electro statics & capacitors · 2020-04-27 · t q q q 2 1 2 0 2 2 0 1 4 1 4 q f l q f l s s...

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ELECTRO STATICS & CAPACITORSJEE-ADV PHYSICS-VOL- II

Charge and its properties Study of characteristics of electric charges at rest

is known as electrostatics. Electric charge is the property associated with a

body or a particle due to which it is able to produceas well as experience the electric and magneticeffects. Charge is a fundamental property of matter andnever found free. The excess or deficiency of electrons in a body givesthe concept of charge. There are two types of charges namely positive andnegative charges. The deficiency of electrons in a body is known aspositively charged body. The excess of electrons in a body is known asnegatively charged body. If a body gets positive charge, its mass slightlydecreases. If a body is given negative charge, its mass slightlyincreases. Charge is relativistically invariant, i.e. it does notchange with motion of charged particle and nochange in it is possible, whatsover may be thecircumstances. i.e.

static dynamicq q Charge is a scalar. S.I. unit of charge is coulomb(C).

One electrostatic unit of charge

(esu) = 9

1

3 10 coulomb.

One electromagnetic unit of charge(emu) = 10coulomb

Charge is a derived physical quantity withdimensions [AT].

Quantization of Charge : The electric chargeis discrete. It has been verified by Millikan’s oil dropexperiment.

Charge is quantised. The charge on any body is anintegral multiple of the minimum charge or electroncharge, i.e if q is the charge then q ne wheren is an integer, and e is the charge of electron =

191.6 10 C .

The minimum charge possible is 191.6 10 C .

If a body possesses 1n protons and 2n electrons,

then net charge on it will be 1 2 ,n n ei.e. 1 1 22 en e n n n e Law of conservation of charge

The total net charge of an isolated physical systemalways remains constant,

i.e. q q q constant. In every chemical or nuclear reaction, the total

charge before and after the reaction remainsconstant.

This law is applicable to all types of processes likenuclear, atomic, molecular and the like.

Charge is conserved. It can neither be created nordestroyed. It can only be transferred from oneobject to the other.

Like charges repel each other and unlike chargesattract each other.

Charge always resides on the outer surface of acharged body. It accumulates more at sharp points.

The total charge on a body is algebric sum of thecharges located at different points on the body.

Electrification: A body can be charged byfriction, conduction and induction.

By Friction: When two bodies are rubbedtogether, equal and opposite charges are producedon both the bodies.

By Conduction: An uncharged body acquiringcharge when kept in contact with a charged bodyis called conduction. Conduction preceedsrepulsion.

By Induction: If a charged body is brought neara neutral body, the charged body will attractopposite charge and repel like charge present inthe neutral body. Opposite charge is induced at thenear end and like charge at the farther end. Inducingbody neither gains nor loses charge. Inductionalways preceeds attraction.

Repulsion is the sure test of electrification.

Induced charge 1 1

1q qK

where K is

Dielectric constant

SYNOPSIS

ELECTRO STATICS & CAPACITORS

SR-MAIN-CHEM-VOL-IIJEE-ADV PHYSICS-VOL- IIELECTRO STATICS & CAPACITORS

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W.E-1: Can two similarly charged bodies attrackeach other?

Sol: Yes, when the charge on one body (q1) is much

greater than that on the other (q2) and they are close

enough to each other so that force of attractionbetween q

1 and induced charge on the other

exceeds force of repulsion between q1 and q

2

Coulomb’s Law: ‘The force of attraction orrepusion between two stationary electric chargesis directly proportional to the product of magnitudeof the two charges and is inversely proportional tothe square of the distance between them and thisforce acts along the line joining those two charges’

0

1

4 r

F 1 22

q q

r

0 - permittivity of free space or vacuum or air..

r - Relative permittivity or dielectric constant ofthe medium in which the charges are situated.

120 8.857 10 2

2

C

Nmor ,

farad

metre

and 9 2 2

0

19 1 0 /

4N m C

Permitivity of Medium: Permitivity is themeasure of degree of the medium which resist theflow of chargesIn SI. for medium other than free space, the constant

10

1

4K so that we can write the equation for

the force between the charges as

1 22

1

4

q qF

r 0

0r

F

F

r is known as the relative permitivity of the

medium. It is a constant for a given medium and itcharges separated by a medium decreasescompared with the force between the same chargesin free space separated by the same distance.

