of uniformly

charged disc having surface charge density �/and R which also contains a concentric hole

of radius r.

Solution : Consider a ring of radius y (r < y < R)and width dy concentric with disc and in the plane

of the disc. Due to this ring, the electric field at the point P :

dE = 2 2 3/2

( )

[ ]

K dq x

X Y�

34

Enet

= 2 2 3/2

. (2 )

[ ]

R

r

Kx y dy

x y

� �� [1dq = �2�ydy]

Enet

= 2

2

kx��2 2

2 2

3/2

x R

x r

dt

t

�

�� , put x2 + y2 = t, 2y. dy = dt

= 02

x�� 2 2 2 2

1 1

x r x R

� � �

� �� away from centre

Alternate method

We can also use superposition principle to solve this problem.

(i) Assume a disc without hole of radius R having surface charge density + �.

(ii) Also assume a concentric disc of radius r in the same plane of first disc having charge

density – �.

Now using derived formula in last example the net electric field at the centreis :

netE�

=RE

� +

rE�

= 02

x�� 2 2 2 2

1 1

r x R x

� � �

� �� away from centre.

Example 21. A point charge q is placed at a distance r from a very long charged thread of uniform linear

charge density l. Find out total electric force experienced by the line charge due to the point

charge. (Neglect gravity).

Solution : Force on charge q due to the thread,

F =2K

r

��

.q

By Newton’s III law, every action has equal

and opposite reaction So, force on the

thread =2

.K

qr

� (away from point

charge)

Example 22. Figure shows a long wire having uniform charge density/�/as shown in figure.

Calculate electric field intensity at point P.

37°

�

r

Solution : q1 = 90º and q

2 = 360º – 37º So

Ex =

K

r

� [sinq

1+sinq

2] ;

Ey =

K

r

�[cosq

2– cosq

1]

35

Example 23. Find electric field at point A, B, C, D due to infinitely long uniformly charged wire with linear

charge density l and kept along z-axis (as shown in figure).Assume that all theparameters are

in S.I. units.

Solution : EA

= 2 K ˆ( )

3i

�� E

B =

2 K ˆ( )4

j�

EC

=2 K

5

�OC

2

=ˆ ˆ2 K 3 4

5 5

i j� � ��

ED

= ˆ ˆ2 K 3 4

5 5

i j� � ��

� ED

= EC

Example 24. An infinitely large plate of surface charge density +� is lying in horizontal xy-plane. A

particle having charge –qo and mass m is projected from the plate with velocity u making

an angle � with sheet. Find :

(i) The time taken by the particle to return on the plate..

(ii) Maximum height achieved by the particle.

(iii)At what distance will it strike the plate (Neglect gravitational force on the particle)

Solution :

Electric force acting on the particle Fe = q

oE : F

e = (q

o)

2 o

��

� ��

downward

So, acceleration of the particle : a = eF

m =

2

o

o

q

m

�� = uniform

This acceleration will act like ‘g’ (acceleration due to gravity)

So, the particle will perform projectile motion.

36

(i) T =2 sinu

g

� =

2 sin

2o

o

u

q

m

��

��

(ii) H =

2 2sin

2

u

g

� =

2 2

2

sin

2

o

o

u

q

m

��

��

(iii) R =

2 sin 2u

g

� =

2 sin 2

2

o

o

u

q

m

��

��

Example 25. If an isolated infinite sheet contains charge Q1 on its one surface and charge Q

2 on its

other surface then prove that electric field intensity at a point in front of sheet will be

2 O

Q

A� , where Q = Q1 + Q

2

Solution : Electric field at point P :

E�

= 1 2Q QE E��

=1

0

ˆ2

Qn

A� +2

0

ˆ2

Qn

A�

= 1 2

0

ˆ2

Q Qn

A��

= 0

ˆ2

Qn

A�

[This shows that the resultant field due to a sheet depends only on the total charge of the

sheet and not on the distribution of charge on individual surfaces].

Example 26. Three large conducting parallel sheets are placed at a finite distance from each other as

shown in figure. Find out electric field intensity at points A, B, C & D.

Solution : For point A :

3 –2net Q Q QE E E E� � ��

= –0 0 0 0

Q 3Q 2Q Qˆ ˆ ˆ ˆi – i i – i2A 2A 2A A

� ��

37

For point B:

3 –2net Q Q QE E E E� � ��

= –0 0 0

3 2ˆ ˆ ˆ2 2 2

Q Q Qi i i

A A A� � �� = 0

�

For point C :

3 –2net Q Q QE E E E� � ��

= +0 0 0 0

3 2ˆ ˆ ˆ ˆ– –2 2 2

Q Q Q Qi i i i

A A A A� � � ��

For point D :

3 –2net Q Q QE E E E� � ��

= +0 0 0 0

3 2 2ˆ ˆ ˆ ˆ–2 2 2

Q Q Q Qi i i i

A A A A� � � ��

Example 27. Determine and draw the graph of electric field due to infinitely large non-conducting sheet of

thickness t and uniform volume charge density r as a function of distance x from its symmetry

plane.

(a) x 3/2

t(b) x 4/

2

t

Solution : We can consider two sheets of thickness2

tx

� � � ��

and 2

tx

� ��

Where the point P lies inside the sheet.

Now, net electric field at point P :

E = E1 – E

2 =

1

02

Q

A� –2

02

Q

A� [Q1 : charge of left sheet; Q

2 : charge of right sheet.]

38

=

0

2 2

2

t tA x A x

A

� �

�

� � � �� =

0

x��

For point which lies outside the sheet we can consider a complete sheet of

thickness t

55�02

Q

A�

E =02

tA

A

�� =

02

t��

Alternate : We can assume thick sheet to be made of

large number of uniformly charged thin sheets. Consider

an elementary thin sheet of width dx at a distance x

from symmetry plane.

Charge in sheet = rAdx (A : assumed area of sheet)

Surface charge density, � = Adx

A

�

so, electric field intensity due to elementary sheet : dE = 02

dx��

(a) When x <2

t/� E

Net =

0/22

x

t

dx��

� –

/2

02

t

x

dx�� =

0

x��

(b) When x >2

t� E

Net =

/2

0/22

t

t

dx��

� = 02

t��

Example 28. Thin infinite sheet of width wcontains uniform charge distribution s. Find out electric

field intensity at following points :

(a) A point which lies in the same plane at a distance d from one of its edge.

(b) A point which is on the symmetry plane of sheet at a perpendicular distance d

from it.

39

Solution :

(a) Consider a thin strip of width dx. Linear charge density of strip : �/�/�dx

So, electric field due to this strip at point P

dE =2k dx

x

�

Enet

=02

d w

d

dx

x

��

�

�

= 02

d wn

d

��

��

�

(b) Consider a thin strip of width dx. Linear charge density of strip :

�/�/�dx

� Ep = 2 cosdE ��

or Ep= 2.

/2

2 2 2 20

0

.2

w dx d

d x d x

�

��

= /2

2 200

wd dx

d x

�� =

0

�� tan–1

Example 29. Figure shows a uniformly charged sphere of radius R and total

charge Q. A point charge q is situated outside the sphere at a

distance r from centre of sphere. Find out the following

(i) Force acting on the point charge q due to the sphere.

(ii) Force acting on the sphere due to the point charge.

Solution : (i) Electric field at the position of point charge 2ˆ

KQE r

r��

So, 2ˆ

KqQF r

r��

2| |

KqQF

r�

�

(ii) Since we know that every action has equal and opposite reaction so

2ˆ

sphere

KqQF r

r�

�

2sphere

KqQF

r�

�

Example 30. Figure shows a uniformly charged thin sphere of total charge

Q and radius R. A point charge q is also situated at the centre

of the sphere. Find out the following

(i) Force on charge q

(ii) Electric field intensity at A.

(iii) Electric field intensity at B.

40

Solution : (i) Electric field at the centre of the uniformly charged hollow sphere = 0

So force on charge q = 0

(ii) Electric field at A

AE�

= Sphere qE E��

= 0 + 2

Kq

r; r = CA

E due to sphere = 0 , because point lies inside the charged hollow sphere.

(iii) Electric field B

E�

at point B = Sphere qE E��

= 2 2ˆ ˆ. .

KQ Kqr r

r r� = 2

( )ˆ.

K Q qr

r

� ; r = CB

Example 31. Two concentric uniformly charged spherical shells of radius

R1 and R

2 (R

2> R

1) have total charges Q

1 and Q

2 respectively.

Derive an expression of electric field as a function of r for

following positions.

(i) r < R1

(ii) R1 3/r < R

2(iii) r 4/R

2

Solution : (i) For r < R1, therefore, point lies inside both the spheres

Enet

= EInner

+ Eouter

= 0 + 0

(ii) For R1 3 r < R

2, point lies outside inner sphere but inside outer sphere:

� Enet

= Einner

+ Eouter

1

2

KQ

r r + 0 = 1

2ˆ

KQr

r

(iii) For r 4/R2

point lies outside inner as well as outer sphere.

Therefore, ENet

= Einner

+ Eouter

=1

2ˆ

KQr

r +

2

2ˆ

KQr

r =

1 2

2

( )ˆ

K Q Qr

r

�

Example 32. A spherical shell having charge +Q (uniformly distributed) and a point charge + q0 are

placed as shown. Find the force between shell and the point charge(r >> R).

(i) Force on the point charge + q0 due to the shell = q E

�0 shell

= (q0) 2

ˆKQ

rr

� ��

= 0

2ˆ

KQqr

r where r , is unit vector along OP..

From action - reaction principle, force on the shell due to the point charge will

also be :Fshell

=0

2ˆ( )

KQqr

r

Conclusion : To find the force on a hollow sphere due to outside charges, we can

replace the sphere by a point charge kept at centre.

41

Example 33. A Uniformly charged solid non-conducting

sphere of uniform volume charge density r and

radius R is having a concentric spherical cavity

of radius r. Find out electric field intensity at

following points, as shown

in the figure :

(i) Point A (ii) Point B (iii) Point C (iv) Centre of the sphere

Solution : Method-I :

(i) For point A :We can consider the solid part of sphere to be made of large number of

spherical shells which have uniformly distributed charge on its surface. Now, since

point A lies inside all spherical shells so electric field intensity due to all shells will be

zero.

AE��

= 0

(ii) For point B :All the spherical shells for which point B lies inside will make electric

field zero at point B. Soelectric field will be due to charge present from radius r to OB.

So,B

E��

=

3 3

3

4( )

3K OB r

OB

� � OB��

=03

��

3 3

3

[ ]OB r

OB

OB��

(iii)For point C, similarly we can say that for all the shell points C lies outside the

shell

So,CE��

=

3 343

3

[ ( )]

[ ]

K R r

OC

� OC��

= 03

��

3 3

3[ ]

R r

OC

OC��

Method-II :We can consider that the spherical cavity is filled with charge density r and

also –r, thereby making net charge density zero after combining. We can consider two

concentric solid spheres: One of radius R and charge density r and other of radius r and

charge density –r. Applying superposition principle :

+ ®

(i)A

E��

= E�

��+ E �

��=

0

( )

3

OA��

��

+0

[ ( )]

3

OA��

��

= 0

(ii)BE��

= E�

�� + E �

�� =

0

( )

3

OB��

��

+

3

3

4( )

3

( )

K r

OBOB

� ��

=

3

3

0 03 3 ( )

rOB

OB

� ��

� � �

� �

��

=

3

3

0

13

rOB

OB

��

� � �

� �

��

42

(iii)CE��

= E�

�� + E �

�� =

3

3

4

3K R

OC

� �� OC

�� +

3

3

4( )

3K r

OC

� �� OC

��

= 3

03 ( )OC

��

3 3[ ]R r OC ��

(iv)O

E��

= E�

��+ E �

�� = 0 + 0= 0

Example 34. In above question, if cavity is not concentric and centered at point P then repeat all the

steps.

Solution : Again assume r and –r in the cavity, (similar to the previous example) :

(i)AE��

= E�

�� + E �

��

=0

[ ]

3

OA��

��

+0

( )

3

PA��

��

03

�� [ OA��

– PA��

] =0

3

�� [ ]OP��

Note :Here, we can see that the electric field intensity at point A is independent of position

of point Ainside the cavity. Also the electric field is along the line joining the centres of

the sphere and the spherical cavity.

(ii) B

E��

= E�

�� + E �

�� =

0

( )

3

OB��

��

+

343

3

[ ( )]

[ ]

K r

PB

� � PB��

(iii) CE��

= E�

��+ E �

�� =

343

3

[ ]

[ ]

K R

OC

� �OC��

+

343

3

[ ( )]

[ ]

K r

PC

� � PC��

(iv) O

E��

= E�

�� + E �

�� = 0 +

343

3

[ ( )]

[ ]

K r

PO

� � PO��

Example 35. A non-conducting solid sphere has volume charge density that varies as r = r0 r, where r

0 is

a constantand r is distance from centre. Find out electric field intensities at following

positions.

(i) r < R (ii) r/4 R

Solution : Method I :

(i) For r< R :

The sphere can be considered to be made of large number of spherical shells. Each

shell has uniform charge density on its surface. So the previous results of the spherical

shell can be used. Consider a shell of radius x and thickness dx as an element. Charge

on shell dq = (4px2dx)r0x.

� Electric field intensity at point P due to shell, dE = 2

Kdq

x

Since all the shell will have electric field in same direction

43

�//0

R

E dE� � =0

r

dE� +

R

r

dE�Due to shells which lie between region r < x 3 R, electric field at point P will be

zero.

� 2

0

0

rKdq

Er

� ��

=

2

0

2

0

.4r

K x dx x

r

� �� =

0

2

4 K

r

� � 4

04

r

x� ��

=

2

0

04

r��

(ii) For r 4 R, E�

=0

R

dE� =

2

0

2

0

.4R

K x dx x

r

� �� =

4

0

2

0

ˆ4

Rr

r

��

Method II :

(i) The sphere can be considered to be made of large number of spherical shells. Each

shell has uniform charge density on its surface. So the previous results of the spherical

shell can be used. we can say that all the shells for which point lies inside will make

electric field zero at that point,

So ( )r RE 6

� =

2

0

0

2

(4 )

r

K x dx x

r

� �� =

2

0

0

ˆ4

rr

��

(ii) Similarly, for r/4 R, all the shells will contribute in electric field.Therefore :

( )r RE 6

�=

2

0

0

2

(4 )

R

K x dx x

r

� ��=

4

0

2

0

ˆ4

Rr

r

��

Example 36 A charge 2-C is taken from infinity to a point in an electric field, without changing

its velocity. If work done against electrostatic forces is –40-J, then find the potential

at that point.

Solution : V = extW

q=

40

2

J

C

--

= –20 VV

Example 37 When charge 10-C is shifted from infinity to a point in an electric field, it is found that

work done by electrostatic forces is –10-J. If the charge is doubled and taken again from

infinity to the same point without accelerating it, then find the amount of work done by

electric field and against electric field.

Solution : Wext

)'� p = –w

el)'/�p

= wel)

p �' = 10 -J

because DKE = 0

� Vp =

( )

20

ext pW

C-'�

=10

10

J

C

-- = 1V

So, if now the charge is doubled and taken from infinity then

1 =)

20

ext pw

C-'�

� or Wext

)'/� P = 20 -J

� Wel )'/�P

= –20 -J

44

Example 38 A charge 3-C is released from rest from a point P where electric potential is 20 V then its

kinetic energy when it reaches infinity is :

Solution : Wel = DK = K

f – 0

� Wel)

P �/' = qVP = 60 mJ So, K

f = 60 -J

Example 39. A rod of length l is uniformly charged with charge q. Calculate potential at point P.

