E20-1

Electrostatics

Coulomb’s Force

Q.1 When 1410 electrons are removed from a neutral metal sphere, the charge on the sphere becomes

(1) C16 (2) C16 (3) C32 (4) C32

Ans. (1)

Sol. 1914 106.110 neQ CCQ 16106.1 5

Electrons are removed, so chare will be positive.

Q.2 Number of electrons in one coulomb of charge will be

(1) 291046.5 (2) 181025.6 (3) 19106.1 (4) 11109

Ans. (2)

Sol.18

191025.6

106.1

1

e

Qn

Q.3 There are two charges +1 microcoulombs and +5 microcoulombs. The ratio of the forces acting on them will be

(1) 1 : 5 (2) 1 : 1 (3) 5 : 1 (4) 1 : 25

Ans. (2)

Sol. The same force will act on both bodies although their directions will be different.

Q.4 A charge 1q exerts some force on a second charge 2q . If third charge 3q is brought near, the force of 1q exerted on 2q

(1) Decreases

(2) Increases

(3) Remains unchanged

(4) Increases if 3q is of the same sign as 1q and decreases if 3q is of opposite sign

Ans. (3)

Sol. The force will still remain 20

21

4 r

qq

.

Q.5 The ratio of the forces between two small spheres with constant charge )(a in air )(b in a medium of dielectric constant

K is:

(1) 1 : K (2) K : 1 (3) 2:1 K (4) 1:2K

Ans. (2)

Sol. 20

212

0

21

4,

4 rK

qqF

r

qqF ba

1:: KFF ba .

Q.6 Four charges are arranged at the corners of a square ABCD , as shown in the adjoining figure. The force on the charge

kept at the centre O is

O

C

+q

+2q

BA+q

D

– 2q

(1) Zero (2) Along the diagonal AC

(3) Along the diagonal BD (4) Perpendicular to side AB

Ans. (3)

EXERCISES-I

Electrostatics

E20-2

Electrostatics

Sol. We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B

and D is towards D along the diagonal BD.

Q.7 A total charge Q is broken in two parts 1Q and 2Q and they are placed at a distance R from each other. The maximum

force of repulsion between them will occur, when

(1)R

QQQ

R

QQ 12 , (2)

3

2,

412

QQQ

QQ

(3)4

3,

412

QQ

QQ (4)

2,

221

QQ

QQ

Ans. (4)

Sol. QQQ 21 ..... (i) and 221

r

QQkF .....(ii)

From (i) and (ii) 211 )(

r

QQkQF

For F to be maximum 01

dQ

dF

221

QQQ

Q.8 Three charges are placed at the vertices of an equilateral triangle of side ‘a’ as shown in the following figure. The force

experienced by the charge placed at the vertex A in a direction normal to BC is

A

+QCaB

– Q

+ Q

(1) )4/( 20

2 aQ (2) )4/( 20aQ (3) Zero (4) )2/( 2

02 aQ

Ans. (3)

Sol. 2

2

.||||a

QkFF CB

60o

A

+Q

FB

FC

– QCB

a

FC sin 60o

FB sin 60o

60o

60oFC cos 60o

FB cos 60o

60o 60o

Hence force experienced by the charge at A in the direction normal to BC is zero.

Electric field and field lines

Q.9 A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in

equilibrium, if q is equal to

(1)2

Q (2)

4

Q (3)

4

Q (4)

2

Q

Ans. (2)

Sol. Suppose in the following figure, equilibrium of charge B is considered. Hence for it’s equilibrium |||| CA FF

E20-3

Electrostatics

2

2

0 44

1

x

Q

204

1

x

qQ

4

Qq

QA = Q q

x

x1 x2

FCFA

A BC

QB = Q

Short Trick : For such type of problem the magnitude of middle charge can be determined if either of the extreme charge

is in equilibrium by using the following formula.

If charge A is in equilibrium then q = –

2

1

x

xQB

If charge B is in equilibrium then

2

2

x

xQq A

If the whole system is in equilibrium then use either of the above formula.

Q.10 ABC is an equilateral triangle. Charges q are placed at each corner. The electric intensity at O will be

+q +q

+q

r

rr

A

B C

O

(1) 204

1

r

q

(2) r

q

04

1

(3) Zero (4) 20

3

4

1

r

q

Ans. (3)

Sol.

EC =E EB =E

EA = E

120o

120o

120o

EA = E

EBC = E

EC EB

EA

Enet = 0

Q.11 The magnitude of electric field intensity E is such that, an electron placed in it would experience an electrical force equal

to its weight is given by

(1) mge (2)e

mg(3)

mg

e(4) g

m

e2

2

Ans. (2)

Sol. According to the question, mgeE e

mgE

E20-4

Electrostatics

Q.12 An electron and a proton are in a uniform electric field, the ratio of their accelerations will be

(1) Zero (2) Unity

(3) The ratio of the masses of proton and electron (4) The ratio of the masses of electron and proton

Ans. (3)

Sol.m

qEa

e

p

p

e

m

m

a

a

Q.13 Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field

between the plates is mV /102 5 . When the space is filled with dielectric, the electric field becomes mV /101 5 . The

dielectric constant of the dielectric material

(1) 1/2 (2)1 (3)2 (4)3

Ans. (3)

Sol. 2101

1025

5

dielectricwith

dielectricwithout

E

EK

Q.14 The distance between the two charges C25 and C36 is cm11 At what point on the line joining the two, the intensity

will be zero

(1) At a distance of cm5 from C25 (2) At a distance of cm5 from C36

(3) At a distance of cm10 from C25 (4) At a distance of cm11 from C36

Ans. (1)

Sol. Suppose electric field is zero at point N in the figure then

E2 E1N

x1 x2

Q1 = 25C Q2 = 36C

x = 11 cm

At N |E1| = |E

2|

which gives cm

Q

Q

xx 5

125

36

11

11

2

1

Q.15 A charge particle is free to move in an electric field. It will travel

(1) Always along a line of force

(2) Along a line of force, if its initial velocity is zero

(3) Along a line of force, if it has some initial velocity in the direction of an acute angle with the line of force

(4) None of the above

Ans. (2)

Sol. Because E points along the tangent to the lines of force. If initial velocity is zero, then due to the force, it always moves

in the direction of E. Hence will always move on some lines of force.

Q.16 The intensity of the electric field required to keep a water drop of radius cm510 just suspended in air when charged with

one electron is approximately

(1) cmvolt /260 (2) coulombnewton /260 (3) cmvolt /130 (4) coulombnewton /130

)106.1,/10( 19 coulombekgnewtong

Ans. (2)

E20-5

Electrostatics

Sol. For balance eEmg e

mgE

Also kgdrm 1000)10(7

22

3

4

3

4 373

19

37

106.1

101000)10(7/223/4

E = 260 N/C

Q.17 The figure shows some of the electric field lines corresponding to an electric field. The figure suggests

A B C

(1) CBA EEE (2) CBA EEE (3) BCA EEE (4) BCA EEE

Ans. (3)

Sol. At A and C, electric lines are equally spaced and dense that’s why BCA EEE

Q.18 Four charges are placed on corners of a square as shown in figure having side of cm5 . If Q is one microcoulomb, then

electric field intensity at centre will beQ – 2Q

+ 2Q– Q

(1) CN /1002.1 7 upwards (2) CN /1004.2 7 downwards

(3) CN /1004.2 7 upwards (4) CN /1002.1 7 downwards

Ans. (1)

Sol. Side a = 5 10–2 m

Half of the diagonal of the square2

ar

O

2E

E E

2E

q – 2q

+ 2q– q

r

90o

E

E

q – 2q

+ 2q– q

Electric field at centre due to charge q2

2

a

kqE

Now field at O 222 EEE 2.