Relative permitivity r is also known as dielectricconstant K of the medium or specific inductivecapacity.Relative permitivity of a medium is defined as theratio of permitivity of the medium to permitivity offree space (or) air(or)Relative permitivity of a medium is defined as the

electrostatic force (0F ) between two charges in air

to the force (F) between the same two charges keptin the medium at same distance.Dielectric constant (or) Relative permitivity

Pemitivity of themediumK

Permitivity of free space

It has no units and no dimesionsHence, the mathematical form of inverse square lawis given as

1 2 1 22 2

0

1 1 1

4 4

q q q qF

r K r For force or vacuum or air K=1 and for a goodconductor like metals, K Conclusion : 1) The introduction of a glass slabbetween two charges will decrease the magnitudeof rorce between them.

2) The introduction of a metallic slab between twocharges will decrase the magnitude of force to zero.

Note:1 When the some charges are separated by thesome distance in two different media,

1 21 2

1 0

1 1

4

q qF

K r --------(1)

and 1 2

2 22 0

1 1

4

q qF

K r ------(2)

from (1) and (2) 1 1 2 2F K F K Note:2 When the same charges are separated by

different distance in the same medium

Fd2 = constant (or) 2 21 1 2 2F d F d

Note : 3 If different charges are at the same separation

in a given medium 1 111 2

1 2

q qF

F q q

Note : 4 If the force between two charges in twodifferent media is the same for different separations.

1 22

0

1 1

4

q qF

K r = constant

Kr2=constant or 2 21 1 2 2K r K r

If the force between two charges separated by a

distance 0' 'r in vacuum or air is same as the forcebetween the same charges separated by a distance‘r’ in a medium.

2 2 00

rKr r r

K

Here K is dielectric constant of the medium.The effective distance ‘r’ in medium for a distance

0r in vaccum = 0r

K.

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Similarly, the effective distance in vaccum for adielectric slab of thickness ‘x ’ and dielectric

constant K is effx f KCoulomb’s Law in Vector Form

1 21212 2

0 2

1

4

q qF r

r

and 21 12F F

q2q121F 12F

Here 12F is force exerted by 1q on 2q and 21F is

force exerted by 2q on 1q Suppose the position vector of two charges 1q and

2q are 1r

and 2r

, then electric force on charge 1q

due to 2q is,

1 21 1 23

01 2

1

4

q qF r r

r r

Similarly, electric force on 2q due to charge 1q is

1 22 2 13

02 1

1

4

q qF r r

r r

Here 1q and 2q are to be substitued with sign.

1 1 1 1r x i y j z k and 2 2 2 2r x i y j z k

where

1 1 1, ,x y z and 2 2 2, ,x y z are the co-

ordinates of charges 1q and 2q .Limitations of Colulomb’s Law Coulomb’s law holds for stationary charges onlywhich are point sized.This law is valid for all types of charge distributions.This law is valid at distances greater than 1510 .mThis law obeys Newton’s third law.This law represents central forces.This law is analogous to Newton law of gravitationin mechanics. The electric force is an action reaction pair, i.e thetwo charges exert equal and opposite forces oneach other. The electric force is conservative in nature. Coulomb force is central. Coulomb force is much stronger than gravitational

force. 3610 g EF FForces between multiple charges : Force on a charged particle due to a number ofpoint charges is the resultant of forces due toindividual point charges

2 31 .....F F F F

W.E-2: Two point sized identical spheres carryingcharges 1q and 2q on them are seperated by acertain distance. The mutual force betweenthem is F. These two are brought in contactand kept at the same separation. Now, the force

between them is 1F . Then 21

1 2

1 24

q qF

F q q

.

Sol: When charges seperated by certain distance theforce is given by

Then 1 2

20

1

4

q qF

r ------------(1)

When charges brought in contanct and kept at thesame distance the force is given by

2

1 212

0

1

4 4

q qF

r ----------(2)

from (1) and (2) ; 21

1 2

1 24

q qF

F q q

W.E-3: Consider three charges q1,q2 and q3 each

equal to q at the vertices of an equilateraltriangle of side ‘l ’ what is the force on anycharge due to remaining charges.

Sol : The forces acting on the charge ‘q’ are

q

qq

2

1 20

2

2 20

1

4

1

4

qF

l

qF

l

clearly 1 2F F F

The resultant force is1 2 2 02 cos60F F F FF

2

20

13 3

4

qF

l W.E-4: A particle of mass ‘m’ carrying a charge

1q is moving around a fixed charge 2qalong a circular path of radius ' 'r find timeperiod of revolution of charge q1

Sol: Electrostatic force on -q1 to +q

2 will provide the


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necessary centripetal force

Hence 2

1 22

Kq q mv

r r ; 1 2Kq q

vmr

2 r

Tv

3 30

1 2

16 mr

q q

W.E-5: Two identical small charged spheres each

having a mass ‘m’ hang in equilibrium asshown in fig. The length of each string is ' 'land the angle made by any string with verticalis .Find the magnitude of the charge on eachsphere.