Solution : Take a small element of length dx, at a distance x from left end. Potential due to this small element

dV = ( )K dq

x� Total potential /////� V =

0

x

x

kdq

x

�

��

Where dq = q

�dx � V =

x r

x r

qK dx

x

� �

�

� ��

�

� = loge

Kq

�

r

r

��

�

Example 40. Figure shows two rings having charges Q and – 5 Q. Find Potential difference between AA

and B i.e. (VA– V

B).

Solution : VA =

KQ

R + 2 2

5 Q

2

K

R R

�; VV

B =

5 Q

2

K

R

+ 2 2

QK

R R�

From above, we can easily find VA – V

B.

Example 41. Two concentric spherical shells of radius R1 and R

2 (R

2> R

1) are having uniformly

distributed charges Q1 and Q

2 respectively. Find out potential

(i) at point A

(ii)at surface of smaller shell (i.e. at point B)

(iii)at surface of larger shell (i.e. at point C)

45

(iv) at r 3 R1

(v) at R13/r 3/R

2

(vi) at r/4/R2

Solution : Using the results of hollow sphere.

(i) VA =

1

1

KQ

R +

2

2

KQ

R

(ii) VB =

1

1

KQ

R +

2

2

KQ

R

(iii) VC =

1

2

KQ

R +

2

2

KQ

R

(iv) for r £ R1,V =

1

1

KQ

R+

2

2

KQ

R

(v) for R1£ r £ R

2,V =

1KQ

r +

2

2

KQ

R

(vi) for r ³ R2 ,V =

1KQ

r+

2KQ

r

Example 42. 1-C charge is shifted from A to B and it is found that work done by an external force is

40-J in doing so against electrostatic forces, then find potential difference VA – V

B

Solution : (WAB

)ext

= q(VB – V

A) � 40 mJ = 1-C (V

B – V

A) /� V

A – V

B = – 40 V

Example 43. A uniform electric field is present in the positive x-direction. If the intensity of the field is

5N/C then find the potential difference (VB –V

A) between two points A (0m, 2 m) and

B(5 m, 3 m)

Solution : VB – V

A = – E

�. AB

�= – (5 i ).(5 i + j ) = –25V..

The electric field intensity in uniform electric field, E = V

d

++

Where DV = potential difference between two points.

Dd = effective distance between the two points.

(projection of the displacement along the direction of electric field.)

Example 44. Find out following

(i)VA – V

B(ii)V

B – V

C(iii)V

C – V

A(iv)V

D – V

C

(v) VA – V

D(vi)Arrange the order of potential for points A, B, C and D.

Solution : (i) ABV Ed+ � = 20 × 2 × 10–2 = 0.4 so, V

A – V

B = 0.4 V

because In the direction of electric field potential always decreases.

46

(ii) BCV Ed+ � = 20 × 2 × 10–2 = 0.4 so, V

B – V

C = 0.4 V

(iii) CAV Ed+ � = 20 × 4 × 10–2 = 0.8 so, V

C – V

A = – 0.8 V


(iv) DCV Ed+ � = 20 × 0 = 0 so, V

D – V

C = 0

because the effective distance between D and C is zero.

(v) ADV Ed+ � = 20 × 4 × 10–2 = 0.8 so, V

A – V

D = 0.8 V


(vi) The order of potential is :VA> V

B> V

C = V

D.

Example 45. Some equipotential surfaces are shown in figure. What can you say about the magnitude

and the direction of the electric field ?

Solution : Here, we can say that the electric will be perpendicular to equipotential surfaces.

Also, | |E�

= V

d

++

Where, DV = potential difference between two equipotential surfaces.

Dd = perpendicular distance between two equipotential surfaces.

So | |E�

= 2

60

(10sin 37º ) 10 � = 1000 V/m

Now there are two perpendicular directions: either direction 1 or direction 2 as shown in

figure, but since we know that in the direction of electric field, electric potential decreases, so

the correct direction is direction 2.

Hence E = 1000 V/m, making an angle 127° with the x-axis

Example 46. Figure shows some equipotential surfaces produce by some charges. At which point, the

value of electric field is greatest?

Solution : E is larger where equipotential surfaces are closer. ELOF are & to equipotential surfaces.

In the figure, we can see that for point B, they are closer so E at point B is maximum

Example 47. A proton moves from a large distance with a speed u m/s directly towards a free proton originally

at rest. Find the distance of closest approach for the two protons in terms of mass of proton m

and its charge e.

47

Solution : As here the particle at rest is free to move, when one particle approaches the other, due to

electrostatic repulsion other will also start moving and so the velocity of first particle will

decrease while of other will increase and at closest approach, both will move with same

velocity. So, if v is the common velocity of each particle at closest approach, then by

‘conservation of momentum’ of the two proton system.

mu = mv + mv i.e., v = 1

2u

And by conservation of energy, 1

2mu2 =

1

2mv2 +

1

2mv2 +

0

1

4��

2e

r

�1

2mu2 – m

2

2

u� ��

= 0

1

4��

2e

r[as v =

2

u]

�1

4mu2 =

2

04

e

r�� r =

2

2

0

e

m u� �

Example 48. Figure shows an arrangement of three point charges. The total potential energy of this

arrangement is zero. Calculate the ratio q

Q.

Solution : Usys

= 0

1

4��( )( ) ( )

2

qQ q q Q q

r r r

� � � �� = 0

–Q + 2

q – Q = 0 or 2Q =

2

qor

q

Q =

4

1.

Example 49. Three equal charges q each are placed at the corners

of an equilateral triangle of side a.

(i) Find out potential energy of charge system.

(ii) Calculate work required to decrease the side of

triangle to a/2.

(iii)If the charges are released from the shown position and each of them has same mass m

then find the speed of each particle when they lie on triangle of side 2a.

Solution : (i) Method I (Derivation)

Assume all the charges are at infinity initially.

Work done in putting charge q at corner A

� W1 = q(v

f – v

i) = q(0 – 0)

Since potential at A is zero in absence of charges, work done in putting q at corner B

in presence of charge at A :

� W2 = 0

Kq

a

� � � ��

q =

2Kq

a

Similarly work done in putting charge q at corner C in presence of charge at A and B.

48

� W3 = q(v

f – v

i) = q 0

Kq Kq

a a

� �� =

22Kq

a

So, net potential energy PE = W1 + W

2 + W

3 = 0 +

2Kq

a+

22Kq

a=

23Kq

a

Method II (using direct formula):

U = U12

+ U13

+ U23

=

2Kq

a +

2Kq

a +

2Kq

a=

23Kq

a

(ii) Work required to decrease the sides W = Uf – U

i =

23

/ 2

Kq

a –

23Kq

a=

23Kq

a Joules

(iii) Work done by electrostatic forces = Change is kinetic energy of particles.

Ui – U

f = K

f – K

iÞ

23Kq

a –

23

2

Kq

a = 3(

1

2mv2) – 0 Þ v =

2Kq

am

Example 50. A spherical shell of radius R with uniform charge q is expanded to a radius 2R. Find the

work performed by the electric forces and external agent against electric forces in this

process.

Solution : Wext

= Uf – U

i =

2

016

q

R�� –

2

08

q

R�� = –

2

016

q

R��

Welec

= Ui – U

f =

2

08

q

R�� –

2

016

q

R�� =

2

016

q

R��

Example 51. Two non-conducting hollow uniformly charged spheres of radii R1 and R

2 with charge Q

1

and Q2 respectively are placed at adistance r. Find out total energy of the system.

Solution : Utotal

= Uself

+ UInteraction

=

2

1

0 18

Q

R�� +

2

2

0 28

Q

R�� + 1 2

04

Q Q

r��Example 52. Find out energy stored in an imaginary cubical volume of side a in front of a infinitely

large non-conducting sheet of uniform charge density s.

Solution : Energy stored : U= 2

0

1

2E dV�� ; where dV is small volume

� U =2

0

1

2E� dV� � E is constant .

� U =

2

0 2

0

1

2 4

��

� . a3 .=

2 3

08

a��

Example 53. Find out energy stored inside a solid non-conducting sphere of total charge Q and radius

R. [Assume charge is uniformly distributed in its volume.]

Solution : We can consider solid sphere to be made of large number of concentric spherical shells.

Also electric field intensity at the location of any particular shell is constant.

49

� Uinside

=2

00

1

2

R

E dV��Consider an elementary shell of thickness dx and radius x.

Volume of the shell = (4px2dx)

Uinside

=

2

0 30

1

2

R KQx

R� � �

� � �� .4px2 dx = 0

1

2�

2 2

6

4K Q

R

� 4

0

R

x dx�

= 0

6

4

2R

�� 2

2

0(4 )

Q

�� .

5

5

R=

2

040

Q

R�� =

2

10

KQ

R.

Example 54. V = x2 + y. Find E�

.

Solution :V

x

))

c = 2x, 1V

y

)�

) and 0

V

z

)�

)

E�

= –ˆˆ ˆV V V

i j kx y z

� �) ) )� �� ) ) )�

= –(2x i + j )

�/Electric field is non-uniform.

Example 55. For given E�

= ˆ ˆ2 3xi yj� , find the potential at (x, y) if V at origin is 5 volts.

Solution :

5

v

dV E dr� !� ��

= –0

x

xE dx� – 0

y

yE dy�

� V – 5 = –

2 22 3

2 2

x y � V = –

2 22 3

2 2

x y + 5.

Example 56. The electric field due to a short dipole at a distance r, on the axial line, from its mid

point is the same as that of electric field at a distance r’, on the equatorial line, from its

mid-point. Determine the ratio ´

r

r.

Solution :0

1

4�� 3

2 p

r=

0

1

4�� 3'

p

ror 3

2

r = 3

1

´ror

3

3´

r

r = 2 or,

´

r

r = 21/3

Example 57. (i) Find potential at point A and B due to the small charge - system fixed near

origin.(Distance between the charges is negligible).

(ii) Find work done to bring a test charge q0 from point A to point B, slowly. All parameters

are in S.I. units.

Solution : (i) Dipole moment of the system is P�

= (qa) i + (qa) j

Potential at point A due to the dipole

50

VA = K 3

( · )P r

r

� �

= 3

ˆ ˆ ˆ ˆ[( ) ( ) ]·(4 3 )

5

K qa i qa j i j� � =

( )7

125

k qa

� VB = 3

ˆ ˆ ˆ ˆ( ) ( ) · 3 4

(5)

K qa i qa j i j� �� =

( )

125

K qa

(ii) WA ® B

= UB – U

A = q

0 (V

B – V

A) = q

0

( ) ( )(7)

125 125

K qa K qa� ��

� WA� B

= 0–

(8)125

Kqq a

Example 58. Two point masses of mass m and equal and opposite charge of magnitude q are attached

on the corners of a non-conducting uniform rod of mass m and the system is released

from rest in uniform electric field E as shown in figure from � = 53°

(i) Find angular acceleration of the rod just after releasing

(ii) What will be angular velocity of the rod when it passes through stable

equilibrium.

(iii)Find work required to rotate the system by 180°.

Solution : (i) tnet

= PE sin53° = I �

� � = 2 22

4( )

485

35

12 2 2

q EqE

mmm m

� ��

� � � ��

�

��

(ii) From energy conservation : Ki + U

i = K

f + U

f

� 0 + (– PE cos 53°) = 1

2Iw2 + (–PE cos 0°)

where I =

2

12

m� + m

2

2

� ��

� + m

2

2

� ��

�=

27

12

m�

�1

2Iw2 = PE (1– 3/5) =

2

5PE

�1

2 ×

27

12

m� ×w2 =

2

5qlE or w =

48

35 m

qE

�

(iii) � Wext

= Uf – U

i

� Wext

= (–PE cos(180° + 53°)) – (–PEcos 53°)

51

or Wext

= (ql)E3

5

� ��

+ (ql)E3

5

� ��

�/Wext

= 6

5

� ��

qlE

Example 59. Find force on short dipole P2 due to short dipole P

1 if they are placed at a distance r a part

as shown in figure.

Solution : Force on P2 due to P

1

F2 = (P

2)

1dE

dr

� ��

\ F2 = (P

2)

1

3

2 kPd

dr r

� ��

or F2 = –

1 2

4

6KPP

r

Here – sign indicates that this force will be attractive (opposite to r)

Example 60. Find force on short dipole P2 due to short dipole P

1 if they are placed a distance r apart as

shown in figure.

Solution :

Fnet

= q2 E (x + dx) - q

2 E (x)

Fnet

= q2

( ) ( )E x dx E x

dx

� � ��

dx

or Fnet

= (P2)

dE

dx

� ��

(Usually this formula is valid when the dipole is placed along E�

. However, in this case

also, we are getting the same formula)

52

Type (I) : Very Short Answer Type Questions : [01 Mark Each]

1. What kind of charges are produced one ach, when (i) a glass rod is rubbed with silk and (ii) an

ebonite rod is rubbed with wool ?

2. Given two point charges q1 and q

2, such that q

1q

2 < 0. What is the nature of force between them ?

3. Electrostatic force between two charges is called central force. Why ?

4. The test charge used to measure electric field at a point should be vanishingly small. Why ?

5. Two point electric charges of unknown magnitude and sign are placed a distance apart. The elec-

tric field intensity is zero at a point not between the charges but on the line joining them. Write

two essential conditions for this to happen.

6. Why two electric lines of force cannot intersect each other ?

7. Write a relation between electric field at a point and its distance from a short dipole.

8. Does an electric dipole always experience a torque, when placed in a uniform electric field ?

9. When is the torque acting on an electric dipole maximum, when placed in uniform electric field?

10. A charged particle is free to move in an electric field. Will it always move along an electric line of

force?

11. Define electric flux.

Or

What do you understand by electric flux ?

12. Suppose a gaussian surface does not include any net charge. Does it necessarily mean that E is

equal to zero for all points on the surface ?

13. What is the use of gaussian surface ?

14. How does electric field at a point charge with distance r from an inifinte thin sheet of charge ?

15. Two infinite parallel planes have uniform charge densities ± � . What is the electric field

(a) in the region between the planes and

(b) outside the plates ?

In what way does the infinite extension of the planes simplify your derivation ?

16. What would be the work done, if a point charge +q is taken from a point a to a point B on the

circumference of a circle drawn with another point charge +q at the centre ?

17. If a point charge +q is taken first from A to C and then from C to B of a circle drawn with another

point charge +q as centre [fig.], then along which path more work will be done ?

53

18. How is electric field at a point related to potential gradient ?

19. Why the electric field is always at right angles to the equipotential surface ? Give reason.

Type (II) : Short Answer Type Questions : [02 Marks Each]

20. Why can one ignore quantization of electric charge, when dealing with macroscopic i.e., large

scale charges?

21. Can two balls having same kind of charge on them attract each other ? Explain.

22. Fig. shows tracks of three charged particles crossing a uniform electrostatic field with same ve-

locities along horizontal. Give the signs of the three charges. Which particle has the highest

charge to mass ratio?

23. What is electric line of force ? What is its importance ?

24. Electric field intensity within a conductor is always zero. Why ?

25. A small metal sphere carrying charge +Q is located at the centre of a spherical cavity in a large

uncharged metal sphere as shown in the fig.

Use Gauss’ theorem to find electric field at points P1 and P

2.

26. If a conductor with a cavity having charge Q. Now another charge q is inserted into the cavity

show that the total charge on the outside surface of the conductor is Q + q.

27. Explain the term electrostatic shielding. Or

What is the principle of electrostatic shielding ?

28 Is it possible to transfer all the charge from a conductor to another insulated conductor ?

29. work done in moving a test charge between two points in an electric field is independent of the

path followed comment.

30. Figure (a) and (b) show the field lines of a single positive and negative charge respectively :

(i) Given the sign of the potential difference Vp – V

Q and V

B – V

A .

(ii) Give the sign of the potential energy difference of a small negative charge between the

points Q and P; A and B.

(iii) Give the sign of the work done by the field in moving a small positive charge from point

Q to P.