2

2

a

kq

22

69

)105(

2210109

CN /1002.1 7 (upward)

E20-6

Electrostatics

Q.19 Figures below show regular hexagons, with charges at the vertices. In which of the following cases the electric field at the

centre is not zero

(1)

q q

q

qq

q(2)

q – q

q

– qq

q(3)

2q 2q

q

2q2q

q

(4)

2q q

2q

q2q

q

Ans. (2)

Sol. Electric field at a point due to positive charge acts away from the charge and due to negative charge it act’s towards the

charge.q q

qq

q

E

E

EE

EE

Enet = 0q

q

2E

2E

2E

120o

Enet = 2E

q – q

– qq

q E

E

E

E E

E

2q 2q

2q2q

q q2E

2E2E

E2E

EEnet = 0

2q q

q2q

q 2q2E

E2E

2E

E

E

E

2E

E

2E

Enet = 0

Gauss law and application

Q.20 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

(1) ER 22 (2) ER /2 (3) ERR /)( 2 (4) Zero

Ans. (4)

Sol. Flux through surface A 2REA and 2REB

E

A BC

Flux through curved surface C odsEdsE 90cos. = 0

Total flux through cylinder CBA = 0

Q.21 Electric field at a point varies as 0r for

(1) An electric dipole (2) A point charge

(3) A plane infinite sheet of charge (4) A line charge of infinite length

Ans. (3)

Sol. E = / (20)

E20-7

Electrostatics

Q.22 An electric charge q is placed at the centre of a cube of side a. The electric flux on one of its faces will be

(1)06

q(2) 2

0 a

q

(3) 204 a

q

(4)0

q

Ans. (1)

Sol. By Gauss’s theorem.

Q.23 For a given surface the Gauss’s law is stated as 0 dsE . From this we can conclude that

(1) E is necessarily zero on the surface (2) E is perpendicular to the surface at every point

(3) The total flux through the surface is zero (4) The flux is only going out of the surface

Ans. (3)

Q.24 A cube of side l is placed in a uniform field E , where iE ˆE . The net electric flux through the cube is

(1) Zero (2) El 2 (3) El 24 (4) El 26

Ans. (1)

Sol. As there is no charge residing inside the cube, hence net flux is zero.

Q.25 A charge q is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface

of the vessel is

(1) Zero (2)0

q(3)

02

q(4)

0

2

q

Ans. (3)

Sol. To apply Gauss’s theorem it is essential that charge should be placed inside a closed surface. So imagine another similar

cylindrical vessel above it as shown in figure (dotted).

q

Q.26 Electric charge is uniformly distributed along a long straight wire of radius 1mm. The charge per cm length of the wire is

Q coulomb. Another cylindrical surface of radius 50 cm and length 1m symmetrically encloses the wire as shown in the

figure. The total electric flux passing through the cylindrical surface is

50cm

1m

++

+

++

+

(1)0

Q(2)

0

100

Q(3) )(

10

0

Q(4) )(

100

0

Q

Ans. (2)

Sol. Charge enclosed by cylindrical surface (length 100 cm) is .100 QQenc By applying Gauss’s law )100(1

)(1

0.

0

QQenc

E20-8

Electrostatics

Q.27 321 ,, qqq and 4q are point charges located at points as shown in the figure and S is a spherical Gaussian surface of

radius R. Which of the following is true according to the Gauss’s lawS

q1 R

q2 q3

q4

(1)0

321321

2).(

qqqAdEEE

s

(2)0

321321

)().(

qqqAdEEE

s

(3)0

4321321

)().(

qqqqAdEEE

s

(4) None of the above

Ans. (2)

Sol. By using )(1

0encQdAE

Q.28 If the electric flux entering and leaving an enclosed surface respectively is 1 and 2 the electric charge inside the

surface will be

(1) 021 )( (2) 012 )( (3) 021 /)( (4) 012 /)(

Ans. (2)

Sol. encnet Q0

1

012 )( encQ

Q.29 Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is

+q

S+q

+q

(1) 0/3 q (2) 0/2 q (3) 0/ q (4) Zero

Ans. (2)

Sol. )2(11

00

qQenc

Q.30 Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of

the electric field over the spherical surface the electric field will be due to

q2+q1

–q1

(1) 2q (2) Only the positive charges

(3) All the charges (4) 1q and 1q

Ans. (3)

Sol. The electric field is due to all charges present whether inside or outside the given surface.

E20-9

Electrostatics

Q.31 An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct(1) Electric flux is coming towards sphere(2) Electric flux is coming out of sphere(3) Electric flux entering into sphere and leaving the sphere are same(4) Water does not permit electric flux to enter into sphere

Ans. (3)Sol. In electric dipole, the flux coming out from positive charge is equal to the flux coming in at negative charge i.e. total charge on sphere = 0. From Gauss

law, total flux passing through the sphere = 0.

Q.32 Two infinite plane parallel sheets separated by a distance d have equal and opposite uniform charge densities . Electric field at a point between

the sheets is

(1) Zero (2)0

(3)02

(4) Depends upon the location of the point

Ans. (2)Sol. According to Gauss’s applications.

Q.33 Which of the following graphs shows the variation of electric field E due to a hollow spherical conductor of radius R as a function of distance from thecentre of the sphere

(1)

R

E

r

(2)

R

E

x

(3)

R r

E

(4)

R r

E

Ans. (1)

Sol. Electric field due to a hollow spherical conductor is governed by following equation ,0E for r < R ...(i)

and 204 r

QE

for Rr ....(ii)

i.e. inside the conductor field will be zero and outside the conductor will vary according to 2

1

rE

Q.34 A conducting sphere of radius 20R cm is given a charge CQ 16 . What is E at centre

(1) CN /106.3 6 (2) CN /108.1 6 (3) Zero (4) CN /109.0 6

Ans. (3)Sol. Electric field inside a conductor is always zero.

Q.35 At a point 20 cm from the centre of a uniformly charged dielectric sphere of radius 10 cm, the electric field is 100 V/m. The electric field at 3 cm fromthe centre of the sphere will be(1) 150 V/m (2) 125 V/m (3) 120 V/m (4) Zero

Ans. (3)

Sol. Electric field outside of the sphere 2r

kQEout ...(i)

Electric field inside the dielectric sphere 3R

kQxEin ...(ii)

From (i) and (ii),R

xrEE outin

2

Þ At 3 cm, mVE /12010

)20(3100

3

2

E20-10

Electrostatics

Electrostatic potential and potential energy

Q.36 A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of

the sphere is

(1)0 V (2)10V

(3) Same as at point 5 cm away from the surface (4) Same as at point 25 cm away from the surface

Ans. (2)

Sol. Since potential inside the hollow sphere is same as that on the surface.

Q.37 If a unit positive charge is taken from one point to another over an equipotential surface, then

(1) Work is done on the charge (2) Work is done by the charge

(3) Work done is constant (4) No work is done

Ans. (4)

Sol. On the equipotential surface, electric field is normal to the charged surface (where potential exists) so that no work will

be done.