So l :The forces acting on the sphere are tension in thestring T, force of gravity ‘mg’ and repulsice force

Fe.

x

ll T cos

T sinW W

T T

FF

cosT mg ----(1)2

2sin e

KqT F

r ---(2)

From (1) and (2)

taneF mg ; 2

2tan

Kqmg

r

from fig 2 sinr l ; 2

20

1tan

4 2 sin

qmg

l

2 2016 tan sinq l mg

W.E-6: Two identical balls each having density are suspended from a common point by twoinsulating strings of equal length Both theballs have equal mass and charge. Inequilibrium, each string makes an angle with the vertical. Now both the ball areimmersed in a liquid. As a result, the angle does not change. The density of the liquid is . Find the dielectric constant of the liquid.

Sol:

x

ll T cos

T sinW air W

T T

FF x

ll T cos + Vg

T sinW Liquid W

T T

FF

Let v is the volume of each ball, then mass of eachball is m v ; When balls are in air

cosT mg ; sinT F tan tanF mg vg ---------(1)

When balls are suspended in liquid. The coulumbic

force is reduced to 1 F

FK

and apparent weight

= weight - upthrust ; 1W vg vg According to the problem, angle isuncharged-Therefore 1 1 tan tanF W vg vg ------(2)

From (1) and (2) ; 1

FK

F

Test charge: That small positive charge, whichdoes not influence the other charges and by the helpof which we determine the effect of other charges,is defined as test charge.

Linear charge density is defined as the

charge per unit length.

dq

dl where dq is the charge on an infinitesimal length dl.Units of are Coulomb / meter (C/m)Examples:-Charged straight wire, circular chargedring

Surface charge density is defined as the

charge per unit area.

dq

ds where dq is the charge on an infinitesimal surface

area ds. Units of are 2/coulomb meter (C /2m ) .

Examples:-Plane sheet of charge, conductingsphere.

Volume charge density is defined as charge

per unit volume.

dq

dv where dq is the charge on an infinitesimal volume

element dv. Units of are 3/coulomb meter

(C/ 3m )Examples:- Charge on a dielectric sphere etc.,

Charge given to a conductor always resides on itsouter surface.

If surface is uniform then the charge distributes

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uniformly on the surface. In conductors having nonspherical surfaces, the

surface charge density will be larger when the

radius of curvature is small The working of lightening conductor is based on

leakage of charge through sharp point due to highsurface charge density.

W.E-7: A ring of radius R is with a uniformlydistributed charge Q on it. A charge q is nowplaced at the centre of the ring. Find theincrement in tension in the ring

Sol: Consider an element of the ring. Its enlarged viewis as shown. For equilibrium of this segment, wecan write.

d Q

q d /2

q

F

d /2TTd

2 sin2

dF T

Here F is the repulsive force between q andelemental charge dQ

2

QdQ Rd

R

The electric outward force on element is

20

1

4

qdQF

R From the above three equations, we can write

20

12

4 2 2

q QRd dT

R R

sin for small angle

W.E-8: A thin fixed ring of radius ‘a’ has a positivecharge ‘q’ uniformly distributed over it. Aparticle of mass ‘m’ having a negative charge‘Q’ is placed on the axis at a distance of

x x a from the centre of the ring. Show

that the motion of the negatively charged

particle is approximately simple harmonic.Calculate the time period of oscillation.

Sol: The force on the point charge Q due to the elementdq of the ring is

20

1

4

dqQdF

r along AB

For every element of the ring, there is symmetricallysituated diametrically opposite element, thecomponents of forces along the axis will add upwhile those perpendicular to it will cancel each other.

Hence, net force on the charge Q is -ve signshows that this force will be towards the centre ofring.

cos cosF dF dF 2

0

1

4

x Qdq

r r

so,

32 20 2

0 3

1 1

44

QqxF dq

Qxa x

r

----(1)

(as 12 2 2r a x and 1

2 2 2r a x )

As the restoring force is not linear, the motion willbe oscillatory. However, if x a , then

30

1

4

QqF x kx

a with 3

04

Qqk

a i.e., the restoring force will become linear and sothe motion is simple harmonic with time period

3042

2 2mam

Tk qQ

W.E-9: A point charge q is situated at a distance‘r’ from one end of a thin conducting rod oflength L having a charge Q (uniformlydistributed a long its length). find themagnitude of electric force between the two.