(iv) Give the sign of the work done by an external agency in moving a small negative charge

from point B to A.

54

(v) Does the kinetic energy of a small negative charge increase or decrease in going from

point B to A?

Type (III) : Long Answer Type Questions: [04 Mark Each]

31. Explain the terms : (i) quantization of charge(ii) conservation of charge (iii) additive nature of

charge.

32. State Coulomb’s law in electrostatics. Express it in vector form. What is importance of express-

ing it in vector form ?

33. Find the electric field intensity at any point on the axis of a uniformly charged ring or hoop. What

happens, if ring is far away from the point ?

34. Give important properties of electric lines of force .

35. Explain the terms electric dipole and dipole moment. Derive a relation for the intensity of electric

field at an equatorial point of an electric dipole.

36. Show that in a uniform electric field, a dipole experiences only a torque but no net force. Derive

expression for the torque.

37. Distinguish between linear, surface and volume charge density. Obtain expression for force on a

charge q due to continuous distribution of charge over a volume.

38. State Gauss’ theorem in electrostatics. Apply this theorem to calculate the electric field due to a

solid sphere of charge at a point (a) outside the sphere (b) on the sphere and (c) inside the sphere.

39. Define electric potential at a point. Derive an expression for the potential at a point due to a point

charge.

40. What do you mean by potential energy of the configuration of two electric charges ? Derive

expression for a system of two point charges lying at distance r.

41. Obtain expression for potential energy of the configuration of three charges. Hence, generalise

the result for a system of n point charges.

55

PROPERTIES OF CHARGE AND COULOMB'S LAW

1. Relative permittivity of mica is :

(A) one (B) less than one (C) more then one (D) infinite

2. Two identical metallic sphere are charged with 10 and -20 units of charge. If both the spheres

are first brought into contact with each other and then are placed to their previous positions,

then the ratio of the force in the two situations will be :-

(A) -8 : 1 (B) 1 : 8 (C) -2 : 1 (D) 1 : 2

3. Two equal and like charges when placed 5 cm apart experience a repulsive force of 0.144 newton.

The magnitude of the charge in microcoloumb will be :

(A) 0.2 (B) 2 (C) 20 (D) 12

4. Two charges of +1 �C & + 5 �C are placed 4 cm apart, the ratio of the force exerted by both

charges on each other will be -

(A) 1 : 1 (B) 1 : 5 (C) 5 : 1 (D) 25 : 1

5. A negative charge is placed at some point on the line joining the two +Q charges at rest. The

direction of motion of negative charge will depend upon the :

(A) position of negative charge alone

(B) magnitude of negative charge alone

(C) both on the magnitude and position of negative charge

(D) magnitude of positive charge.

6. A body has 80 microcoulomb of charge. Number of additional electrons on it will be :

(A) 8 x 10–5 (B) 80 x 1015 (C) 5 x 1014 (D) 1.28 x 10–17

7 Coulomb’s law for the force between electric charges most closely resembles with :

(A) Law of conservation of energy (B) Newton’s law of gravitation

(C) Newton’s 2nd law of motion (D) The law of conservation of charge

8. A charge Q1 exerts force on a second charge Q

2. If a 3rd charge Q

3 is brought near, the force

of Q1 exerted on Q

2.

(A) Will increase

(B) Will decrease

(C) Will remain unchanged

(D) Will increase if Q3 is of the same sign as Q

1 and will decrease if Q

3 is of opposite sign

9. A charge particle q1 is at position (2, - 1, 3). The electrostatic force on another charged

particle q2 at (0, 0, 0) is :

(A)0

21

56

qq

�� )k3ji2( �� (B)0

21

1456

qq

�� )k3ji2( ��

(C) 0

21

56

qq

�� )k3i2j( �� (D) 0

21

1456

qq

�� )k3i2j( ��

10. Three charge +4q, Q and q are placed in a straight line of length l at points distance 0, l/2 and

l respectively. What should be the value of Q in order to make the net force on q to be zero?

(A) –q (B) –2q (C) –q/2 (D) 4q

11. 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 dielectric constant

K = 16 is :

(A) r (B) r/4 (C) r/8 (D) 2r

56

11. Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of

permittivity is:

(A) ML–2T2A (B) M–1L–2T4A2 (C) MLT–2A (D) ML2T–1A2

12. Two point charges +9e and +e are kept 16 cm. apart from each other. Where should a third charge q

be placed between them so that the system is in equilibrium state :

(A) 6 cm from + 9e (B) 12 cm from +9e (C) 6 cm from +e (D) 12 cm from +e

13. The ratio of the forces between two small spheres with constant charge (a) in air (b) in a medium of

dielectric constant K is

(A) 1 : K (B) K : 1 (C) 1 : K2 (D) K2 : 1

14. A total charge Q is broken in two parts Q1 and Q

2 and they are placed at a distance R from each other.

The maximum force of repulsion between them will occur, when

(A) 2 1,

Q QQ Q Q

R R� � � (B) 2 1

2,

4 3

Q QQ Q Q� � �

(C) 2 1

3,

4 4

Q QQ Q� � (D) 1 2,

2 2

Q QQ Q� �

15. Three charges 4q, Q and q are in a straight line in the position of 0, l/2 and l respectively. The resultant

force on q will be zero, if Q =

(A) – q (B) –2q (C) 2

q� (D) 4q

16. ABC is a right angled triangle in which AB = 3cm and BC = 4 cm. And 2

ABC�

� � . The three

charges +15. + 12 and -20 e.s.u. are placed respectively on A, B and C. The force acting on B is

(A) 125 dynes (B) 35 dynes (C) 25 dynes (D) Zero

17. Two similar spheres haying +q and -q charge are kept at a certain distance, F force acts between the

two. If in the middle of two spheres; another similar sphere having +Q charge is kept, then it experience

a force in magnitude and direction as

(A) Zero having no direction (B) 8F towards +q charge

(C) 8F towards -q charge (D) 4F towards +q charge

18. Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not

experience any force. The value of q

m is

(A) l (B) 0

G

��(C)

04

G

�� (D) 0

4 G��

ELECTRIC FIELD

1. If an electron is placed in a uniform electric field, then the electron will :

(A) experience no force.

(B) moving with constant velocity in the direction of the field.

(C) move with constant velocity in the direction opposite to the field.

(D) accelerate in direction opposite to field.

2. If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be

(A) 100 N/C (B) 50 N/C (C) 200 N/C (D) 10 N/C

3. Two infinite linear charges are placed parallel at 0.1 m apart. If each has charge density of 5�

C/m, then the force per unit length of one of linear charges in N/m is :

(A) 2.5 (B) 3.25 (C) 4.5 (D) 7.5

57

4. The electric field intensity due to a uniformly charged sphere is zero :

(A) at the centre (B) at infinity

(C) at the centre and at infinite distance (D) on the surface

5. Two spheres of radii 2 cm and 4 cm are charged equally, then the ratio of charge density on

the surfaces of the spheres will be -

(A) 1 : 2 (B) 4 : 1 (C) 8 : 1 (D) 1 : 4

6. Total charge on a sphere of radii 10 cm is 1 �C. The maximum electric field due to the sphere

in N/C will be -

(A) 9 x 10–5 (B) 9 x 103 (C) 9 x 105 (D) 9 x 1015

7. A charged water drop of radius 0.1 mm is under equilibrium in some electric field. The

charge on the drop is equivalent to electronic charge. The intensity of electric field is (g = 10

m/s2)-

(A) 1.61 NC–1 (B) 26.2 NC–1 (C) 262 NC–1 (D) 1610 NC–1

8. Two large sized charged plates have a charge density of + and -. The resultant force on the

proton located midway between them will be -

(A) 0/e � (B) 02/e � (C) 0/e2 � (D) zero

9. Two parallel charged plates have a charge density + and -. The resultant force on the proton

located outside the plates at some distance will be -

(A) 0/e2 � (B) 0/e � (C) 0/ 2e � (D) zero

10. The charge density of an insulating infinite surface is (e/�) C/m2 then the field intensity at a

nearby point in volt/meter will be -

(A) 2.88 x 10–12 (B) 2.88 x 10–10 (C) 2.88 x 10–9 (D) 2.88 x 10–19

11. There is a uniform electric field in x-direction. If the work done by external agent in moving a

charge of 0.2 C through a distance of 2 metre slowly along the line making an angle of 60º with

x-direction is 4 joule, then the magnitude of E is:

(A) C/N3 (B) 4 N/C (C) 5 N/C (D) 20 N/C

12. A simple pendulum has a length l, mass of bob m. The bob is given a charge q coulomb. The

pendulum is suspended in a uniform horizontal electric field of strength E as shown in

figure, then calculate the time period of oscillation when the bob is slightly displace from its

mean position is :

(A) g

2�

� (B) �

�

�

�

�

��

m

qEg

2�

(C) �

�

�

�

�

��

m

qEg

2�

(D) 2

2

m

qEg

2

��

��

��

��

58

13. Charge 2Q and –Q are placed as shown in figure. The point at which electric field intensity is

zero will be:

(A) Somewhere between –Q and 2Q

(B) Somewhere on the left of –Q

(C) Somewhere on the right of 2Q

(D) Somewhere on the right bisector of line joining –Q and 2Q

14. The maximum electric field intensity on the axis of a uniformly charged ring of charge q and

radius R will be :

(A) 04

1

�� 2R33

q(B)

04

1

�� 2R3

q2(C)

04

1

�� 2R33

q2(D)

04

1

�� 2R32

q3

15. A charged particle of charge q and mass m is released from rest in an uniform electric field E.

Neglecting the effect of gravity, the kinetic energy of the charged particle after time ‘t’ seconds

is

(A) t

Eqm(B)

m2

tqE 222

(C) mq

tE2 22

(D) 2

2

t2

mEq

16. Figures below show regular hexagon, the charges are placed at the vertices. In which of the

following cases the electric field at the centre is zero.

(A) IV (B) III (C) I (D) II

17. A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell

having charge –Q. The radius of the sphere is a and that of the spherical shell is b(b > a). What is the

electric field at a distance R(a < R < b) from the centre ?

(A) 2

0

4

2

Q

R�� (B) 2

0

3

4

Q

R�� (C) 2

0

3

2

Q

R�� (D) 2

02

Q

R��

18. Two charge particles A and B kept in air, electric force between them is 100 N. If they are placed in

medium having dielectric constant k = 5, at same separation then magnitude of electric force on one

of the charge particle due to medium is.

(A) 20N (B) 100N (C) 80N (D) 500N

19. Find the electric field at centre of semicircular ring shown in figure (charge distribution is uniform)

(A) 2

4 ˆkqi

R��

(B) 2

2 ˆkqi

R��

(C) 2ˆkqi

R�(D) None of these

59

20. The linear charge density on upper half of a segment of ring is � and at lower half, it is �� . The

direction of electric field at centre O of ring is :

(A) along OA (B) along OB (C) along OC (D) along OD

21. The given figure gives electric lines of forces due to two charges q1 and q

2. What are the signs of the

two charges?

(A) Both are negative (B) Both are positive

(C) q1 is positive but q

2 is negative (d) q

1 is negative but q

2 is positive

22. A rod of length L has a total charge Q distributed uniformly along its length. It is bent in the shape of

a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

(A) 2

02

Q

L� (B) 2

0

2Q

L� (C) 2

0

Q

L� (D) None of these

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

1. If we move in a direction opposite to the electric lines of force :

(A) electrical potential decreases. (B) electrical potential increases.

(C) electrical potential remains uncharged (D) nothing can be said.

2. The distance between two plates is 2 cm, when an electric potential of 10 volt is applied to

both the plates, then the value of electric field will be -

(A) 20 N/C (B) 500 N/C (C) 5 N/C (D) 250 N/C

3. Two objects A and B are charged with equal charge Q. The potential of A relative to B will

be -

(A) more (B) equal (C) less (D) indefinite

4. In electrostatics the potential is equivalent to -

(A) temperature in heat (B) height of levels in liquids

(C) pressure in gases (D) all of the above

5. The potential due to a point charge at distance r is -

(A) proportional to r. (B) inversely proportional to r.

(C) proportional to r2. (D) inversely proportional to r2

6. The dimensions of potential difference are -

(A) ML2T–2Q–1 (B) MLT–2Q–1 (C) MT–2Q–2 (D) ML2T–1Q–1

7. An object is charged with positive charge. The potential at that object will be -

(A) positive only (B) negative only

(C) zero always (D) may be positive, negative or zero.

8. Two points (0, a) and (0, -a) have charges q and -q respectively then the electrical potential

at origin will be-

(A) zero (B) kq/a (C) kq/2a (D) kq/4a2

60

9. The charges of same magnitude q are placed at four corners of a square of side a. The valueof potential at the centre of square will be -

(A) 4kq/a (B) a/kq24 (C) a2kq4 (D) 2a/kq

10. Three equal charges are placid at the three corners of an isoscelestriangle as shown in the figure. The statement which is true forelectric potential V and the field intensity E at the centre of thetriangle -(A) V = 0, E = 0 (B) V = 0, E � 0(C) V � 0, E = 0 (D) V � 0, E � 0

11. The potential at 0.5 Å from a proton is -(A) 0.5 volt (B) 8m volt (C) 28.8 volt (D) 2 volt

12. A wire of 5 m length carries a steady current. If it has an electric field of 0.2 V/m, the potentialdifference across the wire in volt will be -(A) 25 (B) 0.04 (C) 1.0 (D) none of the above

13. An infinite number of charges of equal magnitude q, but alternate charge of opposite sign areplaced along the x-axis at x = 1, x = 2, x = 4, x =8,... and so on. The electric potential at thepoint x = 0 due to all these charges will be -(A) kq/2 (B) kq/3 (C) 2kq/3 (D) 3kq/2

14 The electric potential inside a uniformly positively charged non conducting solid spherehas the value which -(A) increase with increases in distance from the centre.(B) decreases with increases in distance from the centre.(C) is equal at all the points.(D) is zero at all the points.

15. Two metallic spheres which have equal charges, but their radii are different, are made to toucheach other and then separated apart. The potential the spheres will be -(A) same as before (B) more for bigger (C) more for smaller (D) equal

16. Two spheres of radii R and 2R are given source equally positive charged and then connectedby a long conducting wire, then the positive charge will(A) flow from smaller sphere to the bigger sphere.(B) flow from bigger sphere to the smaller sphere(C) not flow.(D) oscillate between the spheres.

17. The potential difference between two isolated spheres of radii r1 and r

2 is zero. The ratio of their

charges Q1/Q

2 will be-

(A) r1/r

2(B) r

2/r

1(C) r

12/r

22 (D) r

13/r

23

18. The potential on the conducting spheres of radii r1 and r

2 is same, the ratio of their charge densities

will be-(A) r

1/r

2(B) r

2/r

1(C) r

12/r

22 (D) r

22/r

12

19. 64 charged drops coalesce to from a bigger charged drop. The potential of bigger drop will betimes that of smaller drop -(A) 4 (B) 16 (C) 64 (D) 8

20. The electric potential outside a uniformly charged sphere at a distance ‘r’ is (‘a’ being the radiusof the sphere)-(A) directly proportional to a3 (B) directily proportional to r.(C) inversely proportional to r. (D) inversely proportional to a3.