Q.38 Charges of C9103

10 are placed at each of the four corners of a square of side cm8 . The potential at the intersection

of the diagonals is

(1) volt2150 (2) volt21500 (3) volt2900 (4) volt900

Ans. (2)

Sol. Potential at the centre O,2/

.4

14

0 a

QV

where CQ 9103

10 and mcma 21088 2

a

A+q +q

B

O

+qD C+qSo

2

108

103

10

10952

9

9

V volt21500

Q.39 In the electric field of a point charge q , a certain charge is carried from point A to B , C , D and E . Then the work done

+q

B

C D

E

A

(1) Is least along the path AB (2) Is least along the path AD

(3) Is zero along all the paths ADACAB ,, and AE (4) Is least along AE

Ans. (3)

Sol. ABCDE is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to

another.

Q.40 If E is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

(1) E (2) 2E (3) 2/1 E (4) 3E

Ans. (2)

Sol. Electrostatic energy density2

02

1EK

dV

dU

2E

dV

dU

E20-11

Electrostatics

Q.41 Four equal charges Q are placed at the four corners of a square of each side is '' a . Work done in removing a charge – Q from its centre to infinity

is

(1) 0(2)a

Q

0

2

4

2

(3)

a

Q

0

22

(4)

a

Q

0

2

2

Ans. (3)

Sol. Potential at centre O of the squareQ Q

QQ

2

aa

O

)2/(44

0 a

QVO

Work done in shifting (– Q) charge from centre to infinity 0)( QVVVQW O a

Q

0

2

4

24

a

Q

0

22

Q.42 A particle A has charge q and a particle B has charge q4 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 speedB

A

v

vwill become

(1) 1:2 (2) 2:1 (3) 4:1 (4) 1:4

Ans. (2)

Sol. Usingm

QVv

2 Þ Qv Þ 2

1

4

q

q

Q

Q

v

v

B

A

B

A

Q.43 Two charge q and q are situated at a certain distance. At the point exactly midway between them

(1) Electric field and potential both are zero (2) Electric field is zero but potential is not zero

(3) Electric field is not zero but potential is zero (4) Neither electric field nor potential is zero

Ans. (3)

Sol. At O, E ¹ 0, V = 0

r

+q – q

r

O

E+ E–

Q.44 Four identical charges C50 each are placed, one at each corner of a square of side m2 . How much external energy is required to bring another

charge of C50 from infinity to the centre of the square

2

29

0

1094

1Given

C

Nm

(1) J64 (2) J41 (3) J16 (4) J10

Ans. (1)

Sol. Potential at the centre of square

VV 469

102902/2

10501094

Work done in bringing a charge (q = 50 mC) from ¥ to centre (O) of the square is 00 )( qVVVqW

Þ JW 64102901050 46

E20-12

Electrostatics

Q.45 In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference V2400

between the plates. To keep a drop of half the radius stationary the potential difference had to be made V600 . What is

the charge on the second drop

(1)4

Q(2)

2

Q(3) Q (4)

2

3Q

Ans. (2)

Sol. In balance condition

mgQE grd

VQ

3

3

4

V

rQ

3

1

2

3

2

1

2

1

V

V

r

r

Q

Q

; 2

2400

600

2/

3

2

r

r

Q

Q Q

2= Q/2

Q.46 Two point charges C100 and C5 are placed at points A and B respectively with cmAB 40 . The work done by

external force in displacing the charge C5 from B to C , where cmBC 30 , angle2

ABC and

229

0

/1094

1CNm

(1) J9 (2) J20

81(3) J

25

9(4) J

4

9

Ans. (4)

Sol. Work done in displacing charge of 5 m C from B to C is

)(105 6BC VVW where

100 CA

50 cm40 cm

30 cmB

+5 CC

VVB6

69 10

4

9

4.0

10100109

and VVC6

69 10

5

9

5.0

10100109

So JW4

910

4

910

5

9105 666

Q.47 Three charges qQ , and q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic

energy of the configuration is zero if Q is equal to

+q+q

Q

a

(1)21

q(2)

22

2

q(3) q2 (4) q

Ans. (2)

Sol. Net electrostatic energy 02

2

a

kQq

a

kq

a

kQqU

02

QqQ

a

kq

22

2

qQ

E20-13

Electrostatics

Q.48 An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed

of the electron will be

(1) meV / (2) meV / (3) meV /2 (4) meV /2

Ans. (3)

Sol. Kinetic energy eVmvK 2

2

1

m

eVv

2

Q.49 Three charges )(, qQ and )( q are placed at the vertices of an equilateral triangle of side l as shown in the figure. If the

net electrostatic energy of the system is zero, then Q is equal to

Q

+q

l

l +q

l

(1)

2

q(2) )( q (3) )( q (4) Zero

Ans. (1)

Sol. Potential energy of the system

02

l

kqQ

l

kq

l

QqkU

0)( QqQl

kq

2

qQ

Q.50 A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. Let

CBA VVV ,, be the potentials at points BA, and C respectively. Then

C P

A

B

(1) BC VV (2) CB VV (3) BA VV (4) CA VV

Ans. (4)

Sol. Conducting surface behaves as equipotential surface.

Q.51 If an insulated non-conducting sphere of radius R has charge density . The electric field at a distance r from the centre

of sphere )( Rr will be

(1)03

R(2)

0

r(3)

03

r(4)

0

3

R

Ans. (3)

Sol. For non-conducting sphere0

3 3

.

r

R

QrkEin

E20-14

Electrostatics

Q.52 A hollow metallic sphere of radius R is given a charge Q. Then the potential at the centre is

(1) Zero (2) R

Q.

4

1

0 (3) R

Q2.

4

1

0 (4) R

Q

2.

4

1

0

Ans. (2)

Sol. Potential V any where inside the hollow sphere, including the centre is0

1 QV

4 R

.

Q.53 The variation of potential with distance R from a fixed point is as shown below. The electric field at mR 5 is

5

4

3

2

1

1 2 3 4 5 60

Po

ten

tial

invo

lts

Distance R in metres

(1) mvolt /5.2 (2) mvolt /5.2 (3) mvolt /5/2 (4) mvolt /5/2

Ans. (1)

Sol. Intensity at 5m is same as at any point between B and C because the slope of BC is same throughout (i.e., electric field

between B and C is uniform). Therefore electric field at R = 5m is equal to the slope of line BC hence bydr

dVE

;

O 1 2 3 4 5 6

1

2

3

4

5A B

C

Distance R in meters

Po

ten

tial

invo

lts

m

VE 5.2

46

)50(

Q.54 In a hollow spherical shell potential (V) changes with respect to distance (r) from centre

(1)

V

r

(2)

V

r

(3)

V

r

(4)

V

r

Ans. (2)

Sol. R

QVinside

04 for Rr ....(i)

and r

QVout

04 for Rr ....(ii)

i.e. potential inside the hollow spherical shell is constant and outside varies according tor

V1

.

E20-15

Electrostatics

Relation between field and potential

Q.55 The electric potential V at any point O (x, y, z all in metres) in space is given by voltxV 24 . The electric field at the point

)2,0,1( mm in metrevolt / is

(1) 8 along negative X axis (2) 8 along positive X axis

(3) 16 along negative X axis(4) 16 along positive Z axis

Ans. (1)

Sol. The electric potential voltxzyxV 24),,(

Now

z

Vk

y

Vj

x

ViE ˆˆˆ

Now 0,8

y

Vx

x

Vand 0

z

V

Hence ixE ˆ8 , so at point (1m, 0, 2m)

volt/metreiE ˆ8 or 8 along negative X-axis.