Sol: dE

dxP

x

rL


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Consider a small element of the rod of length dx,at a distance ‘x’ from the point charge q. Tratingthe element as a point charge, the force between

‘q’ and charge element will be 20

1

4

qdQdF

x ; But,

QdQ dx

L

So, 20

1

4

qQdxdF

Lx 2

0

1

4

r L

r

qQ dxF dF

L x

0 0

1 1 1 1 1

4 4r L

r

qQ

L x r r L

0

1

4

qQF

r r L Lines of Force: Line of force is an imaginarypath along which a unit +ve test charge would tendto move in an electric field. Lines of force start from +ve charge and end at–ve charge. Lines of force in the case of isolated +ve chargeare radially outwards and in the case of isolated –ve charge are radially inwards. The tangent at any point to the curve gives thedirection of electric field at that point. Lines of force do not intersect. Lines of force tend to contract longitudinally andexpand laterally.

(A) Radially outward (B) Radially inward

(D)(C)

–+ + +

Electric Field: The space around electriccharge upto which its influence is felt is known aselectric field.

Electric field is a conservative field.

Intensity of Electric Field: The intensity ofelectric field or electric field strength E at a pointin space is defined as the force experienced by

unit positive test charge placed at that point”.The intensity of electric field is also ofted called aselectric field strength.Consider an electric field in a given region. Bring acharge q

0 to a given point in that field without

disturbing any other charge that has produced thefield.

Let F

be the electric force experienced by 0q and

it is found to be proportional to 0q

F

0 0F q F Eq . Here E

is proportionality

constant called electric field strength

0

FE

q

Electric field strength is a vector quantity. Its directionis the direction along which a free positive chargeexperiences the force in the electric field.The S.I unit of elctric field strength is newton percoulomb (NC-1). It can also be expressed in voltper metre (Vm-1).

Electric field internsity due to an isolatedpoint charge : Consider a point charge ‘Q’placed at point A as shown. Let us find the electricfield E

at a point P at a distance ‘r’ from charge

Q. Imagine a positive test charge 0q P. The charge

Q produces a field E

at P..

Q0q

The force applied by Q and 0q is given by

02

0

1

4

QqF

r . This acts along Ap.

According to definition

20 0

1ˆ

4

F QE E r

q r

If ‘q’ is positive, E is along AP and if ‘q’ is negative

E will be along PA.If the charge ‘q’ is in a medium of p is in medium ofpermititivity , and dielectric constnat K,

0

K

the intensity of electic field in a medium

(Emed

) is given by

2

1

4med

QE

r freespacemec

EE

K

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NULL POINT OR NEUTRAL POINTIn the case of a system of charges if the net electricfield is zero at a point, it is knwon as null point.Application : Two point (like) charge q

1 and q

2are separated by a distance ‘r’ and fixed, We canlocate the point on the line joining those chargeswhere resultant or net field is zero.

Case 1: If the charges are like, the neutral point will bebetween the charges.

x (r – x)q1 q2

Let P be the null point where 0netE

1 2 0E E (due to those charges)

or 1 2E E and 1 2E E

1 222

0 0

1 1

4 4

q q

x r x or 1 2

22

q q

x r x

on solving we get 2

1

1

rx

q

q

Case 2 : If the charges are unlike, the neutral point willbe outside the charge on the lime joining them.

rq1 q2 x

In this case 1 222

q q

x r x

On solving we get 2

1

1

rx

q

q

If instead of a single charge, field is producedby no.of charges, by the principle of super positionresultant electric field intensity

1 2 3 ....E E E E

If 0q is positive charge then the force acting on it isin the direction of the field.

If 0q is negative then the direction of this force isopposite of the field direction.

F Eq

E

+ F Eq

E

Motion of a charged particle in a uniformelectric field :a) A charged body of mass ‘m’ and charge ‘q’ isinitially at rest in a uniform electric field of intensityE. The force acting on it, F Eq .

Here the direction of F is in the direction of field if‘q’ is ve and opposite to the field if ‘q’ is ve .

The body travels in a straight line path with uniform

acceleration, F Eq

am m

, initial velocity, 0u .

At an instant of time t.