61

21. A conducting shell of radius 10 cm is charged with 3.2 x 10–19 C. The electric potential at a

distance 4cm from its centre in volt be -

(A) 9 x 10–9 (B) 288 (C) 2.88 x 10–8 (D) zero

22. 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) 6 m (B) 12 m (C) 36 m (D) 144 m

23. Figure represents a square carrying charges +q, +q, –q, –q at its

four corners as shown. Then the potential will be zero at points

(A) A, B, C, P and Q

(B) A, B and C

(C) A, P, C and Q

(D) P, B and Q24. Two equal positive charges are kept at points A and B. The electric potential at the points

between A and B (excluding these points) is studied while moving from A to B. The potential(A) continuously increases (B) continuosly decreases(C) increases then decreases (D) decreases than increases

25. A semicircular ring of radius 0.5 m is uniformly charged with a total charge of 1.5 × 10–9

coul. The electric potential at the centre of this ring is :(A) 27 V (B) 13.5 V (C) 54 V (D) 45.5 V

26. The kinetic energy which an electron acquires when accelerated (from rest) through a potentialdifference of 1 volt is called :(A) 1 joule (B) 1 electron volt (C) 1 erg (D) 1 watt

27. The potential difference between points A and B in the givenuniform electric field is : (A) Ea

(B) E 2 2( )a b�(C) Eb

(D) )2/Eb(

28. A particle of charge Q and mass m travels through a potential difference V from rest. Thefinal momentum of the particle is :

(A) mV

Q(B) 2Q mV (C) 2m QV (D)

2QV

m

29 If a uniformly charged spherical shell of radius 10 cm has a potential V at a point distant 5cm from its centre, then the potential at a point distant 15 cm from the centre will be :

(A) 3

V(B)

2

3

V(C)

3

2V (D) 3V

30. Uniform electric field of magnitude 100 V/m in space is directed parallel to the line y = 3 + x.Find the potential difference between point A(3, 1) & B(1, 3):

(A) 100 V (B) 200 2 V (C) 200V (D) 031. The figure below shows two equipotential lines in XY plane for an electric field. The scales are

marked. The X-component Ex and Y-component E

y of the electric field in the space between

these equipotential lines are respectively:

(A) +100 V/m, –200 V/m (B) +200 V/m, +100 V/m(C) –100 V/m, +200 V/m (D) –200 V/m, –100 V/m

62

32. Which of the following is true for the figure showing electric lines of force ?

(E is magnitude of electrical field, V is potential)

(A) EA

> EB

(B) EB > E

A(C)V

A >V

B(D) None of these

33. There are 27 identical drops of a conducting fluid. Each has a radius r and they are charged to a

potential V0. They are then combined to form a bigger drop. Find its potential.

(A) 9 V0

(B) 5V0

(C) 3V0

(D) None of these

34. Three charges 2q, –q, –q are located at the vertices of an equilateral triangle. At the centre of the

triangle

(A) The field is zero but potential is non-zero (B) The field is non-zero but potential is zero

(C) Both field and potential are zero (D) Both field and potential are non-zero

35. A particle A has charge +q and a particle B has charge +4q with each of them having the same

mass m. When allowed to fall from rest through the same electric potential difference, the ratio of

their speed A

B

�� will become

(A) 2 : 1 (B) 1 : 2 (C) 1 : 4 (D) 4 : 1

36. A charge 3 coulomb experiences a force 300 N when placed in a uniform electric field. The

potential difference between two points separated by a distance of 10 cm along the field line is :

(A)10V (B) 90V (C) 1000 V (D) 9000 V

ELECTRIC POTENTIAL ENERGY OF A PARTICLE

1 A nucleus has a charge of + 50e. A proton is located at a distance of 10-12 m. The potential at

this point in volt will be -

(A) 14.4 x 104 (B) 7.2 x 104 (C) 7.2 x 10–12 (D) 14.4 x 108

2 Under the influence of charge, a point charge q is carried along

different paths from a point A to point B, then work done will be

(A) maximum for path four.

(B) maximum for path one.

(C) equal for all paths

(D) minimum for path three.

3 An electron moving in a electric potential field V1 enters a higher electric potential field V

2,

then the change in kinetic energy of the electron is proportional to -

(A) (V2 — V

1)1/2 (B) V

2 — V

1(C) (V

2 — V

1)2 (D)

2 1

2

( )V V

V

�

4 In the electric field of charge Q, another charge is carried from

A to B. A to C, A to D and A to E, then work done will be -

(A) minimum along path AB.

(B) minimum along path AD.

(C) minimum along path AE.

(D) zero along all the paths.

63

5. The work done to take an electron from rest where potential is – 60 volt to another point where

potential is – 20 volt is given by -

(A) 40 eV (B) –40 eV (C) 60 eV (D) –60 eV

6. If a charge is shifted from a low potential region to high potential region. the electrical potential

energy:

(A) Increases (B) Decreases

(C) Remains constant (D) May increase or decrease.

7. Three point charges 1C, 2C and 3C are placed at the comers of an equilateral triangle of side 1m.

The work required to move these charges to the comers of a smaller equilateral triangle of side

0.5m in two different ways as in fig. (A) and fig. (B) are Wa and W

b then:

(A) Wa > W

b(B) W

a < W

b(C) W

a = W

b(D) W

a = 0 and W

b = 0

8. Two isolated spherical shell having charges Q1 and Q

2 and radii r

1 and r

2 are kept at very large

separation. If they are joined by a conducting wire, then choose incorrect statement:

(A) Electrostatic potential energy of system must decrease.

(B) Electrostatic potential energy of system may decrease.

(C) If 1 2 2 1Q r Q r� , then no charge flow through the conducting wire.

(D) If 1 2 2 1Q r Q r� , then electrostatic potential energy of system already minimum.

9. Consider the configuration of a system of four charges each of value +q. Find the work done by

external agent in changing the configuration of the system from figure (i) to figure, (ii) very

slowly.

(A) 2

(3 2)Kq

a

�� (B)

2

( 3 2)Kq

a

�� (C)

2

( 3 2)Kq

a

�� (D) None of these

10. Three charges +q, –q, and +2q are placed at the vertices of a right angled triangle (isosceles

triangle) as shown. The net electrostatic energy of the configuration is

(A) 2

( 2 1)Kq

a� � (B)

2

( 2 1)Kq

a� (C)

2

( 2 1)Kq

a� � (D) None of these

64

11. You are given an arrangment of three point charges q, 2q and 3

Xq� separated by equal finite

distances so that electric potential energy of the system is zero. Calculate the value of x.

(A) 2 (B) 3 (C) 1 (D) None of these

12. A sphere of radius 1cm has potential of 8000 V, then energy density near its surface will be

(A) 564 10� J/m3 (B) 38 10� J/m3 (C) 32 J /m3 (D) 2.83J/m3

13. In the rectangle, shown below, the two corners have charges 1

5q C�� and 2

2.0q C�� . The

work done in moving a charge 3.0 C�� from B to A is (take 0

1/ 4�� =1010N-m2/C2)

(A) 2.8 J (B) 3.5J (C) 4.5 J (D) 5.5J

14. A charge (-q) and another charge (+Q) are kept at two points A and B respectively. Keeping the charge

(+Q) fixed at B, the charge (–q) at A is moved to another point C such that ABC forms an equilateral

triangle of side l. The net work done in moving the charge (-q) is

(A) 0

1

4

Qq

l�� (B) 2

0

1

4

Qq

l�� (C) 0

1

4Qql

�� (D) Zero

15. A sphere of radius 1 cm has potential of 8000 V. The energy density near 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

16. If ' n ' identical water drops assumed spherical each charged to a potential energy U coalesce

to 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) n4/3 U (D) n5/3 U

POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGE

1 In H atom, an electron is rotating around the proton in an orbit of radius r. Work done by an

electron in moving once around the proton along the orbit will be -

(A) ke/r (B) ke2/r2 (C) 2pre (D) zero

2 You are given an arrangement of three point charges q, 2q and xq separated by equal finite

distances so that electric potential energy of the system is zero. Then the value of x is :

(A) 3

2� (B)

3

1� (C)

3

2(D)

2

3

RELATION BETWEEN E�

AND V :

1. A family of equipotential surfaces are shown. The direction of the electric field at point A is along

(A) AB (B) AC (C) AD (D) AF

65

2. Some equipotential surfaces are shown in the figure.The magnitude and direction of the electric field is-(A) 100 V/m making angle 1200 with the x-axis(B) 100 V/m making angle 600 with the x-axis(C) 200 V/m making angle 1200 with the x-axis

(D) none of the above3. The variation of potential with distance r from a fixed point is shown in Figure. The electric

field at r = 5 cm, is :

(A) (2.5) V/cm (B) (–2.5) V/cm (C) (–2/5) cm (D) (2/5) V/cm4 The electric field and the electric potential at a point are E and V respectively

(A) If E = 0, V must be zero (B) If V = 0, E must be zero(C) If E ¹ 0, V cannot be zero (D) None of these

5. The electric field in a region is directed outward and is proportional to the distance r from theorigin. Taking the electric potential at the origin to be zero, the electric potential at a distance r :(A) is uniform in the region (B) is proportional to r(C) is proportional to r2 (D) increases as one goes away from the origin.

DIPOLE

1. If an electric dipole is kept in a non-uniform electric field, then it will experience -

(A) only torque (B) no torque

(C) a resultant force and a torque (D) only a force

2. The force on a charge situated on the axis of a dipole is F. If the charge is shifted to doublethe distance, the acting force will be -

(A) 4F (B) F/2 (C) F/4 (D) F/8

3. A dipole of dipole moment p, is placed in an electric field �E and is in stable equilibrium. The

torque required to rotate the dipole from this position by angle q will be -

(A) pE cos q (B) pE sin q (C) pE tan q (D) –pE cosq

4. The electric potential at a point due to an electric dipole will be -

(A) 3

( . )k p r

r

� �

(B) 2

( . )k p r

r

� �

(C) k p r

r

��

(D) 2

k p r

r

��

5. The ratio of electric fields due to an electric dipole on the axis and on the equatorial line atequal distance will be -

(A) 4 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 1

6. An electric dipole is made up of two equal and opposite charges of 2 x 10–6 coulomb at a distanceof 3 cm. This is kept in an electric field of 2 � 105 N/C, then the maximum torque acting onthe dipole -

(A) 12 � 10–1 Nm (B) 12 � 10–3 Nm (C) 24 � 10–3 Nm (D) 24 � 10–1 Nm

66

7. The distance between two singly ionised atoms is 1Å. If the charge on both ions is equal andopposite then the dipole moment in coulomb-metre is -(A) 1.6 × 10–29 (B) 0.16 × 10–29 (C) 16 × 10–29 (D) 1.6 × 10–29/ 4pe

0

8. The electric potential in volt at a distance of 0.01 m on the equatorial line of an electric dipoleof dipole moment p is -

(A) 0

4p / 4 10� �� (B) zero (C)

0

44 p 10� �� (D) 0

44 /p 10� ��

9. The electric potential in volt due to an electric dipole of dipole moment 2 x 10–8 C-m at a distanceof 3m on a line making an angle of 600 with the axis of the dipole is -(A) 0 (B) 10 (C) 20 (D) 40

10 A dipole of electric dipole moment P is placed in a uniform electric field of strength E. If �is the angle between positive directions of P and E, then the potential energy of the electricdipole is largest when � is :(A) zero (B) � /2 (C) � (D) � /4

11. The work done in rotating an electric dipole of dipole moment p in an electric field E through an angle0 from the direction of electric field, is :(A) pE(1 – cos� ) (B) pE (C) zero (D) –pEcos�

12. An electric dipole moment ˆ ˆ(2.0 3.0 )p i j C�� is placed in a uniform electric field

5ˆˆ(3.0 2.0 ) 10E i k� � ��

NC–1.

(A) The torque that exerts on P is ( ˆˆ ˆ0.6 0.4 0.9i j k� � )Nm.

(B) The potential energy of the dipole is 0 J.(C) The potential energy of the dipole is 0.6 J.(D) None of these

13. The dipole moment of a system of charge +q distributed uniformly on an arc of radius Rsubtending an angle / 2� at its centre where another charge -q is placed is :

(A) 2 2qR

�(B)

2qR

�(C)

qR

�(D)

2qR

�

14. A large sheet carries uniform surface charge density . A rod of length 2l has a linear chargegedensity � on one half and �� on the second half. The rod is hinged at mid point O and makesan angle � with the normal to the sheet. The torque experienced by the rod is(A) 0

(B) 2

0

sin2

l��

�

(C) 2

0

sinl�

��

(D) 02

l��

15. For the situation shown in the figure below (assume r > > length of dipole) mark out the correctstatement(s).(A) Force acting on the dipole is zero

(B) Force acting on the dipole is 304

pQ

r�� and is acting upward

(C) Torque acting on the dipole is 204

pQ

r�� in anti-clockwise direction

(D) None of these

67

16. An electric dipole consisting of two opposite charges of 62 10 C�� each separated by a distance of 3

cm is placed in an electric field 52 10� N/C. The maximum torque on the dipole will be

(A) 112 10�� Nm (B) 312 10�� Nm (C) 124 10�� Nm (D) 324 10�� Nm

17. An electric dipole of moment p�

is placed normal to the lines of force of electric intensity E�

, then

the work done in deflecting it through an angle of 180° is

(A) pE (B) + 2pE (C) –2pE (D) Zero

18. The distance between the two charges +q and –q of a dipole is r. On the axial line at a distance d from

the centre of dipole, the intensity is proportional to

(A) 2

q

d(B) 2

qr

d(C) 3

q

d(D) 3

qr

d

FLUX CALCULATION AND GAUSS'S LAW

1 For an electrostatic system which of the statement is always true :

(a) electric lines are parallel to metallic surface.

(b) electric field inside a metallic surface is zero.

(c) electric lines of force are perpendicular to equi-potential surface.

(A) (a) and (b) only (B) (b) and (c) only

(C) (a) and (c) only (D) (a), (b) and (c)

2. Total flux coming out of some closed surface is :

(A) 0/q � (B) q/0� (C) 0q� (D) 0/q �

3. Three charges q1 = 1 × 10–6 , q

2 = 2 × 10–6 , q

3 = –3 × 10–6 C have been placed, as shown in

figure, in four surfaces S1, S

2, S

3 and S

4 electrical flux emitted from the surface S

2 in N–m2/

C will be -

(A) 36p × 103 (B) –36p × 103 (C) 36p × 109 (D) –36p × 109

4. The intensity of an electric field at some point distant r from the axis of infinite long pipe having

charges per unit length as q wil be :

(A) proportional to r2 (B) proportional to r3

(C) inversely proportional to r. (D) inversely proportional to r2.

5. Eight charges, 1�C,. -7�C, -4�C, 10�C, 2�C, -5�C, -3�C and 6�C are situated at the eight corners

of a cube of side 20 cm. A spherical surface of radius 80 cm encloses this cube. The centre of the

sphere coincides with the centre of the cube. Then the total outgoing flux from the spherical

surface (in unit of volt meter) is-

(A) 36� x 103 (B) 684� x 103 (C) zero (D) none of the above

6. A closed cylinder of radius R and length L is placed in a uniform electric field E, parallel to the

axis of the cylinder. Then the electric flux through the cylinder must be -

(A) 2�R2E (B) (2�R2 + 2�RL)E (C) 2�RLE (D) zero

68

7. A sphere of radius R and charge Q is placed inside an imaginary sphere of radius 2R whose centre

coincides with the given sphere. The flux related to imaginary sphere is:

(A) 0

Q

� (B) 0

2

Q

� (C) 0

4Q

� (D) 0

2Q

�

8. The length of each side of a cubical closed surface is � . If charge q is situated on one of the vertices

of the cube as shown then find the flux passing through shaded face of the cube.

(A) 0

24

q

� (B) 0

4

q

� (C) 0

24q

� (D) None of these

9. Electric charges are distributed in a small volume. The flux of the electric field through a spherical

surface of radius 10 cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of

radius 20 cm will be

(A) 25 V-m (B) 50 V-m (C) 100 V-m (D) 200 V-m.

10. A charge q is placed at the centre of the open end of a cylindrical vessel (figure). The flux of the

electric field through the surface of the vessel is

(A) zero (B) q / 0� (C) q/2 0

� (D) 4q / 0�

11. Mark the correct options:

(A) Gauss’s law is valid only for symmetrical charge diatriburions.