Q.56 A uniform electric field having a magnitude 0E and direction along the positive X axis exists. If the potential V is zero

at 0x , then its value at xX will be

(1) 0)( xEV x (2) 0xEVx (3) 02 ExVx (4) 0

2 ExVx

Ans. (2)

Sol.dX

dVE 0xEVx

Q.57 Figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are AE and BE

respectively and if the displacement between A and B is r then

A Br

(1) BA EE (2) BA EE (3)r

EE B

A (4) 2r

EE B

A

Ans. (1)

Sol. In non-uniform electric field. Intensity is more, where the lines are more denser.

Q.58 Two plates are cm2 apart, a potential difference of volt10 is applied between them, the electric field between the plates

is

(1) CN /20 (2) CN /500 (3) CN /5 (4) CN /250

Ans. (2)

Sol. CNd

VE /500

102

102

E20-16

Electrostatics

Q.59 There is an electric field E in X-direction. If the work done on moving a charge C2.0 through a distance of m2 along

a line making an angle 60 with the X-axis is 4.0, what is the value of E

(1) CN /3 (2) CN /4 (3) CN /5 (4) None of these

Ans. (4)

Sol.60°

d

2m

E

X

dqEqVW .

4 = 0.2 E (2 cos 60o)

= 0.2 E (2 0.5)

120

2.0

4 NCE

Q.60 A charge of C5 experiences a force of N5000 when it is kept in a uniform electric field. What is the potential difference

between two points separated by a distance of cm1

(1) V10 (2) V250 (3) V1000 (4) V2500

Ans. (1)

Sol.d

QVQEF 210

55000

V voltV 10

Q.61 Equipotential surfaces are shown in figure. Then the electric field strength will be

10V

30°

10 20 30O

(cm

)

Y 20V 30V

(cm)

X

(1) 100 Vm–1 along X-axis (2) 100 Vm–1 along Y-axis

(3) 200 Vm–1 at an angle 120o with X-axis (4) 50 Vm–1 at an angle 120o with X-axis

Ans. (3)

Sol.

E

10V

30° 30°

20V

r

Using rdEdV .

cos. rEV

cosr

VE

120cos1010

)1020(2

E

mV /2002/1

10

)30sin(1010

10 2

2

Direction of E be perpendicular to the equipotential surface i.e. at 120° with x-axis.

E20-17

Electrostatics

Q.62 The points resembling equal potentials are

R

Q

S

P

(1) P and Q (2) S and Q (3) S and R (4) P and R

Ans. (3)

Q.63 Figure shows three points A, B and C in a region of uniform electric field E . The line AB is perpendicular and BC is parallel to

the field lines. Then which of the following holds good. Where BA VV , and CV represent the electric potential at points A, B and

C respectively

C

A

B

(1) CBA VVV (2) CBA VVV (3) CBA VVV (4) CBA VVV

Ans. (2)

Sol. In the direction of electric field potential decreases.

Electric dipole

Q.64 An electric dipole when placed in a uniform electric field E will have minimum potential energy, if the positive direction

of dipole moment makes the following angle with E

(1) (2) 2/ (3) Zero (4) 2/3

Ans. (3)

Sol. Potential energy = – pE cos

When = 0. Potential energy = – pE (minimum)

Q.65 An electric dipole of moment p is placed in the position of stable equilibrium in uniform electric field of intensity E . It

is rotated through an angle from the initial position. The potential energy of electric dipole in the final position is

(1) cospE (2) sinpE (3) )cos1( pE (4) cospE

Ans. (4)

Sol. Potential energy of dipole in electric field cosPEU ; where is the angle between electric field and dipole.

Q.66 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

(1) pE (2) pE2 (3) pE2 (4) Zero

Ans. (4)

Sol. Work done 0]cos[sin 27090

270

90 pEdpE

+q

– q

E

E20-18

Electrostatics

Q.67 The electric field due to a dipole at a distance r on its axis is

(1) Directly proportional to 3r (2) Inversely proportional to 3r

(3) Directly proportional to 2r (4) Inversely proportional to 2r

Ans. (2)

Sol. 30

2.

4

1

r

pE

Q.68 The torque acting on a dipole of moment P in an electric field E is

(1) EP (2) EP (3) Zero (4) PE

Ans. (2)

Q.69 Electric potential at an equatorial point of a small dipole with dipole moment P (r, distance from the dipole) is

(1) Zero (2) 204 r

P

(3) 3

04 r

P

(4) 3

04

2

r

P

Ans. (1)

Q.70 The potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the

dipole and the line joining the point to the dipole are respectively

(1) o90 and o180 (2) o0 and o90 (3) o90 and o0 (4) o0 and o180

Ans. (4)

Sol. 2

cos

r

pV

If = 0° then .maxaV

If 180 then mineV .

Q.71 The value of electric potential at any point due to any electric dipole is

(1) 2.

r

rpk

(2) 3

.r

rpk

(3) 2

.r

rpk

(4) 3

.r

rpk

Ans. (4)

Sol. Potential due to dipole in general position is given by

2

cos.

r

pkV

33

).(.cos.

r

rpk

r

rpkV

Q.72 Two charges C19102.3 and C9102.3 kept 2.4 Å apart forms a dipole. If it is kept in uniform electric field of

intensity mvolt/104 5 then what will be its electrical energy in equilibrium

(1) J23103 (2) J23103 (3) J23106 (4) J23102

Ans. (2)

Sol. Potential energy of electric dipole

cospEU cos)2( Elq

cos104)104.2102.3( 51019 U

23103 U (approx.)

E20-19

Electrostatics

Conductors and self-energy and electrostatic potential

Q.73 An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force look like

– – – – – – –

+ + + + + + +

A

– – – – – – –

+ + + + + + +

B C– – – – – – –

+ + + + + + +

D– – – – – – –

+ + + + + + +

(1) A (2) B (3) C (4) D

Ans. (3)

Sol. Electric lines of force never intersect the conductor. They are perpendicular and slightly curved near the surface of

conductor.

Q.74 Conduction electrons are almost uniformly distributed within a conducting plate. When placed in an electrostatic field

E , the electric field within the plate

(1) Is zero

(2) Depends upon E

(3) Depends upon E

(4) Depends upon the atomic number of the conducting element

Ans. (1)

Sol. Electric field inside a conductor is zero.

Q.75 The electric field near a conducting surface having a uniform surface charge density is given by

(1)0

and is parallel to the surface (2)

0

2

and is parallel to the surface

(3)0

and is normal to the surface (4)

0

2

and is normal to the surface

Ans. (3)

Sol. Electric field near the conductor surface is given by0

and it is perpendicular to surface.

Electrostatics

E20-20

(1) 1.6 × 10–17 C. (2) 1.6 × 10–19 C. (3) 1.6 × 10–10 C. (4) 4.8 × 10–10 C.

Ans. (2)

Q.2 An electron at rest has a charge of 1.6 × 10–19 C. It starts moving with a velocity v = c/2, where c is the speed of light, then

the new charge on it is -

(1) 1.6 × 10–19 Coulomb (2) 1.6 × 10–19

2

2

11

Coulomb

(3) 1.6 × 10–19 11

22

Coulomb (4)

2

19

2

11

106.1

Coulomb

Ans. (1)

Q.3 Which one of the following statement regarding electrostatics is wrong ?