Its final velocity, Eq

v u at tm

Displacement

2 21 1

2 2

Eqs ut at t

m

Momentum, P mv Eq t Kinetic energy,

2 22 21 1

.2 2

E qK E mv t

m

When a charged particle enters perpendicularly into

a uniform electric field of intensity E with a velocity‘v’ then it describes parabolic path as shown infigure.

+ ++ + + +

u

qy

+

x

Along the horizontal direction, there is noacceleration and hence x ut .Along the vertical direction, acceleration

F Eqa

m m (here gravitational force is not

considered)

Hence vertical displacement, 21

2

Eqy t

m

22

21

2 2

qE x qEy x

m u mu


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At any instant of time t, horizontal component of

velocity, xv u vertical componet of velocity

yEq

v at tm

2 2 2

2 2 22x y

E q tv v v v u

m

Two charges +Q each are separated by a distance'd'. The intensity of electric field at the mid pointof the line joining the charges is zero.

W.E-10 : Two charges +Q each are placed at thetwo vertices of an equilateral triangle of sidea. The intensity of electric field at the thirdvertex is

Sol: E E

aa

+Q +Qa

1 2 2 2 osE E E EEC 2 22 2E E Cos 22 1 cosE

= 22

ECos

; E= 20

13

4

Q

a W.E-11: Two charges +Q, -Q are placed at the

two vertices of an equilateral triangle of side‘a’, then the intensity of electric field at thethird vertex is

Sol :E1 = 2E cos2

= E ( 0120 )

E

120°

E

a

a

–Q+Q a

E1 = 20

1

4

Q

a .

Oblique projection of charged particlein an uniform elctric field (Neglectinggravitational force) : Consider a uniformelectric field E in space along Y-axis. A negativecharged particle of mass ‘m’ and charge ‘q’ beprojected in the XY plane from a point ‘O’ with avelocity u making an angle with the X-axis.

(Neglecting gracitational force).

u

Ox

E j

Initial velocity of the particle is

ˆ ˆcos sinu u i u j

Force acting on the particle is

F qE (along-ve Y axis)

ˆqEa j

m

Velocity of the particle after time ‘t’ is

v u at ; ˆ ˆcos sinv u i u at j

If the point of projection is taken as origin, itsposition vector after time ‘t’ is

ˆ ˆr xi yj where x=(ucos ) t

21sin

2y u t at If the charged particle is projected along the x-axis,then 00

ˆ ˆEqv ui tj

m Here x ut and

21

2

Eqy t

m

Direction of motion of particle after time ‘t’ makes

an angle with x-axis, where tanEqt

mu

A charged particle of charge Q is projected withan initial velocity u in a vertically upward electricfield making an angle to the horizontal. ThenIf gravitational force is considered

Net force mg F mg Eq

Net acceleration = Eq

gm

The negative sign is used when electric field is inupward direction where as positive sign is usedwhen electric field is in downward direction forpositively charged projected particle.

a. Time of flight 2 sinu

EQg

m

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b. Maximum height 2 2sin

2

u

EQg

m

c. Range 2 sin 2u

EQg

m

Intensity of electric field inside a charged hollowconducting sphere is zero. A hollow sphere of radius r is given a charge Q.Intensity of electric field at any point inside it is zero. Intensity of electric field on the surface of the sphere

is 20

1

4

Q

r

Q

r

Intensity of electric field at any point outside the

sphere is (at a distance 'x' from the centre) 20

1

4

Q

x Q

x

Time period of oscillation of a charged body The bob of a simple pendulum is given a +ve charge

and it is made to oscillate in a vertically upwardelectric field, then the time period of oscillation is

2lEQ

gm

T E

mg

In the above case, if the bob is given a -ve charge

then the time period is given by 2lEQ

gm

T E

mg

A sphere is given a charge of 'Q' and is suspendedin a horizontal electric field. The angle made by the

string with the vertical is, 1tanEQ

m g

The tension in the string is 2 2EQ mg

Hence effective acceleration

22

eff

F Eqg g

m m

Time period of oscillation is given by

22

12 2

eff

lT

g Eqg

m

W.E-12: An infinite number of charges each ‘q’

are placed in the x-axis at distances of1,2,4,8...meter from the origin. If the chargesare alternately positive and negative find theintensity of electric field at origin.

Sol: The electric field intensities due to positive chargesand due to -ve charges the field intensity is towardsthe charges

x=0 x=1 x=2 x=4 x=8

q q qq

E4

E2E1

E3

The resultant intensity at the origin

1 3 4E E E E 2 2 2

0

1 1 11 ........