(B) Gauss’s law is valid only for charges placed in vacuum.

(C) The electric field calculated by Gauss’s law is the field due to the charges inside the Gaussian

surface.

(D) The flux of the electric field through a closed surface due to all the charges is equal to the flux due

to the charges enclosed by the surface.

12. A positive point charge Q is brought near an isolated metal cube.

(A) The cube becomes negatively charged.

(B) The cube becomes positively charged.

(C) The interior becomes positively charged and the surface becomes negatively charged.

(D) The interior remains charge free and the surface gets nonuniform charge distribution.

13. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder

axis. The total flux for the surface of the cylinder is given by

(A) 22 R E� (B) 2 /R E� (C) 2( ) /R R E� �� (D) Zero

CONDUCTOR, IT'S PROPERTIES & ELECTRIC PRESSURE

1 The electric field near the conducting surface of a uniform charge density will be -

(A) 0/ � and parallel to surface. (B) 0/2 � and parallel to surface.

(C) 0/ � and perpendicular to surface. (D) 0/2 � and perpendicular to surface.

2 An uncharged conductor A is brought close to another positive charged conductor B,

then the charge on B -

(A) will increase but potential will be constant.

(B) will be constant but potential will increase

(C) will be constant but potential decreases

. (D) the potential and charge on both are constant.

69

3 The fig. shows lines of constant potential in a region in which an electric field is present. The

value of the potential are written in brackets of the points A, B and C, the magnitude of the

electric field is greatest at the point -

(A) A (B) B (C) C (D) A & C

4 The electric charge in uniform motion produces -

(A) an electric field only (B) a magnetic field only

(C) both electric and magnetic fields (D) neither electric nor magnetic fields

5 Which of the following represents the correct graph for electric field intensity and the distance

r from the centre of a hollow charged metal sphere or solid metallic conductor of radius R :

(A) (B) (C) (D)

6 A neutral metallic object is placed near a finite metal plate carrying a positive charge. The

electric force on the object will be :

(A) towards the plate (B) away from the plate

(C) parallel to the plate (D) zero

7 Figure shows a thick metallic sphere. If it is given a charge +Q,

then

electric field will be present in the region

(A) r < R1 only

(B) r > R1 and R

1 < r < R

2

(C) r � R2 only

(D) r � R2 only

8 An uncharged sphere of metal is placed in a uniform electric field produced by two large

conducting parallel plates having equal and opposite charges, then lines of force look like

(A) (B) (C) (D)

9 You are travelling in a car during a thunder storm, in order to protect yourself from lightening

would you prefer to :

(A) Remain in the car (B) Take shelter under a tree

(C) Get out and be flat on the ground (D) Touch the nearest electrical pole

70

10 The amount of work done in Joules in carrying a charge +q along the closed path PQRSP

between the oppositely charged metal plates is (where E is electric field between the plates)

(A) zero (B) q (C) qE (PQ + QR + SR + SP) (D) 0/q �

11 Figure shows a closed surface which intersects a conducting sphere. If a positive charge is

placed at the point P, the flux of the electric field through the closed surface

(A) will remain zero (B) will become positive

(C) will become neagative (D) will become undefined

12. A solid conducting sphere having a charge Q is surroundsd by an uncharged concentric conducting

hollow spherical shell, Let the potential difference between the surface of the solid sphere and that of

the outer surface of the hollow shell be V, If the shell is now given a charge of -3Q, the new potential

difference between the same two surfaces is ;

(A)V (B) 2V (C) 4V (D) –2V

13. A cube of metal is given a charge (+ Q), which of the following statements is true

(A) Potential at the surface of cubt is zero

(B) Potential within the cube is zero

(C) Electric field is normal to the surface of the cube

(D) Electric field varies within the cube

14. A hollow closed conductor of irregular shaps is given some charge, Which of the following statements

are correct?

(A) The entire charge will appear on its outer surface.

(B) All points on its surfaee will have the same charge density.

(C) All points near its surface and outside it will have the same electric intensity.

(D) None of these

15. Two large thin conducting plates with small gap in between are placed in a uniform electric field ‘E’

perpendicular to the plates). Area of each plate is A and charges +Q and –Q are given to these plates as

shown in the figure. If points R,S and T as shown in the figure are three points in space, then the

(A) field at point R is E (B) field at point S is E

(C) field at point T is 0

QE

A��

��

(D) field at point S is 0

QE

A��

��

71

16. A positive point charge Q is kept (as shown in the figure) inside a neutral condueting shell whose

centre is at C. An external uniform electric field E is applied. Then :

(A) Force on Q due to E is zero

(B) Net force on Q is zero

(C) Net force acting on Q and conducting shell considered as a system Is zero

(D) Net force acting on the shell due to E is zero.

17. For the situation shown in the figure below, mark out the correct statement (s)

(A) Potential of the conductor is 0

4 ( )

q

d R��

(B) Potential of the conductor is 0

4

q

d��

(C) Potential of the conductor can’t be determined as nature of distribution of induced charges is

not known

(D) None of these

18. A charge +q is placed at a distance ‘d’ from the centre of the uncharged metallic cube of side ‘a’,

The electric field at the centre of the cube due to induced charges on the cube will be

(A)zero (B) 2

0

ˆ( )4

qj

d�� (C) 2

0

ˆ( )4

qj

d�� (D) 2

0

ˆ( )

42

qj

ad��

��

19. A charge Q is distributed over two concentric hollow spheres of radii r and R (R > r) such that the

surface densities are equal. Find the potential at the common centre.

(A) 2 2

0

( )

4 ( )

Q R r

r R��

� (B) 2 2

0

( )

( )

Q R r

r R��

� (C) 2 2

0

4 ( )

( )

Q R r

r R

��

�

� (D) None of these

20. Two thin conducting shells of radii R and 3R are shown in figure.The outer shell carries a charge

+Q and the inner shell is neutral. The inner shell is earthed with the help of switch S. Find the

charge attained by the inner shell.

(A) 3

Q�(B)

7

Q�(C)

4

Q(D) None of these

72

PROPERTIES OF CHARGE AND COULOB’S LAW :

1. Force between two charges, when placed in free space is 10 N. If they are in a medium of relative

permittivity 5, the force between them will be

(A) 2N (B) 50 N (C) 0.5 N (D)none of these

2. Two charges of C�2 and C�5 are placed 2.5 cm apart. The ratio of the Coulomb’s force experienced

by them is

(A) 1 : 1 (B) 2 : 5 (C) 5:2 (D)4 : 25

3. The dielectric constant of an insulator cannot be

(A) 1.5 (B) 3 (C) 4.5 (D) �4. A cylindrical conductor is placed near another positively charged conductor. The net charge acquired

by the cylindrical conductor will be

(A) positive only (B) negative only

(C) zero (D)either positive or negative depending upon the distance

5. The electric force between two charges each equal to C�3 , when placed in vacuum is 12 N. What will

be the force between them, if the charges are embedded in a medium of dielectric constant 6?

(A) 36 N (B) 18 N (C)2 N (D)1 N

6. A charge q is distributed over two spheres of radii R and r such that their surface densities are equal.

What is the ratio of the charges on the spheres ?

(A) R

r(B) 2

2

R

r(C) 3

3

R

r(D) 4

4

R

r

7. Three charges 4q, Q and q are in a straight line in the positions 2/,0 l and l respectively. Resultant

force at q will be zero if Q =

(A) -q (B) -2q (C) -q/2 (D)4q

8. Mid way between the two equal and similar charges, we place the third equal and similar charge. Which

of the following statements is correct ?

(A) The third charge experiences a net force inclined to the line joining the charges

(B) The third charge is in stable equilibrium

(C) The third charge is in unstable equilibrium

(D) The third charge experiences a net force perpendicular to the line joining the charges.

9. Two charges 1q and 2q repel each other with a force of 0.1 N. What will be the force exerted by 1q

on 2q , when a third charge is placed near them ?

(A) Less then 0.1 N

(B) More than 0.1 N

(C) 0.1 N

(D) Less than 0.1 N if 1q and 2q are similar and more than 0.1 N if 1q and 2q are dissimilar

10. A charge q is placed at the mid point of the line joining two similar and equal charges each equal to

C�� 2 . The system will be in equilibrium if �q

(A) C�� 5.0 (B) C�� 0.1 (C) C�� 0.1 (D) C�� 5.0

11. The charge Q is placed at each of the two opposite corners of a square. A charge q is placed at each of

the other two opposite corners. The charges Q are in equilibrium, what is the value of q?

(A) 2

Q� (B)

2

Q� (C)

22

Q� (D)

4

Q�

73

12. Three equal and similar charges are placed at )0,0,0(),0,0,( a� and )0,0,( a� . What is the nature of

equilibrium of the charge at the origin ?

(A) stable when moved along the Y-axis (B) stable when moved along Z-axis

(C) stable when moved along X-axis (D) unstable in all of the above cases.

13. Figure shows two charges. If both the charge q and Q are in equilibrium, then what is the ratio Q/q ?

(A) 4� (B) 4� (C)4

1� (D)

4

1�

14. Given that qqq �� 21 . For what ratio 1/ qq , will be force between 21 & qq is maximum ?

(A) 0.25 (B) 0.5 (C) 1 (D)2

15. Six charges each equal to +Q are placed at the corners of a regular hexagon of each side x. What is the

electric field at the intersection of its diagonals ?

(A) 20

36

4

1

x

Q

�� (B) 20

6

4

1

x

Q

�� (C) 204

1

x

Q

�� (D)zero

16. Positive charges of C�2 and C�8 are placed 15 cm apart. At what distance from the smaller chargege

the electric field due to them will be zero ?

(A) 3 cm (B)5 cm (C) 7 cm (D)10 cm

17. Two positively charged particles X and Y are initially far away from each other and at rest. X begins to

move towards Y with some initial velocity. The total momentum and energy of the system are p and E.

(A) If Y is fixed, both p and E are conserved.

(B) If Y is fixed, E is conserved, but not p.

(C) If both are free to move, p is conserved but not E.

(D) If both are free, E is conserved, but not p.

18. Two particles X and Y, of equal mass and with unequal positive charges, are free to move and are initially

far away from each other. With Y at rest, X begins to move towards it with initial velocity u. After a long

time, finally

(A) X will stop, Y will move with velocity u.

(B) X and Y will both move with velocities u/2 each.

(C) X will stop, Y will move with velocity < u.

(D) both will move with velocities < u/2.

19. Four charges are arranged at the corners of a square ABCD, as shown. The

force on a +ve charge kept at the centre of the square is

(A) zero (B) along diagonal AC

(C) along diagonal BD (D) perpendicular to the side AB

20. Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieve equilibrium

for the entire system and where should it be placed form charge q?

(A) Q = 9

4q (negative) at

3

l(B) Q =

9

4q (positive) at

3

l

(C) Q = q (positive) at 3

l(D) Q = q (negative) at

3

l

74

21. Six charges are placed at the corner of a regular hexagon as shown. If an electron

is placed at its centre O, force on it will be:

(A) Zero (B) Along OF

(C) Along OC (D) None of these

ELECTRIC FIELD :

1. The electric field strength at a distance x from a charge Q is E. What will be electric field strength if the

distance of the observation point is increased by x2 ?

(A) 2/E (B) 3/E (C) 4/E (D)none of the above

2. A one coulomb charge is placed on an insulated stand at the centre of a spherical conductor of radius1m. The sphere is given a charge of 1C. The electrostatic force experienced by the charge at the centrewill be

(A) zero (B) 1 N (C) N9

109� (D)none of the above

3. A charged spherical conductor has potential of 6V and its radius is 2m. The electric field intensity at itscentre is

(A) zero (B) 13

�NC (C) 1

12�

NC (D)none of the above

4. An electron moves with a velocity v�

in an electric field E�

. If the angle between v�

and E�

is neither 0nor � , the path followed by the electron is

(A) straight line (B) circle (C) ellipse (D)parabola

5. A charged spherical conductor of radius R carries a charge Q. A point test charge 0q is placed at a

distance x from the surface of the conductor . The force experienced by the test charge will be propor-

tional to

(A) 2)( xR � (B) 2

)(

1

xR �(C) 2

)( xR � (D) 2)(

1

xR �

6. The electric field strength due to a ring of radius R at a distance x from its centre on the axis of ring

carrying charge Q is given by 2/322

0 )(4

1

xR

xQE

��

At what distance from the centre will the electric field be maximum ?

(A) Rx � (B) 2/Rx � (C) 2/Rx � (D) Rx 2�

7. A ring of radius R is carrying uniformly distributed charge +Q. A test charge 0q� is placed on its axis at

a distance 2R from the centre and released. The motion of the particle on the axis will be

(A) Periodic (B) Non periodic (C) Simple Harmonic (D) Random

8. Five equal and similar charges are placed at the corners of a regular hexagon as shown in the fig. What

is the electric field and potential at the centre of the hexagon ?

(A) 200 4

5,

4

5

l

q

l

q

�� (B) 200 4

5,

4

1

l

q

l

q

��

(C) 200 4

1,

4

5

l

q

l

q

�� (D) 200 4

1,

4

1

l

q

l

q

��9. Two point charges Q and -3Q are placed certain distance apart. If the electric field at the location of Q

be E�

, then that at the location of -3Q will be

(A) E�

3 (B) E�

3� (C) 3/E�

(D) 3/E�

�

75

10. A small particle of mass m and charge –q is placed at point P and released. If R >> x, the particle will

undergo oscillations along the axis of symmetry with an angular frequency that is equal to

(A) 3

0mR4

qQ

�� (B) 4

0mR4

qQx

��

(C) 30mR4

qQ

�� (D) 40mR4

qQx

��

11. A point charge 50mC is located in the XY plane at the point of position vector �r i j

02 3� �� . What is the

electric field at the point of position vector �r i j� �8 5� �

(A) 1200V/m (B) 0.04V/m (C) 900V/m (D) 4500 V/m

12. A point charge q is placed at origin. Let AE�

, BE�

and CE�

be the electric field at three points

A (1, 2, 3), B (1, 1, – 1) and C (2, 2, 2) due to charge q. Then

[i] AE�

<BE�

[ii] |BE�

| = 4 |CE�

|

select the correct alternative

(A) only [i] is correct (B) only [ii] is correct

(C) both [i] and [ii] are correct (D) both [i] and [ii] are wrong

13. Two identical point charges are placed at a separation of L P is a point on the line joining the charges, at

a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to

zero to slightly less than l. Which of the following best represents the resulting curve?

(A) (B) (C) (D)

14. A particle of mass m and charge Q is placed in an electric field E which varies with time t as

E = E0 sint. It will undergo simple harmonic motion of amplitude

(A) 2

20

m

QE

(B) 2

0

m

QE

(C) 2

0

m

QE

(D)

m

QE0

15. The charge per unit length of the four quadrant of the ring is 2� – 2� and – respectively. The

electric field at the centre is

(A) – iR2 0��

(B) j

R2 0��

(C) iR4

2

0��

(D) None

16. The direction (�) of �E at point P due to uniformly charged finite rod will be

(A) at angle 300 from x-axis

(B) 450 from x - axis

(C) 600 from x-axis

(D) none of these

76

17. Two equal negative charges are fixed at the points [0, a ] and [0, –a] on the y-axis. A positive charge Q

is released from rest at the points [2a, 0] on the x-axis . The charge Q will

(A) execute simple harmonic motion about the origin

(B) move to the origin and remain at rest

(C) move to infinity

(D) execute oscillatory but not simple harmonic motion.

18. A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly

charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium

of the particle will be stable

(A) for all values of H (B) only if H > 2

R(C) only if H <

2

R(D) only if H =

2

R

19. Find the force experienced by the semicircular rod charged with a charge

q, placed as shown in figure. Radius of the wire is R and the line of

charge with linear charge density l is passing through its centre and

perpendicular to the plane of wire.