(1) Charge is quantized

(2) Charge is conserved

(3) There is an electric field near an isolated charge at rest

(4) A stationary charge produces both electric and magnetic fields

Ans. (4)

Q.4 When the distance between two charged particle is halved, the force between them becomes -

(1) One fourth (2) One half (3) Double (4) Four times

Ans. (4)

Sol. 221

r

qKqF

F4r

qKq.4

)2/r(

qKqF

221

221

1

Q.5 Five balls, numbered 1 to 5, are suspended using separate threads. Pairs (1, 2), (2, 4), (4, 1) show electrostatic

attraction, while pairs (2, 3) and (4, 5) show repulsion. Therefore ball 1 :

(1) Must be positively charged (2) Must be negatively charged

(3*) May be neutral (4) Must be made of metal

Ans. (3)

Sol. Attraction is possible between a charged and a neutral object.

Q.6 Two point charges in air at a distance of 20 cm. from each other interact with a certain force. At what distance from each

other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction –

(1) 8.94 × 10–2 m (2) 0.894 × 10–2 m (3) 89.4 × 10–2 m (4) 8.94 × 102m

Ans. (1)

Sol. 21

212

21

rr

qKq

r

qKqF

21

2 r5

1

)cm20(

1

44

21 1080

5

102020r

EXERCISES-II

Q.1 One quantum of charge should be at least be equal to the charge in coloumb:

r 8.94 10 2m1

Electrostatics

E20-21

Q.7 Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal

length 'L' metre, from a hook fixed to a stand. The whole set up is taken in a satellite in to space where there is no gravity

(state of weight lessness) Then the angle () between the two strings is -

(1) 0º (2)90º (3)180º (4) 0º < < 180º

Ans. (3)

Sol.

221

)L2(

qKqF

221

)L2(

qKqF

Q.8 Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero

(1) 10 cm. away from q and between the charge

(2) 20 cm. away from q and between the charge

(3) 10 cm. away from q and out side the line joining the charge.

(4) 10 cm. away from 4q and out side the line joining them.

Ans. (1)

Sol.x (30-x)

E-04q q

22 )x30(

Kq

x

)q4(K

x = 20 cm from 4q

10 cm away from q

Q.9 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 ?

(1)4

l

(2)4

Q q(positive)at9 3

l

(3) Q = q (positive) atl

3(4) Q = q (negative) at

l

3

Ans. (1)

Sol. Negative charge is placed to achieve equilibrium.

–Q4q

x

q

Net force on Q is zero

2)x(

qQ4K

= 2x

kqQ

x = /3

Net force on q is also zero

2( /3)

kQq

= 2

k4qq

; Q =

9

4q

Q q(negative)at9 3

Electrostatics

E20-22

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

from one end of line. What should be the value of Q in order to make the net force on q to be zero?

(1*) –q (2) –2q (3) –q/2 (4) 4q

Ans. (1)

Sol.

Charges are placed as shown on line AC.

For net force on q to be zero, Q must be of –ve sign. If F1is force on q due ot 4q & F

2due to Q

Then, F1= F

2(magnitudewise)

or 2

qq4k

= 2

2

qQk

4q = 4Q

or Q = q (in magnitude)

Q = –q (with sign)

Q.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 :

(1) r (2) r/4 (3*) r/8 (4) 2r

Ans. (3)

Sol. Initially, F = 221

r

qqk....(1)

Finally, 4F = 221

R16

qqk....(2)

221

221

R16

qkq4

r

qkq4 or R =

8

r

Q.12 A total charge of 20 C is divided into two parts and placed at some distance apart. If the charges experience

maximum coulombian repulsion, the charges should be :

(1) 5 C , 15 C (2*) 10 C , 10 C (3) 12 C , 8 C (4) C3

20,C

3

40

Ans. (2)

Sol.

Let the two charges are q & (20 – q)C

Fe

= 2r

)q20()q(K

Fe

will be max, whendq

dFe= 0

ordq

dFe= 2r

K(20 – 2q) = 0

q = 10 C.

Electrostatics

E20-23

(1) 100 N/C (2) 50 N/C (3) 200 N/C (4) 10 N/C

Ans. (2)

Sol. F= qE

C/N502

100E

Q.14 Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square

due to any one charge is E, then the resultant electric field intensity at centre of square will be :

(1) Zero (2) 4E (3)E (4) 1/2E

Ans. (1)

Sol.

q

q

q

q

E

E

EE

ENet

= 0

Q.15 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 ?

(1)

O

E

Lx

(2)

O

E

Lx

(3) O

E

L x (4) O

E

L x

Ans. (4)

Sol.

L

qx

q

E

L/2

Q.13 If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be -

Electrostatics

E20-24

Q.16 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

(1)Eqm

t(2)

2 2 2E q t

2m(3)

2 22E t

mq(4)

2

2

Eq m

2t

Ans. (2)

Sol.m

qEa

After time t

qEv t

m

2mv2

1KE =

m2

tqE 222

Q.17 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:

(1) C/N3 (2) 4 N/C (3) 5 N/C (4*) 20 N/C

Ans. (4)

Sol. W = Fr cos 4 = (0.2) E (2) cos 60º E = 20 N/C.

Q.18 The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be :

(1)04

1

2R33

q(2)

04

1

2R3

q2(3*)

04

1

2R33

q2(4)

04

1

i 2R32

q3

Ans. (3)

Sol. At point P on axis, E = 2/322 )xR(

kqx

Px

E+R + x2 2

For max E,dx

dE= 0 or x =

2

R

Putting x in (i)Emax

= 2R33

kq2

Q.19 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 -

(1) directed towards the gap, inversely proportional to R3.

(2) directed towards the gap, inversely proportional to R2.

(3) directed away from the gap, inversely proportional to R3.

(4) directed away from the gap, inversely proportional to R2.

Ans. (1)

Sol. 2

KdqE

R

ddq

2 R

.d

d

E

dq

3

KE .d

2 R

3

1E

R

Electrostatics

E20-25

Q.20 The direction () of E

at point P due to uniformly charged finite rod will be -

(1) at angle 30° from x-axis (2) 45° from x-axis

(3) 60° from x-axis (4) none of these

Ans. (1)

Sol.

d2/k

d

k2/3

1= 0,

2= 60°

d

k

2

3]0sin60[sin

d

kE

d2

k–]0cos60[cos

d

kE||

tan =3

1 = 30°

Q.21 Two point charges a & b, whose magnitudes are same are positioned at a certain distance from each other with a at

origin. Graph is drawn between electric field strength at points between a & b and distance x from a. E is taken

positive if it is along the line joining from a to b. From the graph, it can be decided that

(1*) a is positive, b is negative (2) a and b both are positive

(3) a and b both are negative (4) a is negative, b is positive

Ans. (1)

Sol. a & b can't be both +ve or both – ve otherwise field would have been zero at their mid point.

b can't be positive even, otherwise the field would have been in –ve direction to the right of mid point

answer is (1)

Electrostatics

E20-26

Q.22 The given figure gives electric lines of force due to two charges q1

and q2. What are the signs of the two charges?

(1*) Both are negative (2) Both are positive

(3) q1

is positive but q2

is negative (4) q1

is negative but q2

is positive

Ans. (1)

Sol. By definition

Q.23 If electric field is uniform, then the electric lines of forces are:

(1) Divergent (2) Convergent (3) Circular (4*) Parallel

Ans. (4)

Q.24 Select the correct statement :

(1) The electric lines of force are always closed curves

(2) Electric lines of force are parallel to equipotential surface

(3*) Electric lines of force are perpendicular to equipotential surface

(4) Electric line of force is always the path of a positively charged particle.