4 2 4 8

QE

Since the expression in the bracket is in GP with a

common ratio = 2

1 1

2 4

0 0

1 4

4 4 511

4

Q QE

0

4

5 4

QE ;

05

QE

W.E-13: A point mass ‘m’ and charge ‘q’ isconnected with a spring of negligible masswith natural length L., Initially spring is innatural length. Now a horizontal uniformelectric field E is switched on as shown. Finda) The maximum separation between the massand the wallb) Find the separation of the point mass andwall at the equilibrium position of massc) Find the energy stored in the spring at theequilibrium position of the point mass.


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K

L

wallm

q

E

Sol: At maximum separation, velocity of point mass iszero. From work energy theorem,

0spring fieldW W 2

0 0

10

2qEx kx (x

0 is maximum elongation)

0

2qEx

K ; separation =

2qEL

k

b) At equilibrium position. EqqE

Eq kx xk

separation =

qEL

k

c) 2 2 2

21 1

2 2 2

qE q EU kx k

k k

W.E-14: A block having mass ’m’ ad carge ‘q’ isresting on a frctionless plane at distance Lfrom the wall as shown inf fig. Discuss themotion of the block when a uniform electricfield E is applied horizontally towards the wallassuming that collision of the block with thewall is perfectly elastic.

Sol: The situation is shown in fig. Electric forece F qE

will accelerate the block towards the wall producingan acceleration

F qEa

m m

21

2L at

i.e., 2 2L mL

ta qE

R

E

qm

mg

L

As collision with the wall is perfectly elastic, theblock will rebound with same speed and as now ismotion is oppisite to the acceleration,, it will cometo rest after travelling same distance L in same timet. After stopping it will beagain accelerated towards

the wall and so the block will execute oscillatorymotion with ‘spain’ L and time period

22

mLT t

qE

However, as the restoring force F(=qE) when theblock is moving away from the wall is constanceand not proportional to displacement x, the motionis not simple harmonic.

W.E-15: Six charges are placed at the vertices of aregular hexagon as shown in thg figure. Theelectric field on the line passing through pointO and perpendicular to the plane of the figure

at a distance of x a from O isa

–Q+Q

+Q –Q

–Q

O

+Q

Sol: This is basically a problem of finding the electricfield due to three dipoles. The dipole moment of

each dipole is 2P Q aElectric field due to each dipole will be 3

KPE

x

The direction of electric field due to each dipole isas shown below:

02 cos60 2netE E E E

30 0

1 22

4

Qa Qa

x x

a

–Q+Q

+Q –Q

–Q

E

60°

60°

E

E

+Q

W.E-16: The field lines for two point charges are

shown in fig.A D B C E

i. Is the field uniform?

ii. Datermine the ratio /A Bq q .

iii. What are teh sing of Aq and Bq ?

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iv. If Aq and Bq are separated by a distance 10 2 1 cm, find the position of neutral

point.Sol: i. No

ii. Number of lines coming from or coming to acharge is proportional to magnitude of charge, so

122

6A

B

q

q

iii. Aq is positive and Bq is negativeiv. C is the other neutral point.

v. For neutral point AE = BE

2 20 0

1 1

4 41A Bq q

xx A B

EBEA

Cl2

2 10A

B

ql xx

x q

cm

ELECTRIC FIELD STRENGTH DUE TO ACHARGED CIRCULAR ARC AT ITS

CENTREW.E-17. Consider a circular arc of radius R which

subtends an angle at its centre. Let uscalculate the electric field strength at C.

Sol:

d

R

X+++

++

++

++

C

Y

dEcosdEsindE

R

Consider a polar segment on arc of angular widthd at an angle from the angular bisector XY asshown. The length of elemental segment is Rd .The charge on this element dq is

Qdq dDue to this dq, electric field at centre of arc C isgiven as

204

dqdE

R

The electric field component dE to this segment dEsin which is perpendicular to the angle bisectorgets cancelled out on integration.The net electric field at centre will be along anglebisector which can be calculated by integratingdEcos within limits from / 2 to / 2Hence net elctric field strength at centre C is

coscE dE /2

20/2

cos4

Qd

R

/2

20 /2

cos4

Qd

R

/2

2 /20

sin4

Q

R

20

sin / 2 sin / 24

Q

R

20

2 sin / 2

4c

QE

R

for a semi circular ring . So at centre

2 2 2 2

0 0 0

2 sin / 2 2 sin / 2 2

4 4 4c

Q Q QE

R R R

ElECTRIC FIELD STRENGTH DUE TO AUNIFORMLY CHARGED ROD

At an axial point :

P

L

+ + + + + + + + + rConsider a rod of length L, uniformly charged witha charge Q. To calculate the electric field strengthat a pont P situated at a distance ‘r’ from one endof the rod, consider an element of length dx on therod as shown in the figure.

dx x

L

P

rdE

Charge on the elemental length dx is Q

dq dxL

2 20 04 4

dq QdxdE

x Lx The net electric field at point P can be given byintegrating this expression over the length of the rod.