(A) R2

q

02��

(B)

R

q

02��

(C)

R4

q

02��

(D)

R4

q

0��

20. An equilateral triangle wire frame of side L having 3 point charges at its vertices

is kept in x-y plane as shown. Component of electric field due to the configuration

in z direction at (0, 0, L) is [origin is centroid of triangle]

(A) 2L8

kq39(B) zero (C) 2L8

kq9(D) None

21. A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A

and B are far away from the dipole and at equal distance from it. The field at A and B are BA

EandE��

.

(A) BA

EE��

� (B) BA

E2E��

�

(C)BA

E2E��

�� (D) |E|2

1|E|

AB� , and

BE�

is perpendicular to A

E�

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE

1. Electric potential V due to a dipole is related to the distance r of the observation point as

(A) rV (B) 1� rV (C) 2rV (D) 2� rV

2. Two identical metallic balls carry charges of C�� 20 and C��10 . They are put in contact and again

separated to the same distance as before. What will be the ratio of initial to final force between them ?

Ignore the nature of force.

(A) 2 : 1 (B) 4 : 1 (C)8 : 1 (D)16 : 1

3. A conductive sphere of radius r is given a charge Q. Another uncharged sphere of radius R is put in

contact with it and after they have shared the charge, they are separated. What should be the value of

R so that the force between the two charged spheres be the maximum ?

(A) rR 2� (B) rR 2� (C) 2/Rx � (D) 2rR �

77

4. In the above case if the initial potential of the first sphere be V, then what will be the potential after the

sharing of charges ?

(A) V/2 (B) 2/V (C) V/3 (D) 3/V

5. The electric potential at a certain distance from a source charge is 600 V and electric field strength at

that point is 150 N/C. What is the distance of the observation point from the source charge ?

(A) 2 m (B) 3 m (C)4 m (D)6 m

6. Which of the following is a volt :

(A) Erg per cm (B) Joule per coulomb

(C) Erg per ampere (D) Newton / (coulomb x m2)

7. n small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then

its potential will be

(A) V/n (B) Vn (C) Vn1/3 (D) Vn2/3

8. 1000 identical drops of mercury are charged to a potential of 1 V each. They join to form a single drop.

The potential of this drop will be

(A) 0.01 V (B) 0.1 V (C) 10 V (D) 100 V

9. Potential difference between centre & the surface of sphere of radius R and uniform volume

charge density ��within it will be :

(A) �R2

06�(B)

�R2

04�(C) 0 (D)

�R2

02�

10. When a negative charge is released and moves in electric field, it moves toward a position of

(A) lower electric potential and lower potential energy

(B) lower electric potential and higher potential energy

(C) higher electric potential and lower potential energy

(D) higher electric potential and higher potential energy

11. A solid sphere of radius R is charged uniformly. At what distance from its surface is the electrostatic

potential half of the potential at the centre?

(A) R (B) R/2 (C) R/3 (D) 2R

12. In a uniform electric field, the potential is 10V at the origin of coordinates, and 8V at each of the points

(1, 0, 0), (0, 1, 0) and (0, 0, 1). The potential at the point (1, 1, 1) will be

(A) 0 (B) 4 V (C) 8 V (D) 10 V

13. A charge 3 coulomb experiences a force 3000 N when placed in a uniform electric field. The potential

difference between two points separated by a distance of 1 cm along the field lines is

(A) 10 V (B) 90 V (C) 1000 V (D) 9000V

ELECTRIC POTENTIAL ENERGY OF PARTICLES

1. The electrostatic potential energy of a charge of 5 C at a point in the electrostatic field is 50 J. Thepotential at that point is

(A) 0.1 V (B) 5 V (C)10 V (D)250 V

2. What is the electric potential at the centre of a charged shell of radius 0.1m if the potential at itssurface is 10 V

(A) 10 V (B) 1 V (C) 0.1 V (D)zero

78

3. A mass of 1g carrying charge q falls through a potential difference V. The kinetic energy acquired by

it is E. When a mass of 2g carrying the charge q falls through a potential difference V, what will be

the kinetic energy acquired by it ?

(A) 0.25 E (B) 0.50 E (C) 0.75 E (D)E

4. A mass of 1kg carrying a charge of 2 C falls through a potential of 1V. What will be the velocityacquired by it ?

(A) 12

�ms (B) 1

2�

ms (C)1

)2/1(�

ms (D) 1)2/1(

�ms

5. An � -particle is shot at the uranium nucleus. Its distance of closest approach is 9.6nm. What is themaximum value of repulsion experienced by the � -particle ?

(A) N9

1046.0�� (B) N

91092.0

�� (C) N10

106.9�� (D) N

10102.9

��

6. A charge of C�5 is placed at the corner A of an equilateral triangle ABC of each side 1m. What will

be the work done in moving a charge of C�1 from B to C ?

(A) J3

1045�� (B) J

3105.22

�� (C) J3

1025.11�� (D)none of the above

7. A charged particle is released from rest and moves under the combined influence of electric andgravitational fields. Which of the following quantities connected with the particle must increase ?

(A) Electric potential energy (B) Kinetic energy

(C) Gravitational potential energy (D) Total energy

8. An electron carrying charge -e is located at O and another charge q is located at P. The electron is

moved from O to Q such that OPQ is an equilateral triangle. What is the work done in doing so ?

(A) zero (B)OQ

qe

04

1

�� (C)OQ

e2

04

1

��(D)

OQ

q2

04

1

��

9. A ring of radius R carries a charge +q. A test charge 0q� is released on its axis at a distance R3

from its centre. How much kinetic energy will be acquired by the test charge when it reaches the

centre of the ring ?

(A) R

qq0

04

1

�� (B)R

qq

24

1 0

0�� (C)R

qq

34

1 0

0�� (D)R

qq

34

1 0

0��

10. Two identical rings of radii 0.1 m are placed co-axially at a distance 0.5 m apart. The charges on the

rings are C�2 and C�4 respectively. The work done in transferring C�5 charge from the centre of

one ring to that of the other will be nearest to

(A) 0.50 J (B) 0.75 J (C) 1.00 J (D)1.50 J

11. A ring of radius 6 cm is given a charge C�10 . How much work will be done in transporting a

charge of C�6 from its centre to a point 8 cm along its axis ?

(A) 63 J (B) 84 J (C) 105 J (D)126 J

RELATION BETWEEN E & V :

1. A uniform electric field E is directed along +ve x-axis. If the potential V is zero at 0�x , then the

potential at a point +x will be

(A) xE� (B) Ex2

(C) Ex2� (D) xE

79

2. A point charge q is rotated along a circle in the electric field generated by another point charge Q. The

work done by the electric field on the rotating charge in one complete revolution is

(A) zero

(B) positive

(C) negative

(D) zero if the charge Q is at the centre and nonzero otherwise

3. What is the angle between maximum value of potential gradient and the equipotential surface?

(A) zero (B) 4/� (C) 2/� (D) �

4. A uniform electric field of 400 V/m is directed at 45° above the x-axis as shown in the figure. The

potential difference BA VV � is given by

(A) 0

(B) 4V

(C) 6.4 V

(D) 2.8V

5. The variation of electric potential with distance from a fixed point is shown in the figure. What is the

electric field at 3�x ?

(A) 5/3 (B) 5/2 (C) 5 (D)zero

6. An infinite nonconducting sheet of charge has a surface charge density of 10–7 C/m2. The separation

between two equipotential surfaces near the sheet whose potential differ by 5V is

(A) 0.88 cm (B) 0.88 mm (C) 0.88 m (D) 5 × 10–7 m

7. A bullet of mass m and charge q is fired towards a solid uniformly charged

sphere of radius R and total charge + q. If it strikes the surface of sphere with

speed u, find the minimum speed u so that it can penetrate through the sphere.

(Neglect all resistance forces or friction acting on bullet except electrostatic forces)

(A) mR2

q

0�� (B) mR4

q

0�� (C) mR8

q

0�� (D)

mR4

q3

0��

8. In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are

placed at (n – 1) corners. At the centre, the intensity is E and the potential is V. The ratio V/E has

magnitude.

(A) r n (B) r (n – 1) (C) (n – 1)/r (D) r(n – 1)/n

80

9. The equation of an equipotential line in an electric field is y = 2x, then the electric field strength vector at

(1, 2) may be

(A) 4 3� �i j� (B) 4 8� �i j� (C) 8 4� �i j� (D) � �8 4� �i j

10. The electric field in a region is given by : E = kz/axjzax2izaxy4 22 �� , where a is a

positive constant. The equation of an equipotential surface will be of the form

(A) z = constant / [x3y2] (B) z = constant / [xy2]

(C) z = constant / [x4y2] (D) None

11. A uniform electric field having strength E�

is existing in x-y plane as

shown in figure. Find the p.d. between origin O & A(d, d, 0)

(A) Ed (cos�� + sin�) (B) –Ed (sin��– cos��)

(C) 2 Ed (D) none of these

12. In a certain region of space, the potential is given by : V = k[2x2 – y2 + z2]. The electric field at the point

(1, 1, 1) has magnitude =

(A) 6k (B) 6k2 (C) 2k 3 (D) 3k4

13. Uniform electric field of magnitude 100 V/m in space is directed along the line y = 3 + x. Find the

potential difference between point A (3, 1) & B (1, 3)

(A) 100 V (B) 200 2 V (C) 200 V (D) 0

14. A, B, C, D, P and Q are points in a uniform electric field. The potentials

a these points are V (A) = 2 volt. V (P) = V (B) = V (D) = 5 volt.

V (C) = 8 volt. The electric field at P is

(A) 10 Vm–1 along PQ (B) 215 V m–1 along PA

(C) 5 V m–1 along PC (D) 5 V m–1 along PA

15. Four equal positive charges are fixed at the vertices of a square of side L. Z-axis is perpendicular to the

plane of the square. The point z = 0 is the point where the diagonals of the square intersect each other.

The plot of electric field due to the four charges, as one moves on the z-axis.

(A) (B) (C) (D)

16. A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of

length d, is removed (d << R). The electric field at the centre of the ring will now be

(A) directed towards the gap, inversely proportional to R3.

(B) directed towards the gap, inversely proportional to R2.

(C) directed away from the gap, inversely proportional to R3.

(D) directed away from the gap, inversely proportional to R2.

17. Four equal charges +q are placed at four corners of a square with its centre at origin and lying in yz

plane. The electrostatic potential energy of a fifth charge +q’ varies on x-axis as:

(A) (B) (C) (D)

81

18. Two identical thin rings, each of radius R meter are coaxially placed at distance R meter apart. If

Q1 and Q

2 coulomb are respectively 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

(A) zero (B) 1 2 0( )( 2 1) / ( 2 4 )q Q Q R��

(C) 1 2 02( ) / 4q Q Q R�� (D) 1 2 0( )( 2 1) / ( 2 4 )q Q Q R��

19. A circular ring of radius R with uniform positive charge density �per unit length is located in the y-z

plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the

point P(R 3 , O, O) on the positive x-axis directly towards O, with an initial kinetic energy 0

4

q

�

.

(A) The particle crosses O and goes to infinity.

(B) The particle returns to P.

(C) The particle will just reach O.

(D) The particle crosses O and goes to –R 3 .

20. In space of horizontal EF (E = (mg)/q) exist as shown in figure and a

mass m attached at the end of a light rod. If mass m is released from the

position shown in figure find the angular velocity of the rod when it

passes through the bottom most position

(A) l

g(B)

l

g2(C)

l

g3(D)

l

g5

21. Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal

plane and the other is projected along the plane directly towards first particle from a large distance with

speed n. The closed distance of approach be

(A) �� m

Q

4

1 2

0

(B) 2

2

0 m

Q4

4

1

��(C) 2

2

0 m

Q2

4

1

��(D) 2

2

0 m

Q3

4

1

��

22. The diagram shows a small bead of mass m carrying charge q. The bead can

freely move on the smooth fixed ring placed on a smooth horizontal plane. In the

same plane a charge +Q has also been fixed as shown. The potential atthe point

P due to +Q is V. The velocity with which the bead should projected from the

point P so that it can complete a circle should be greater than

(A) m

qV6(B)

m

qV(C)

m

qV3(D) none

23. Electric field given by the vector jyixE ��

is present in the XY plane.

A small ring carrying charge +Q, which can freely slide on a smooth non

conducting rod, is projetced along the rod from the point (0, L) such

that it can reach the other end of the rod. What minimum velocity should

be given

to the ring?(Assume zero gravity)

(A) (QL2/m)1/2 (B) 2(QL2/m)1/2

(C) 4(QL2/m)1/2 (D) (QL2/2m)1/2

82

24. A unit positive point charge of mass m is projected with a velocity V inside the

tunnel as shown. The tunnel has been made inside a uniformly charged non

conducting sphere. The minimum velocity with which the point charge should be

projected such it can it reach the opposite end of the tunnel, is equal to

(A) [�R2/4m�0]1/2

(B) [�R2/24m�0]1/2

(C) [�R2/6m�0]1/2

(D) zero because the initial and the final points are at same potential.

25. A particle of mass 1 kg & charge 3

1mC is projected towards a

non conducting fixed spherical shell having the same charge

uniformly distributed on its surface. Find the minimum initial

velocity of projection required if the particle just grazes the shell.

(A) 3

2m/s (B) 2

3

2m/s (C)

3

2m/s (D) none of these

26. The diagram shows three infinitely long uniform line charges placed on

the X, Y and Z axis. The work done in moving a unit positive charge

from (1, 1, 1) to (0, 1, 1) is equal to

(A) (l ln 2) / 2��0

(B) (l ln 2) /��0

(C) (3l ln 2) / 2��0

(D) None

27. A charged particle of charge Q is held fixed and another charged particle of mass m and charge q (of the

same sign) is released from a distance r. The impulse of the force exerted by the external agent on the

fixed charge by the time distance between Q and q becomes 2r is

(A) mr4

Qq

0�� (B) r4

Qqm

0�� (C) r

Qqm

0�� (D) r2

Qqm

0��

28. Two point charges of +Q each have been placed at the positions ( –a /2, 0, 0) and (a / 2, 0, 0). The locus

of the points where –Q charge can be placed such the that total electrostatic potential energy of the

system can become equal to zero, is represented by which of the following equations?

(A) Z2 + (Y–a)2 = 2a (B) Z2 + (Y–a)2 = 27a2/4

(C) Z2 + Y2 = 15a2 /4 (D) None

29. Figure shows equi-potential surfaces for a two charges

system. At which of the labeled points point will an

electron have the highest potential energy?

(A) Point A (B) Point B

(C) Point C (D) Point D

DIPOLE

1. An electric dipole of moment p�

placed in a uniform electric field E�

has minimum (maximum negative)

potential energy when the angle between p�

and E�

is

(A) 2/� (B)zero (C) � (D) 2/3�

2. What is the angle between the electric dipole moment and the electric field strength due to it on the axialline

(A) 0° (B) 90° (C) 180° (D)none of the above

83

3. Electric field strength E due to a short dipole at a distance r from it are related as

(A) 1� rE (B) 2� rE (C) 3� rE (D) 4� rE

4. An electric dipole when placed in a uniform electric field will have minimum potential energy. The anglebetween dipole moment and electric field is

(A) zero (B) 2/� (C) � (D) 2/3�

5. If the radius of the hydrogen atom be 50 pm and the charge on the electron or proton be C19

106.1�� ,

what will be the dipole moment of the hydrogen atom in the ground state ?

(A) Cm7

10160�� (B) Cm

71080

�� (C) Cm7

1040�� (D)zero

6. A wheel having mass m has charges +q and –q on diametrically opposite points.

It remains in equilibrium on a rough inclined plane in the presence of uniform

vertical electric field E =

(A) q

mg(B)

q2

mg(C)

q2

tanmg �(D) none

7. Figure shows the electric field lines around an electric dipole. Which

of the arrows best represents the electric field at point P ?