Q.25 Electric flux through a surface of area 100 m2 lying in the xy plane is (in V-m) ifE i j k 2 3

(1)100 (2)141.4 (3)173.2 (4)200

Ans. (3)

Sol. k100A

, k3j2iE

= A•E

=100 3

Q.26 A cylinder of radius (R) and length (L) is placed in a uniform electrical field (E) parallel to the axis of the cyclinder . the

total flux for the surface of the cylinder is given by -

(1) 2R2E (2)R2E (3)E

RR 22 (4) zero

Ans. (4)

Sol. Incoming flux = Outgoing flux

Q.27 A hemisphere (radius R) is placed in electric field as shown in fig. Total outgoing flux is -

(1)R2E (2)2R2E (3)4R2E (4) (R2E)/2

Ans. (1)

Sol. sdE

, ER2

Ans. (3)

Electrostatics

E20-27

Q.28 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

Rx

(1)

R2

0(2)

2 2 2

0

( )R x(3)

0

2)xR(

(4)

( )R x2 2

0

Ans. (4)

Sol. Radius of the cutting

disc = 22 xR

x

R

charge on disc

q = A

q = (R2 – x2)

Now =0

q

=0

22 )xR(

Q.29 Figure shows a charge Q placed at the centre of open face of a cylinder as shown in figure. A second charge q is placed

at one of the positions A, B, C and D, out of which positions A and D are lying on a straight line parallel to open face of

cylinder. In which position(s) of this second charge, the flux of the electric field through the cylinder remains unchanged?

(1*) A (2) B (3) C (4*) D

Ans. (1)

Sol. If charge is at A or D, its all field lines cut the given surface twice which means that net flux due to this charge

remains zero and flux through given surface remains unchanged.

Q.30 If the electric flux entering and leaving a closed surface are respectively of magnitude 1 and 2 , then the electric

charge inside the surface will be :

(1)0

12 –

(2) 021 )( (3*) )–( 120 (4) )( 120

Ans. (3)

Sol. Net flux =2–

1=

0

inq

qin

= 0(

2–

1)

Electrostatics

E20-28

Q.31 Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius

20cm surrounding the total charge is 50 V-m. The flux over a concentric sphere of radius 40 cm will be:

(1*) 50 V-m (2) 75 V-m (3) 100 V-m (4) 200 V-m

Ans. (1)

Sol. since same no of field lines are passing through both spherical surfaces, so flux has same value for both.

Q.32 Three charges q1= 1c , q

2=2c and q

3= –3c and four surfaces S

1, S

2, S

3and S

4are shown. The flux emerging through

surface S2in N-m2 / C is -

(1) 36 × 103 (2) –36 × 103 (3) 36 × 109 (4) –36 × 109

Ans. (2)

Sol.o

32

o

in qqq

= - 31036

Q.33 A surface enclosed an electric dipole, the flux through the surface is-

(1) Infinite (2) Positive (3) Negative (4) Zero

Ans. (4)

Sol. qin= 0

=0

energy K, towards the disc along its axis. The charge q :

(1) may hit the disc at the centre

(2) may return back along its path after touching the disc

(3) may return back along its path without touching the disc

(4*) any of the above three situations is possible depending on the magnitude of K

Ans. (4)

Sol. By M.E. conservation between initial & final point :

Ui+ K

i= U

f+ K

f

Answer is (4)

Q.35 Three equal charges are placed at the three corners of an isosceles triangle as shown in the figure. The statement which

is true for electric potential V and the field intensity E at the centre of the triangle is-

(1) V = 0, E = 0 (2) V = 0, E 0 (3) V 0 , E =0 (4) V 0, E 0

Ans. (3)

Q.34 A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic

Electrostatics

E20-29

Q.36 Electric potential is a -

(1) Vector quantity (2) Scalar quantity (3) Neither vector Nor scalar (4) Fictious quantity

Ans. (2)

Q.37 ABC is equilateral triangle of side 1m. Charges are placed at its corners as shown in fig. O is the mid- point of side BC the

potential at point (O) is-

(1) 2.7 × 103 V (2) 1.52 × 105 V (3) 1.3 × 103 V (4) – 1.52 × 105 V

Ans. (2)

Sol.2/1

)102(KV

6

2/3

)106(K

2/1

)103(K 66

V1052.1 5

Q.38 A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m. When allowed

to fall from rest through same electrical potential difference, the ratio of their speed vA

: vB

will be :

(1) 2 : 1 (2*) 1 : 2 (3) 4 : 1 (4) 1 : 4

Ans. (2)

Sol. 2AmV

2

1= qV,

2BmV

2

1= 4 qV 4

1

V

V2B

2A 2

1

V

V

B

A

Q.39 A charged particle ‘q’ is shot from a large distance with speed v towards a fixed charged particle Q. It approaches

Q upto a closest distance r and then returns. If q were given a speed ‘2v’, the closest distance of approach would

be :

(1) r (2) 2r (3)2

r(4*)

4

r

Ans. (4)

Sol. Comparison can be shown as :

V 2V k 4k PEmax

4PEmax

r 4

r

Q.40 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 ?

(1*) 6 m (2) 12 m (3) 36 m (4) 144 m

Ans. (1)

Sol. V = Er, r =E

V= 6m.

Electrostatics

E20-30

Q.41 Figure represents a square carrying charges +q, +q, –q, –q at its four corners as shown. Then the potential will be

zero at points :

–q –qQ

A BC

+q+q

P

(1) A, B, C, P and Q (2*) A, B and C (3) A, P, C and Q (4) P, B and Q

Ans. (2)

Sol. Apply the formula V =r

kQ

Q.42 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 :

(1*) 27 V (2) 13.5 V (3) 54 V (4) 45.5 V

Ans. (1)

Sol. VC

=r

kQVV

C=

)5.0(

105.1109 9–9 = 27 V..

Q.43 The kinetic energy which an electron acquires when accelerated (from rest) through a potential difference of 1 volt

is called :

(1) 1 joule (2*) 1 electron volt (3) 1 erg (4) 1 watt

Ans. (2)

Q.44 A particle of charge Q and mass m travels through a potential difference V from rest. The final momentum of the

particle is :

(1)Q

mV(2) mVQ2 (3*) QVm2 (4)

m

QV2

Ans. (3)

Sol. K.E. = VQ and momentum = VQm2)KE(m2

Q.45 If a uniformly charged spherical shell of radius 10 cm has a potential V at a point distant 5 cm from its centre, then

the potential at a point distant 15 cm from the centre will be :

(1)3

V(2*)

3

V2(3) V

2

3(4) 3V

Ans. (2)

Sol. Potential at 5cm.

5cm = V =)cm10(

kq

( point lying inside the sphere)

Pontential at 15 cm V15 cm V =3

2

cm15

kq V..

Electrostatics

E20-31

Q.46 If a charge is shifted from a high potential region to low potential region, the electrical potential energy:

(1) Increases (2) Decreases

(3*) May increase or decrease. (4) Remains constant

Ans. (3)

Sol. PE = q (Vfinal

– Vinitial

)

PE = qV PE decreases if q is +ve increases if q is –ve.