12 PINEGROVE

2 20 0

1

4 4

r L r L

p

r r

Q QE dE dx dx

Lx L x

0

1

4

r L

pr

QE

L x

0 0

1 1

4 4p

Q QE

L r r L r r L

At an equatorial point : To find the electricfield due to a rod at a point P situated at a distance‘r’ from its centre on its equatorial line

P

r

L

(a)

+ ++ ++ ++ +

+ + + + + + +

dE dE cos

dE sin P

r

(b)

dx

Consider an element of length dx at a distance ‘x’from centre of rod as in figure (b). Charge on the

element is Q

dq dxL

.

The strength of electric field at P due to this pointcharge dq is dE.

2 204

dqdE

r x The component dEsin will get cancelled and netelectric field at point P will be due to integration of

cosdE only..Net electric field strength at point P can be given as

2

2 2 2 20

2

1cos

4

L

pL

Qdx rE dE

L r x r x

2

3/22 20

2

4

L

pL

Qr dxE

L r x

From the diagram tanx

r

tanx r ; On differentiation; 2secdx r d 2

3 30

sec

4 secp

Qr r dE

L r

;

2

3 30

sec

4 sec

Q r d

Lr r

0 0

cos sin4 4

Q Qd

Lr Lr

Substituting 1 1

2 2tan sin

x x

r x r

2

2 20

2

4

L

PL

Q xE

L x r

; 20 2

1

4

4

Q

r Lr

2 20

2

4 4p

QE

r L r

ELECTRIC FIELD DUE TO A UNIFORMLYCHARGED RING :

The intensity of electric field at a distance ‘x’meters from the centre along the axis:Consider a circular ring of radius ‘a’ having a charge‘q’ uniformly distributed over it as shown in figure.Let ‘O’ be the cetnre of the ring .

xp

(x, 0) dE2

XdE1

A

Bdx

dx

a

O

Consider an element dx of the ring at point A. Thecharge on this element is given by

dq dx charge density 2 2

q qdxdq dx

a a a) The intensity of electric field 1dE at point P due to

the element dx at A is given by

1 20

1

4

dqdE

rThe direction of 1dE is as shown in figure. Thecomponent of intensity along x-axis will be

120

1cos cos

4

dqdE

r

The component of intensity along y-axis will be

120

1sin sin

4

dqdE

r

Similarly if we consider an element dx of the ringopposite to A which lies at B, the component ofintensity perpendicular to the axis will be equal andopposite perpendicular to the axis will be equal andopposite to the component of intensity perpendicularto the axis due to element at A. Hence they canceleach other. Due to symmetry of ring the componentof intensity due to all elements of the ring

13 PINEGROVE


perpendicular to the axis will cancel.So the resultant intensity is only along the axis ofthe ring. The resultant intensity is given by

20

1cos

4

dqE

r

20

1

4 2

qdx xE

ar r (where cos /x r )

32 20 2

1 1

4 2

qxE dx

aa x

3/23 2 2r a x

3/22 20

1 12

4 2

qxE a

a a x

3/22 20

1

4

qxE

a x At its centre x = 0 Electric field at centre is zero.By symmetry we can say that electric field strengthat centre due to every small segment on ring iscancelled by the electric field at centre due to theelement exactly opposite to it. As in the figure theelectric field at centre due to segment A is cancelledby that due to segment B. Thus net electric fieldstrength at the centre of a uniformly charged ring is

0centreE .