(A) (B)

(C) (D)

8. A dipole consists of two particles one with charge +1mC and mass 1kg and the other with charge

–1mC and mass 2kg separated by a distance of 3m. For small oscillations about its equilibrium position,

the angular frequency, when placed in a uniform electric field of 20kV/m is

(A) 0.1rad/s (B) 1.1 rad/s (C) 10 rad/s (D) 2.5rad/s

9. The dipole moment of a system of charge +q distributed uniformly on an arc of radius R subtending an

angle p/2 at its centre where another charge -q is placed is :

(A) �

qR22(B)

�qR2

(C) �

qR(D)

�qR2

10. An electric dipole is kept on the axis of a uniformly charged ring at distance 2R from the centre of the

ring. The direction of the dipole moment is along the axis. The dipole moment is P, charge of the ring is Q

and radius of the ring is R. The force on the dipole is nearly

(A) 2R33

kPQ4(B) 3R33

kPQ4(C) 3R33

kPQ2(D) zero

11. A large sheet carries uniform surface charge density �. A rod of length

2l has a linear charge density on one half and – on the second half.

The rod is hinged at mid point O and makes an angle � with the

normal to the sheet. The torque experienced by the rod is

(A) 0 (B) ��l 2

02

sin�

(C) ��l 2

0

sin� (D) ��l

20

84

12. Two short electric dipoles are placed as shown. The energy of electric

interaction between these dipoles will be

(A) 321

r

cosPkP2 �(B) 3

21

r

cosPkP2 ��(C) 3

21

r

sinPkP2 ��(D) 3

21

r

cosPkP4 ��

13. Point P lies on the axis of a dipole. If the dipole is rotated by 90° anticlock wise, the electric field vector

E�

at P will rotate by(A) 90° clock wise (B) 180° clock wise (C) 90° anti clock wise (D) none

14. 4 charges are placed each at a distance 'a' from origin. The dipole moment of

configuration is

(A) jqa2 (B) jqa3 (C) ]ji[aq2 � (D) none

15. Two identical positive charges are fixed on the y-axis, at equal distances from the origin O. A particle

with a negative charge starts on the x-axis at a large distance from O, moves along the + x-axis, passes

through O and moves far away from O. Its acceleration a is taken as positive along its direction of

motion. The particle’s acceleration a is plotted against its x-coordinate. Which of the following best

represents the plot?

(A) (B) (C) (D)

ELECTRIC FLUX & GAUSS

1. The number of lines of force radiating outward from 1 coulomb of positive charge is

(A) 0� (B) 0/1 � (C) infinite (D)negligible

2. A cylinder of radius r and length l is placed in a uniform electric field E parallel to the axis of the cylinder..

What is the total electric flux through the surface of the cylinder ?

(A) Elr�2 (B) Er2� (C) Erlr )2(

2�� (D)zero

3. Two thin and infinite parallel plates have uniform densities of charge �� and �� . What is the electric

field in the space between them ?

(A) 02�

�(B)

0��

(C)0

2

��

(D)zero

4. Electric flux through a surface of area 100 m2 lying in the xy plane is (in V-m) if k3j2iE ��

(A) 100 (B) 141.4 (C) 173.2 (D) 200

5. An infinite, uniformly charged sheet with surface charge density ��cuts through a spherical Gaussian

surface of radius R at a distance x from its center, as shown in the figure. The electric flux ��through the

Gaussian surface is

(A) 0

2R

��

(B) 0

22 xR2

��

(C) 0

2xR

��

(D) 0

22 xR

��

85

CONDUCTOR AND ITS PROPERTIES :

1. A charge of 10 coulombs is moved along an equipotential surface having a potential of 2 volts. The

work done is

(A) 10 J (B)zero (C) 2J (D)20 J

2. An external agency carries C5� of charge from infinity to a point in an electrostatic field and performs

100 Joules of work. The potential at the given point is

(A) + 10 V (B) - 10 V (C) + 20 V (D)- 20 V

3. The electric field at a distance R due to charge q is E. If the same charge is placed on the copper sphereof radius R, the electric field strength at the surface of the conductor will be

(A) 4/E (B) 2/E (C)E (D) 2 E

4. The surface density on the copper sphere is � . The electric field strength on the surface of the sphereis

(A) � (B) 2/� (C) 0/ �� (D) 02/ ��

5. A circle has been drawn round a point positive charge (+q) on its centre. The work done in taking a unitpositive charge once round it is

(A) 1 J (B) Jq�2 (C) Jq (D)zero

6. Three charges Q, +q and +q are placed at the vertices of a right-angled isoscelestriangle as shown in the figure. The net electrostatic energy of the configurationis zero if Q is equal to

(A) 21

q

�

�(B)

22

q2

�

�(C) q2� (D) q�

7. A spherical charged conductor has � as the surface density of charge. The electric field on its surface

is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be

the electric field on the surface of the new sphere ?

(A) E/4 (B) E/2 (C)E (D)2E

8. A test charge 0q is placed at the centre of a spherical conductor of radius R. A charge Q is placed on

the spherical conductor. What will be the electrostatic force on the conductor due to 0q .

(A) zero (B) 2

0

04

1

R

Qq

�� (C) 2

0

0 24

1

R

Qq

�� (D) 2

0

0 44

1

R

Qq

��

9. The surface density of a spherical conductor is 1� and the electric field on its surface is 1E . The surface

density of an infinite cylindrical conductor is 2� and the electric field on its surface is 2E . Which of the

following relations is correct ?

(A) 1221 �� EE (B) 2211 �� EE (C) 1221 �� EE (D) 2211 �� EE

10. A charge is distributed over two concentric hollow spheres of radii R and r, where R>r, such that the

surface densities of charges are equal )(� . What is the potential at their common centre?

(A) )(0

rR ��

(B) )(0

rR ��

(C) R0��

(D) r0��

11. Two conducting spheres of radii 1r and 2r are charged such that they have the same electric field on

their surfaces. The ratio of the electric potential at their centres is

(A) 21 / rr (B) 21 / rr (C) 22

21 / rr (D)none of the above

86

12. An uncharged metallic hollow sphere is placed in uniform external electric field. The path of the electric

field lines in and around the conductor is represented by

(A) 1 (B) 2 (C) 3 (D)4

13. The energy density in the electric field created by a point charge decreases wih the distance from the

point charge as

(A) 1

r(B) 2

1

r(C) 3

1

r(D) 4

1

r

14. The dimension of 21 2 oE� ( :

o� permittivity of free space; E: electric field) is

(A) 1MLT � (B) 2 2MLT � (C) 2MLT � (D)2 1ML T �

15. Two identical conducting spheres, having charges of opposite sign, attract each other with a force of

0.108 N when separated by 0.5 m. The spheres are connected by a conducting wire, which is then

removed, and thereafter, they repel each other with a force of 0.036 N. The initial charges on the

spheres are

(A) ± 5 ×10-6 C and � 15 × 10-6 C (B) ± 1.0 × 10-6 C and � 3.0 × 10-6 C

(C) ± 2.0 × 10-6 C and � 6.0 × 10-6 C (D) ± 0.5 × 10-6 C and � 1.5 × 10-6 C

16. An uncharged sphere of metal placed inside a charged parallel plate capacitor. The lines of force look

like

(A) (B) (C) (D)

17. If the electric potential of the inner metal sphere is 10 volt & that of the

outer shell is 5 volt, then the potential at the centre will be :

(A) 10 volt (B) 5 volt (C) 15 volt (D) 0

18. Three concentric metallic spherical shell A, B and C or radii a, b and c (a < b < c) have surface charge

densities – �, + �, and –� respectively. The potential of shell A is :

(A) 0�� [a + b – c] (B) 0�� [a – b + c]

(C) 0�� [b – a – c] (D) none

19. An infinite number of concentric rings carry a charge Q each alternately positive

and negative. Their radii are 1, 2, 4, 8...... meters in geometric progression as

shown in the figure. The potential at the centre of the rings will be

(A) zero (B) 0

12

Q

�� (C) 0

8

Q

�� (D) 0

6

Q

��

20. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. The

potential at the centre of the sphere is

(A) 0 V

(B) 10 V

(C) same as at point 5 cm away from the surface out side sphere.

(D) same as a point 25 cm away from the surface.

87

21. Both question (i) and (ii) refer to the system of charges as shown in the figure. A spherical shell with an

inner radius 'a' and an outer radius 'b' is made of conducting material. A point charge +Q is placed at the

centre of the spherical shell and a total charge – q is placed on the shell.

(i). Charge – q is distributed on the surfaces as

(A) – Q on the inner surface, – q on outer surface

(B) – Q on the inner surface, – q + Q on the outer surface

(C) +Q on the inner surface, –q – Q on the outer surface

(D) The charge –q is spread uniformly between the inner and outer surface.

(ii). Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic

potential at a distance R (a < R < b) from the centre of the shell is

(A) 0 (B) a

KQ(C)

R

qQK

�(D)

b

qQK

�(where K =

04

1

��)

22. Two spherical, nonconducting, and very thin shells of uniformly distributed positive charge Q and radius

d are located a distance 10d from each other. A positive point charge q is placed inside one of the shells

at a distance d/2 from the center, on the line connecting the centers of the two shells, as shown in the

figure. What is the net force on the charge q?

(A) 20d361

qQ

�� to the left (B) 20d361

qQ

�� to the right

(C) 20d361

qQ362

�� to the left (D) 20d361

qQ360

�� to the right

23. A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of

sphere and cavity are displaced by a small distance l�

. Force on charge q is :

(A) in the direction parallel to vector l�

(B) in radial direction

(C) in a direction which depends on the magnitude of charge density in sphere

(D) direction can not be determined.

24. There are four concentric shells A, B, C and D of radii a, 2a, 3a and 4a respectively. Shells B and D are

given charges +q and –q respectively. Shell C is now earthed. The potential difference VA – V

C is :

(A) Kq

a2(B)

Kq

a3(C)

Kq

a4(D)

Kq

a6

1. Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1,2), (2,4) and (4,1) show

electrostatic attraction while pairs (2,3) and (4,5) show repulsion. Therefore ball 1 must be

(A) positively charged (B) negatively charged (C) neutral (D) made of metal

2. A negative point charge placed at the point A is

(A) in stable equilibrium along x-axis

(B) in unstable equilibrium along y-axis

(C) in stable equilibrium along y-axis

(D) in unstable equilibrium along x-axis

88

3. Two fixed charges 4Q (positive) and Q (negative) are located at A and B, the distance AB being 3 m.

(A) The point P where the resultant field due to both is zero is on AB outside AB.

(B) The point P where the resultant field due to both is zero is on AB inside AB.

(C) If a positive charge is placed at P and displaced slightly along AB it will execute oscillations.

(D) If a negative charge is placed at P and displaced slightly along AB it will execute oscillations.

4. Two identical charges +Q are kept fixed some distance apart. A small particle P with charge q is placed

midway between them. If P is given a small displacement D, it will undergo simple harmonic motion if

(A) q is positive and D is along the line joining the charges.

(B) q is positive and D is perpendicular to the line joining the charges.

(C) q is negative and D is perpendicular to the line joining the charges.

(D) q is negative and D is along the line joining the charges.

5. A charged cork of mass m suspended by a light string is placed in uniform

electric filed of strength E = )ji( � × 105 NC–1 as shown in the fig. If in equilibrium

position tension in the string is 2mg

(1+ 3) then angle ‘�’ with the vertical is

(A) 60° (B) 30° (C) 45° (D) 18°

6. Which of the following is true for the figure showing electric lines of force?

(E is electrical field, V is potential)

(A) EA > E

B(B) E

B > E

A

(C) VA > V

B(D) V

B > V

A

7. A particle of mass m and charge q is thrown in a region where uniform gravitational field and electric field

are present. The path of particle

(A) may be a straight line (B) may be a circle

(C) may be a parabola (D) may be a hyperbola

8. Four charges of 1 �C, 2 �C, 3 �C, and – 6�C are placed one at each corner of the square of side 1m.

The square lies in the x-y plane with its centre at the origin.

(A) The electric potential is zero at the origin.

(B) The electric potential is zero everywhere along the x-axis only of the sides of the square are parallel

to x and y axis.

(C) The electric potential is zero everywhere along the z-axis for any orientation of the square in the x-

y plane.

(D) The electric potential is not zero along the z-axis except at the origin.

9. Two point charges Q and – Q/4 are separated by a distance x. Then

(A) potential is zero at a point on the axis which is x/3 on the right side of the charge – Q/4

(B) potential is zero at a point on the axis which is x/5 on the left side of the charge – Q/4

(C) electric field is zero at a point on the axis which is at a distance x on the right side of the charge – Q/4

(D) there exist two points on the axis where electric field is zero.

89

10. At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the

potentials are 100V and 75V respectively . Then

(A) potential at its surface is 150V.

(B) the charge on the sphere is (5/3) × 10-10C.

(C) the electric field on the surface is 1500 V/m.

(D) the electric potential at its centre is 225V.

11. Three point charges Q, 4Q and 16Q are placed on a straight line 9 cm long. Charges are placed in such

a way that the system has minimum potential energy. Then

(A) 4Q and 16Q must be at the ends and Q at a distance of 3 cm from the 16Q.

(B) 4Q and 16Q must be at the ends and Q at a distance of 6 cm from the 16Q.

(C) Electric field at the position of Q is zero.

(D) Electric field at the position of Q is 04

Q

�� .

12. Potential at a point A is 3 volt and at a point B is 7 volt , an electron is moving towards A from B.

(A) It must have some K.E. at B to reach A

(B) It need not have any K.E. at B to reach A

(C) to reach A it must have more than or equal to 4 eV 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.

13. Two infinite sheets of uniform charge density +s and –s are parallel to each other as shown in the figure.

Electric field at the

(A) points to the left or to the right of the sheets is zero.

(B) midpoint between the sheets is zero.

(C) midpoint of the sheets is s / e0 and is directed towards right.

(D) midpoint of the sheet is 2s / e0 and is directed towards right.

14. A thin-walled, spherical conducting shell S of radius R is given charge Q. The same amount of charge isalso placed at its centre C. Which of the following statements are correct?

(A) On the outer surface of S, the charge density is 2R2

Q

�.

(B) The electric field is zero at all points inside S.

(C) At a point just outside S, the electric field is double the field at a point just inside S.

(D) At any point inside S, the electric field is inversely proportional to the square of its distance from C.

15. A hollow closed conductor of irregular shape is given some charge. Which of the following statementsare correct?

(A) The entire charge will appear on its outer surface.

(B) All points on the conductor will have the same potential.

(C) All points on its surface will have the same charge density.

(D) All points near its surface and outside it will have the same electric intensity.

16. A conducting sphere A of radius a, with charge Q, is placed concentrically insidea conducting shell B of radius b. B is earthed. C is the common centre of the Aand B.

(A) The field is a distance r from C, where a ��r � b is 20 r

Q

4

1

�� .

(B) The potential at a distance r from C, where a � r ��b, is r

Q

4

1

0�� .

(C) The potential difference between A and B is ��

��

� �� b

1

a

1Q

4

1

0

(D) The potential at a distance r from C, where a � r � b, ��

��

� �� b

1

r

1Q

4

1

0.

90

17. Two thin conducting shells of radii R and 3R are shown in the figure. The outer shell carries a charge +

Q and the inner shell is neutral. The inner shell is earthed with the help of a switch S.

(A) With the switch S open, the potential of the inner sphere is equal to that of the outer.

(B) When the switch S is closed, the potential of the inner sphere becomes zero.

(C) With the switch S closed, the charge attained by the inner sphere is – q/3.

(D) By closing the switch the capacitance of the system increases.

18. X and Y are large, parallel conducting plates closed to each other. Each face has an area A. X is given a

charge Q. Y is without any charge. Points A, B and C are as shown in figure.