Q.47 A particle of mass 2 g and charge 1C is held at rest on a frictionless horizontal surface at a distance of 1 m from a

fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at distance of

10 m from the fixed charge is:

(1) 100 m/s (2*) 90 m/s (3) 60 m/s (4) 45 m/s

Ans. (2)

Sol. By conservation of machenical energy

2

21

1

212

r

qqk–

r

qqkmv

2

1 23– v)102(

2

1

= 9 × 109 × 10–6 × 10–3

10

1–

1

1

or v2 = 9 × 103 ×10

9or v = 90 m/sec

Q.48 When the separation between two charges is decreased, the electric potential energy of the charges

(1) increases (2) decreases

(3*) may increase or decrease (4) remains the same

Ans. (3)

Sol. PE may increase may decrease depending on sign of charges.

Q.49 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 :

(1*)3

2 (2)

3

1 (3)

3

2(4)

2

3

Ans. (1)

Sol. P.E. of system =a

xkq

a

xkq2

a

Kq2 222

= 0 where a is distance between charges.

or 2 + 3x = 0 x = –3

2

Q.50 Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other

is projected along the plane directly towards first particle from a large distance with speed v. The closed distance of

approach be

(1)

2

0

1 Q

4 mv(2)

2

20

1 4Q

4 mv(3)

2

20

1 2Q

4 mv(4)

2

20

1 3Q

4 mv

Ans. (2)

Sol.

m m

Q Q

v

Initially

mu m

Q Q

ufinallyat closestdistance

d

fromE.C.

Electrostatics

E20-32

d

kqmv2/12mv

2

1 222 ...(1)

from

M.C. mv = 2mu u = v/2 ...(2)

from (1) and (2)

2 221 mv kq

mv2 4 d

2

2

mv

kq4d

Q.51 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 A

(1) PointA (2) Point B (3) Point C (4) Point D

Ans. (2)

Sol. U =– QV

Q.52 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.

(1) r n (2) r (n – 1) (3) (n–1)/r (4) r(n–1)/n

Ans. (2)

Sol. E = 2r

Kq; V =

r

)1n(Kq

Vr(n 1)

E

Q.53 When the seperation between two charges is increased, the electric potential energy of the charges

(1) increases (2) decresaes (3) remains the same (4) may increase or decrease

Ans. (4)

Sol. q1 q2d

Seperation increase then

d

qKqU 21

But

q1 –q2d

if d then

d

qkqU 21

Q.54 A point positive charge of Q' units is moved round another point positive charge of Q units in circular path. If the radius

of the circle r is the work done on the charge Q' in making one complete revolution i -

(1) r4

Q

0(2) r4

'QQ

0(3) r4

'Q

0(4)0

Ans. (4)

Electrostatics

E20-33

Q.55 When a negative charge is released and moves in electric field, it moves toward a position of

(1) lower electric potential and lower potential energy

(2) lower electric potential and higher potential energy

(3) higher electric potential and lower potential energy

(4) higher electric potential and higher potential energy

Ans. (3)

Sol.Higher

potential (v )1Lower

potential (v )2

1 1U qV qE –q U2= – qV

2

U1< U

2

Q.56 Two identical thin rings, each of radius R meter are coaxially placed at distance R meter apart. If Q1

and Q2

coulomb are

respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring

to that of the other is

(1) zero (2) q Q Q R( )( ) / ( . )1 2 02 1 2 4

(3) q Q Q R2 41 2 0( ) / (4) q Q Q R( )( ) / ( . )1 2 02 1 2 4

Ans. (2)

Sol.

R R

Q1 Q1

R2

KQ

R

KQv 21

A R2

KQ

R

KQv 12

B

2 1 1 2kQ KQ kQ kQW q

R R2R 2R

2

QQ

2

QQ

4R

qW 2

11

20

)R42(|)12)(QQ(qW 021

Q.57 In an electron gun, electrons are accelerated through a potential difference of V volt. Taking electronic charge and

mass to be respectively e and m, the maximum velocity attained by them is :

(1)m

eV2(2*)

m

eV2(3) 2 m/eV (4) (V2 /2em)

Ans. (2)

Sol.2

1mv2 = eV v =

m

eV2

Electrostatics

E20-34

Q.58 In a cathode ray tube, if V is the potential difference between the cathode and anode, the speed of the electrons,

when they reach the anode is proportional to : (Assume initial velocity = 0)

(1) V (2) 1/V (3*) V (4)

2V

2em

Ans. (3)

Sol.2

1mu2 = eV u =

m

eV2

2

1

Vu

Q.59 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

(1)0 V

(2)10V

(3) same as at point 5 cm away from the surface out side sphere

(4) same as a point 25 cm away from the surface

Ans. (2)

Sol.

5cm

v = const.

R

k= 10V

Q

vin

= 10V

Q.60 Two similar conducting spherical shells having charges 40 C and –20C are some distance apart. Now they are touched

and kept at same distance. The ratio of the initial to the final force between them is :

(1) 8 : 1 (2) 4 : 1 (3) 1 : 8 (4) 1 : 1

Ans. (1)

Sol.

40µC 20µC

F1= 2d

)20)(40(k

After touching the charge on sphere = 10µC

10µC 10µC

F2= 2d

)10)(10(k

F1: F

2= 8 : 1

Electrostatics

E20-35

Q.61 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

b

potential difference between two points separated by a distance of 1 cm is-

(1) 10 Volt (2) 90 Volt (3) 1000 Volt (4) 3000 Volt.

Ans.

Sol. V=E.R

V= 1000×1×10-2 = 10Volt

Q.63 In a region where E = 0, the potential (V) varies with distance r as-

(1)r

1V (2) rV (3) 2r

1V (4) V = const. independent of (r)

Ans. (4)

Sol. When E = 0

dx

dvE

V = constant

Q.64 An equipotential surface and a line of force :

(1) never intersect each other (2) intersect at 45° (3) intersect at 60° (4) intersect at 90°

Ans. (4)

Sol.

line of force

equipotential surface

Angle between both = 90°

Q.65 The potential difference between points A and B in the given uniform electric field is :a

B

bE

E

C

A

(1) Ea (2) E )ba( 22 (3*) Eb (4) )2/Eb(

Ans. (3)

Sol. Since B and C are at same potential (lying on a line to electric field i.e. equipotential surface)

VAB

=VAC

= Eb.

(1) 10 volt (2) 5 volt (3) 15 volt (4)0

Ans. (1)

Q.62 When charge of 3 coulomb is placed in a Uniform electric field , it experiences a force of 3000 newton, within this field,

Electrostatics

E20-36

Q.66 An equipotential surface is that surface

(1) On which each and every point has the same potential

(2) Which has negative potential

(3) Which has positive potential

(4) Which has zero potential

Ans. (1)

Q.67 The equation of an equipotential line in an electric field is y = 2x, then the electric field strength vector at (1, 2) may be

(1) 4 3 i j (2) 4 8 i j (3) 8 4 i j (4) –8 i j 4

Ans. (4)

Sol. e.f is perpendicular to equipotential surface

m for e.f = –2

1

Now check option Ans - D

Q.68 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

(1)10V (2)90V (3)1000V (4)9000 V

Ans. (1)

Sol. F = qE

3000=3E

E = 1000 N/c

V = E. d = 1000 10–2 = 10 volt

Q.69 A, B, C, D, P and Q are points in a uniform electric field. The potentials at these points are V(A) =2 volt. V(P) = V(B) = V(D)

= 5 volt. V(C) = 8 volt. The electric field at P is

0.2

m

0.2 m

C

Q

D

P

B

A

(1) 10 Vm–1 along PQ (2) 15 2 Vm–1 along PAA

(3) 5 Vm–1 along PC (4) 5 Vm–1 along PA

Ans. (2)

Sol.

B

P5v

C8v

A

2vE

D

v = vA

–vP

= 3V

E =d

V=

2 2

3

0.1 0.1= 15 2

Electrostatics

E20-37

Q.70 In the above question, the electric force acting on a point charge of 2 C placed at the origin will be :

(1) 2 N (2) 500 N (3) –5 N (4*) –500 N

Ans. (4)

Sol. At origin, E = –dr

dV= – 2.5 V/cm = – 250 V/m

F = force on 2C = q E = 2 × (– 250) N = – 500 N.

Q.71 The electric potential V as a function of distance x (in metre) is given by

V = (5x2 + 10x – 9) volt.

The value of electric field at x = 1 m would be :

(1*) – 20 volt/m (2) 6 volt/m (3) 11 volt/m (4) –23 volt/m

Ans. (1)

Sol. E = –dx

dV= – 10 x – 10

E(x =1m)

– 10 (1) – 10 = – 20 V/m

Q.72 The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the

electric potential at the origin to be zero, the electric potential at a distance r :

(1) increases as one goes away from the origin. (2*) is proportional to r2

(3) is proportional to r (4) is uniform in the region

Ans. (2)

Sol. E r r

0

v

0

rrddV V (–2

r2

) V r2

Q.73 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

(1)0.88 cm (2)0.88mm (3)0.88 m (4) 5 × 10–7 m

Ans. (2)

Sol.

+++++

+++++

+++++

r2

r1

A B

10

A r2

V

2

0B r

2V

Given VB

– VA

= 5 V

V5)rr(2

120

r2

– r1= 0.88 mm

(1) a force (2) a couple and mover (3) a couple and rotates (4) a force and moves.

Ans. (3)

Q.74 If an electric dipole is kept in a uniform electric field, Then it will experience

Electrostatics

E20-38

Q.75 An electric dipole consists of two opposite charges each of magnitude 1 × 10–6 C separated by a distance 2cm. The

dipole is placed in an external field of 10 × 105N/C. The maximum torque on the dipole is -

(1) 0.2 × 10–3 N-m (2) 1.0 × 10–3 N-m (3) 2 × 10-2 N-m (4) 4 × 10–3 N-m

Ans. (3)

Sol. P = qd

1 × 10-6 × 2 × 10-2= 2×10-8

Maximum torque

= PE = 2 × 10-2Nm

Q.76 The ratio of the electric field due to an electric dipole on its axis and on the perpendicular bisector of the dipole is-

(1) 1 : 2 (2) 2 : 1 (3) 1 : 4 (4) 4 : 1

Ans. (2)

Sol. 3axisr

KP2E

31r

KPE

1

2

E

E

1

axis

Q.77 The region surrounding a stationary electric dipole has-

(1) electric field only (2) magnetic field only

(3) both electric and magnetic fields (4) neither electric nor magnetic field

Ans. (1)

Q.78 Due to an electric dipole shown in fig., the electric field intensity is parallel to dipole axis :

(1) at P only (2) at Q only (3*) both at P and at Q (4) neither at P nor at Q

Ans. (3)

Sol. Since P & Q are axial & equatorial points, so electric fields are parallel to axis at both points.

Q.79 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 electric dipole is largest when is :

(1) zero (2) /2 (3*) (4) /4

Ans. (3)

Sol. max PE position of unstable equilibrium = .

Q.80 Two opposite and equal charges of magnitude 4 × 10–8 coulomb each when placed 2 × 10–2 cm apart form a dipole. If

this dipole is placed in an external electric field of 4 × 108 N/C, the value of maximum torque and the work

required in rotating it through 180º from its initial orientation which is along electric field will be : (Assume

rotation of dipole about an axis passing through centre of the dipole):

(1) 64 × 10–4 N-m and 44 × 10–4 J (2) 32 × 10–4 N-m and 32 × 10–4 J

(3) 64 × 10–4 N-m and 32 × 10–4 J (4*) 32 × 10–4 N-m and 64 × 10–4 J

Ans. (4)

Sol. PEmax = 4 × 10–8 × 2 × 10–4 × 4 × 108 = 32 × 10–4 N-m.

Work done W = (P.E.)f– (P.E.)

i= PE – (–PE) = 2PE = 64 × 10–4 N-m

Electrostatics

E20-39

Q.81 An electric dipole is placed at the centre of a sphere. Mark the correct options.

(1) The electric field is zero at every point of the sphere.

(2) The flux of the electric field through the sphere is non-zero.

(3) The electric field is zero on a circle on the sphere.

(4*) The electic field is not zero anywhere on the sphere.

Ans. (4)

Sol. Since, dipole has net charge zero, so flux through sphere is zero with non-zero electric field at each point of sphere.

force on the object will be :

(1) away from the plate B (2*) towards the plate B

(3) parallel to the plate B (4) zero

Ans. (2)

Sol.+ +

++++

+++

–––

–––––

– –

+ vely charged

BAmetal

plate

The given diagram shows induction on sphere (metallic) due to metal plate.

Since distance between plate and –ve charge is less than that between plate and +ve charge. electric force acts on

object towards plate.

Q.83 A positive point charge q is brought near a neutral metal sphere.

(1) The sphere becomes negatively charged.

(2) The sphere becomes positively charged.

(3*) The interior remains neutral and the surface gets non-uniform charge distribution.

(4) The interior becomes positively charged and the surface becomes negatively charged.

Ans. (3)

Sol. Induction takes place on outer surface of sphere producing non-uniform charge distribution & since external electric field

can not enter the sphere, so interior remains charge free.

Q.84 You are travelling in a car during a thunder storm. In order to protect yourself from lightening, would you prefer to:

(1*) Remain in the car (2) Take shelter under a tree

(3) Get out and be flat on the ground (4) Touch the nearest electrical pole

Ans. (1)

Sol. Car (A conductor) behaves as electric field shield in which a person remains free from shock.

Q.85 A positively charged body 'A' has been brought near a neutral brass sphere B mounted on a glass stand as shown

in the figure. The potential of B will be:

(1) Zero (2) Negative (3*) Positive (4) Infinite

Ans. (3)

Q.82 A neutral spherical metallic object A is placed near a finite metal plate B carrying a positive charge. The electric

Electrostatics

E20-40

Sol. Potential of B = Potintial at the centre of B

= Potential due to induced charges + potential due to A.

= 0 + (+ ve)

Potential of B is +ve.

Q.86 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)

(1*) zero (2) q

(3) qE (PQ + QR + SR + SP) (4) 0/q

Ans. (1)

Sol. Since electric field produced by charge is conservative, so work done in closed path is zero.

Q.87 The net charge given to an isolated conducting solid sphere :

(1) must be distributed uniformly on the surface (2) may be distributed uniformly on the surface

(3) must be distributed uniformly in the volume (4) may be distributed uniformly in the volume.

Ans. (1)

Sol. In a conductor given charge is distributed uniformly on the surface of sphere

Q.88 The net charge given to a solid insulating sphere:

(1) must be distributed uniformly in its volume

(2) may by distributed uniformly in its volume.

(3) must be distributed uniformly on its surface.

(4) the distribution will depend upon whether other charges are present or not.

Ans. (2)

Sol. Depends on body either conductor or non-conducting.