(A) The field at B is A2

Q

0�

(B) The field at B is A

Q

0�

(C) The fields at A, B and C are of the same magnitude.

(D) The field at A and C are of the same magnitude, but in opposite directions.

Question No. 19 to 21 (3 questions)

An empty thick conducting shell of inner radius a and outer radius b is shown in

figure. If it is observed that the inner face of the shell carries a uniform charge

density –��and the surface carries a uniform charge density '�'

19. If a point charge qA is placed at the center of the shell, then choose the correct statement(s)

(A) The charge must be positive

(B) The charge must be negative

(C) The magnitude of charge must be 4��a2

(D) The magnitude of charge must be 4��(b2 – a2)

20. Choose the correct statement related to the potential of the shell in absence of qB

(A) Potential of the outer surface is more than that of the inner surface because it is positively charged

(B) Potential of the outer surface is more than that of the inner surface because it carries more charge

(C) Both the surfaces have equal potential

(D) The potential of the outer surface is 0

b

��

21. If the inner surface of the shell is earthed, then identify the correct statement(s)

(A) The potential of both the inner and outer surface of the shell becomes zero

(B) Charge on the outer surface becomes zero

(C) Charge on the inner surface decreases

(D) Positive charge flows from the shell to the earth

22. An electric dipole moment )j0.3i0.2(p ��

mC. m is placed in a uniform electric field

)k0.2i0.3(E ��

× 105 N C–1.

(A) The torque that E�

exerts on p�

is )k9.0j4.0i6.0( �� Nm.

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

(C) The potential energy of the dipole is 0.6 J.

(D) If the dipole is rotated in the electric field, the maximum potential energy of the dipole is 1.3 J.

91

23. Three points charges are placed at the corners of an equilateral triangle of

side L as shown in the figure.

(A) The potential at the centroid of the triangle is zero.

(B) The electric field at the centroid of the triangle is zero.

(C) The dipole moment of the system is qL2

(D) The dipole moment of the system is qL3 .

24. An electric dipole is placed at the centre of a sphere. Mark the correct answer

(A) the flux of the electric field through the sphere is zero

(B) the electric field is zero at every point of the sphere.

(C) the electric potential is zero everywhere on the sphere.

(D) the electric potential is zero on a circle on the surface.

25. If we use permittivity �0, resistance R, gravitational constant G and voltage V as fundamental physical

quantities, then

(A) [angular displacement] = �0R0G0V0 (B) [Velocity] = �–1R–1G0V0

(C) [dipole moment] = �1R0G0V1 (D) [force] = �1R0G0V2

26. An electric field converges at the origin whose magnitude is given by the expression

E = 100rNt/Coul, where r is the distance measured from the origin.

(A) total charge contained in any spherical volume with its centre at origin is negative.

(B) total charge contained at any spherical volume, irrespective of the location of its centre, is negative.

(C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude

3 ×10–13C.

(D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude

3 × 10–9 Coul.

PARAGRAPH-1

A solid, conducting sphere of radius 6 cm carries a charge 3nC.

This sphere is located centrally inside a thick, conducting sphere

with an inner radius of 18 cm and an outer radius of 27 cm. The

hollow sphere is also given a charge 3nC. Three points A, B and C

are marked on the surfaces as shown.

1. Which one of the following figures shows a qualitatively accurate sketch of the electric field lines in

and around this system ?

(A) (B) (C) (D)

92

2. Suppose VA,V

B and V

C are potentials at points A,B and C respectively then values of potential

differences VC – V

B and V

B – V

A respectively are :

(A) 0 V and – 300 V (B) 0 V and o 300 V

(C) 450 V and o 150 V (D) 0 V and o – 150 V

3. Suppose the shell is given additional charge 3nC. The potential difference VB – V

A will become :

(A) –100 V (B) –200 V (C) 300 V (D) –300 V

PARAGRAPH-2

The sketch below shows cross-sections of equipotential surfaces between two charged conductors

that are shown in solid black. Some points on the equipotential surfaces near the conductors are

marked as A,B,C,........ The arrangement lies in air. (Take��0 = 8.85 × 10–12 C2/N m2]

4. Surface charge density of the plate is equal to

(A) 8.85 × 10–10 C/m2 (B) –8.85 × 10–10 C/m2

(C) 17.7 × 10–10 C/m2 (D) –17.7 × 10–10 C/m2

5. A positive charge is placed at B. When it is released :

(A) no force will be exerted on it. (B) it will move towards A.

(C) it will move towards C. (D) it will move towards E.

6. How much work is required to slowly move a – 1mC charge from E to D ?

(A) 2 × 10–5 J (B) –2 × 10–5 J (C) 4 × 10–5 J (D) –4 × 10–5 J

PARAGRAPH-3

A charged particle is suspended at the centre of two thin concentric spherical charged shells, made

of non conducting material. Figure A shows cross section of the arrangement. Figure B gives the net

flux f through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere.

7. What is the charge on the central particle ?

(A) 0.2 mC (B) 2 mC (C) 1.77 mC (D) 3.4 mC

8. What is the charge on shell A ?

(A) 5.31 × 10–6 C (B) – 5.31 × 10–6 C (C) – 3.54 × 10–6 C (D) – 1.77 × 10–6 C

9. In which range of the values of r is the electric field zero ?

(A) 0 to rA

(B) rA to r

B

(C) for r > rB

(D) for no range of r, electric field is zero.

93

PARAGRAPH-4

An uncharged ball of radius R is placed at a point in space and the region out side (from R to �) the

ball is non uniformly charged with a charge density ��= 3r

C coul/m3 where ‘C’ is a positive constant.

10. Electric potential at the centre of the ball is :

(A) Directly proportional to R (B) Directly proportional to R2

(C) Inversely proportional to R (D) Inversely proportional to R2

11. Electric field intensity at a distance x from centre of the ball (x > R) is :

(A) R

xn

R

C2

0

�� (B) �

�

��

� �

� R

Rxn

R2

C2

0

� (C) 20 R2

C

� (R2 – x2) (D) R

xn

x

C2

0

��

12. As we move away from ball’s surface, electric potential :

(A) decreases. (B) increases.

(C) decreases then increases. (D) increases then decreases.

PARAGRAPH-5

In a certain region, there are non-uniform electrical potential (Ve) as well as gravitational potential

(Vg).

The electrical potential varies only with x as shown in figure (i), and the gravitational potential varies

only with y as shown in figure (ii).

Consider a particle of mass 200 kg and charge 20 mC in this field.

13. If the particle is released from point (5cm, 15 cm), it will try to move toward

(A) +x direction and +y direction (B) +x direction and – y direction

(C) – x direction, +y direction (D) – x direction, – y direction

14. What will be acceleration of the particle at point (25, 35)

(A) )ji2( � × 10–2 m/sec2 (B) )ji2( � × 10–2 m/sec2

(C) )ji2( �� × 10–2 m/sec2 (D) )j2i3( � × 10–2 m/sec2

15. Minimum work required to bring the particle from (5, 15) to (25, 35) is :

(A) 0.2 J (B) 0.1 J (C) – 0.2 J (D) – 0.1 J

94

1. Two conducting plates are placed parallel to each other at certain separation in the uniform electric

field (E0) as shown.

Column I Column II

(A) Charge on surface A (P) must be positive value

(B) Charge on surface B (Q) must be negative value

(C) Charge on surface C (R) may be positive value

(D) Charge on surface D (S) may be negative value

2 Column II lists the dependence of electric field along centroidal axis on the distance r from the centre of

the given uniform charge distribution arrangemen t in Column I.

Column I Column II

(A) Ring of radius R (r >> R) (P) r

(B) Infinite line of charge (Q)r

1

(C) Infinite sheet of charge (R) 2r

1

(D) Uniformly charged non-conducting shell (r > R) (S) None of these

of radius ‘a’ at the point r < a.

3. Column I are the uniform charge distribution arrangements and column II are the standard value of the

potential due to them at any point in space. (Symbols have there usual meaning)

Column I Column II

(A) Ring along the axis ( 0r � ) (P)R

KQ

(B) Hemispherical shell at the centre (Q) 22Rr

KQ

�

(C) Spherical shell on the surface (R) )rR3(R2

KQ 22

3�

(D) Solid non-conducting sphere (r < R) (S)R2

KQ3

95

4 Column II corresponds to the graph of electric field versus distance from centre of charge distribution

in column I.

Column I Column II

(A) Ring along its axis (P)

(B) Solid non conducting sphere (Q)

(C) Spherical shell (R)

(D) Combination of charge +Q and –Q (S)

at the perpendicular bisector

5 Two charged parallel conducting plates X and Y are kept close to each other, are given charge Q1 and

Q2 respectively.

(A) charge on surface A (P) zero

(B) charge on surface B (Q) 2

QQ 21 �

(C) charge on surface C (R) 2

QQ 21 �

(D) charge on surface D (S) 2

QQ 12 �

1. A thin circular wire of radius r has a charge Q. If a point charge q is placed at the centre of the ring, thenfind the increase in tension in the wire.

2. A negative point charge 2q and a positive charge q are fixed at a distance l apart. Where should a

positive test charge Q be placed on the line connecting the charge for it to be in equilibrium? What is the

nature of the equilibrium with respect to longitudinal motions?Q3. A charge + 10-9 C is located at the origin in free space & another charge Q at (2, 0, 0). If the X-component

of the electric field at (3, 1, 1) is zero, calculate the value of Q. Is the Y-component zero at (3, 1, 1)?4. A simple pendulum of length l and bob mass m is hanging in front of a large nonconducting sheet having

surface charge density �. If suddenly a charge +q is given to the bob & it is released from the positionshown in figure. Find the maximum angle through which the string is deflected from vertical.

96

5. A clock face has negative charges -q, -2q, -3q, ........., -12q fixed at the position of the corresponding

numerals on the dial. The clock hands do not disturb the net field due to point charges. At what time

does the hour hand point in the same direction is electric field at the centre of the dial.

6. A cavity of radius r is present inside a solid dielectric sphere of radius R, having

a volume charge density of �� The distance between the centres of the sphere

and the cavity is a . An electron e is kept inside the cavity at an angle ��= 45°

as shown . How long will it take to touch the sphere again?

7. Two identical balls of charges q1 & q

2 initially have equal velocity of the same magnitude and direction.

After a uniform electric field is applied for some time, the direction of the velocity of the first ball changes

by 60º and the magnitude is reduced by half . The direction of the velocity of the second ball changes

there by 90º. In what proportion will the velocity of the second ball changes ?

8. A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge

density, ��0 R

r, where��

0 is a constant and r is the distance from the centre of the sphere. Show that:

(a) the total charge on the sphere is Q = ��0 R3 and

(b) the electric field inside the sphere has a magnitude given by, E =4

2

R

rQK.

9. A particle of mass m and negative charge q is thrown in a gravity free space with

speed u from the point A on the large non conducting charged sheet with surface

charge density �� as shown in figure. Find the maximum distance from A on sheet

where the particle can strike.

10. The electric field in a region is given by ixE

E 0��

l� . Find the charge contained inside a cubical volume

bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103N/C, l = 2cm and

a = 1cm.

11 There are 27 drops of a conducting fluid. Each has radius r and they are charged to a potential V0. They

are then combined to form a bigger drop. Find its potential.

12. A charge + Q is uniformly distributed over a thin ring with radius R. A negative point charge – Q and

mass m starts from rest at a point far away from the centre of the ring and moves towards the centre.

Find the velocity of this particle at the moment it passes through the centre of the ring.

13. A point charge + q & mass 100 gm experiences a force of 100 N at a point at a distance 20 cm from a

long infinite uniformly charged wire. If it is released find its speed when it is at a distance 40 cm from wire

14. Two identical particles of mass m carry charge Q each. Initially one is at rest on a smooth horizontal

plane and the other is projected along the plane directly towards the first from a large distance with

an initial speed V. Find the closest distance of approach.

15. Three charges 0.1 coulomb each are placed on the corners of an equilateral triangle of side 1 m. If the

energy is supplied to this system at the rate of 1 kW, how much time would be required to move one of

the charges onto the midpoint of the line joining the other two?

16. A circular ring of radius R with uniform positive charge density��per unit length is fixed in the Y-Z plane

with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P on

the positive X-axis directly towards O, with initial velocity v . Find the smallest value of the speed v such

that the particle does not return to P.

97

17. Two concentric rings of radii r and 2r are placed with centre at origin. Two

charges +q each are fixed at the diametrically opposite points of the rings

as shown in figure. Smaller ring is now rotated by an angle 90° about Z-axis

then it is again rotated by 90° about Y-axis. Find the work done by

electrostatic forces in each step. If finally larger ring is rotated by 90° about

X-axis, find the total work required to perform all three steps.

18. The figure shows three infinite non-conducting plates of charge perpendicular

to the plane of the paper with charge per unit area + �, + 2� and –�. Find the

ratio of the net electric field at that point A to that at point B.

19. Two thin conducting shells of radii R and 3R are shown in figure. The outer shell carries

a charge +Q and the inner shell is neutral. The inner shell is earthed with the help of

switch S. Find the charge attained by the inner shell.

20. Consider three identical metal spheres A, B and C. Spheres A carries charge + 6q and sphere B carries

charge – 3q. Sphere C carries no charge. Spheres A and B are touched together and then separated.

Sphere C is then touched to sphere A and separated from it. Finally the sphere C is touched to sphere B

and separated from it. Find the final charge on the sphere C.

21. A spherical balloon of radius R charged uniformly on its surface with surface density �. Find work done

against electric forces in expanding it upto radius 2R.

22. Six charges are placed at the vertices of a regular hexagon as shown in the

figure. Find the electric field on the line passing through O and perpendicular

to the plane of the figure as a function of distance x from point O.

(assume x >> a)

23. In the figure shown S is a large nonconducting sheet of uniform charge density s. A

rod R of length l and mass ‘m’ is parallel to the sheet and hinged at its mid point. The

linear charge densities on the upper and lower half of the rod are shown in the figure.

Find the angular acceleration of the rod just after it is released.

24. A dipole is placed at origin of coordinate system as shown in figure, find

the electric field at point P (0, y).

25. Two point dipoles k2

pandkp are located at (0, 0, 0) and (1m, 0, 2m) respectively. Find the resultant

electric field due to the two dipoles at the point (1m, 0, 0).

98

PROPERTIES OF CHARGE AND COULOMB’S LAW :

ELECTRIC FIELD :

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE :


POTENTIAL ENERGY OF SYSTEM OF POINT CHARGE


DIPOLE :

FLUX CALCULATION AND GAUSS'S LAW :


99

PROPERTIES OF CHARGE AND COULOMB’S LAW :

ELECTRIC FIELD :

ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE :



DIPOLE :

FLUX & GAUSS :


100

(1.) a = l(1 + 2 ), stable (2.) –

2/3

11

3�

!"#

$3 × 10–9 C (3.) 0

(4.) 0 (5.) 20

2r8

qQ

�� (6.)0m2

3

��

(7.) 2 tan–1

��

��

�

�

�

mg2

q

0

0(8.)

mR

kQ2 2

(9.)– 0

32R

��

(10.) 2n20 l (11.) 9V0

(12.) Q

m V

2

0

2� �

(13.)�

�q

mu2 20

(14.) 1.8 ´ 105 sec (15.) –Q/3

(16.) 1.125 q (17.) kP

yi j

22

3( � �)� � (18.) kkp

8

7�

(19.) 9.30 (20.)m2

q

0�

(21.) H2 = h

1 + h

2 - g

�

V

��

��

2

(22.)Wfirst step

= r

Kq

5

4

3

82

��

��

�� , W

second step = 0, W

total = 0 (23.)

ae

mr26 0

��

(24.)3

v(25.) 2.2 × 10–12C

electrostatics - Bothra Classes

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