ELECTRIC FIELD STRENGTH DUE TO AUNIFORMLY SURFACE CHARGED DISC

Consider a disc of radius R, charged on its surfacewith a charge density .Let us find electric field strength due to this disc ata distance ‘x’ from the centre of disc on its axis atpoint P as shown in figure.Consider an elemental ring of radius ‘y’ and widthdy in the disc as shown in figure. The charge on thiselemental ring dq can be given as 2dq ydy {Area of elemental ring ds= 2dy ydy }

x

P

dE

dy

Electirc field strength due to a ring of radius Y,charge Q at a distance x from its centre on its axiscan be given as

3/22 204

QxE

x y Due to the lemental ring electric field strength dE atpoint P can be given as

3/2 3/22 2 2 20 0

2

4 4

xdq y dyxdE

x y x y

Net electric field at point P due to whole disc isgiven by integrating above expression within thelimits from 0 to R

3/22 20 0

2

4

R xydyE dE

x y

3/2 2 22 20 00 0

2 2 1

4 4

RRx ydy x

x yx y

2 2

0

12

xE

x R

Electric field strength due to a uniformly chargeddisc at a distance x from its surface is given as

2 20

12

xE

x R

If we put x = 0 we get

02E

Electric dipole: A system of two equal andopposite point charges fixed at a small distanceconstitutes an electric dipole. Electric dipole isanalogous to bar magnet or magnetic dipole inmagnetism. Every dipole has a characteristicproperty called dipole moment, which is similar tomagnetic moment of a bar magnet. If 2a is thedistance between the charges +q and –q, thenelectric dipole moment is p = q.2a.

2a

–q +qPDipole moment is a vector quantity and its directionis from negative charge to positive charge as shown.ELECTRIC FIELD AT ANY POINT DUE TOA DIPOLE : We know that the electric field is the-ve gradiant of potential. In polar form if V is thepotential at r,θ the electric field will have twocomponents radial and transverse componentswhich are represented by

r θE & E respectively..


14 PINEGROVE

–q +qO

Er

E

E

P

Then

r 2

0

V pcos 1E

r 4 r r

r 30

2 p c o sE

4 r

r

VE

r1 V

Er

The tranverse component of electric field

1 VE

r

20

1 psin

r 4 r

30

p s inE

4 r

2 2RE E E

2 2 2 2

2 23 30 0

p sin 4p cosE

4 r 4 r

2 23

0

pE 4cos sin

4 r

2

30

pE 1 3cos

4 rField at a point on the axial line : ( 00 )

axial 3

0

2pE

4 rField at a point on the equitorial line ( 090 )

equ ito ria l 3

0

pE

4 rThe direction of E at any point is given by

3

0

r3

0

psin4 rE

tan2pcosE4 r

1

tan tan2

1tan 1/ 2 tan Note : Electric dipole placed in an uniform electric field

experiences torque is given by pEsin in vector form p E

+q qE

2a sin

–qqE

E

The torque on the dipole tends to align the dipolealong the direction of electric field.The net force experienced by it is zero.

Note : The potential energy of dipole in an electric fieldis

V2V1

+q

–q

2a sin E

U= – pE cos θ .In vector form U p.E

if o0 ; 0 and U pEif o90 ; pE and U 0if o180 ; 0 and U pESo, if p

is parallel to E

then, potential energy is

minimum and torque on the dipole is zero, and thedipole will in stable equilibrium.If p

is anti parallel to E

then, potential energy ismaximum and again torque is zero, but it is inunstable equilibrium

Note : Work done in rotating a dipole in electric

field from an initial angle 1θ with field to final

angle 2θ with field is W 1 2pE(cos cos )

Note : Force on dipole in non-uniform electric field:The force on the dipole due to electric field is givenby F U (Force = negative potential energygradient).If the electric field is along r , we can write

d

F (p.E)dr

If p and E

are along the same direction we can

write

d

F (pEcos )dr

or dE

F pdr

.

15 PINEGROVE


OSCILLATORY MOTION OF DIPOLE INAN ELECTRIC FIELD

When dipole is displaced from its position ofequilibrium. The dipole will then experience a torquegiven by sinpE For small value of , pE ----------(1)Where negative sign shows that torque is actingagainst increasing value of Also, ,I Where, I = moment of inertia and angular acceleration.

2

2

dd

dtdt

2

2

dI

dt

---------(ii)

Hence, from eqs (i) and (ii), we have2

2

dI pE

dt

or 2

2

d pE

dt I

----(iii); 2

2

d

dt

This equation represents simple harmonic motion(SHM). when dipole is displaced from its meanposition by small angle, then it will have SHM.

Eq (iii) can be written as 2

20

d pE

dt I

On comparing above equation with standardequation of SHM.

2

22

0d

ydt

, we have ; 2 pE pE

I I

2I

TpE

, where T is the time period of

oscillations.W.E-18: An electric dipole of dipole moment p is

kept at a distance r from an infinite longcharged wire of linear charge density asshown. Find the force acting on the dipole ?

+++++

r

P

Sol :Field intensity at a distance r from the line of charge

is 0

E2 r

The force on the dipole is dE

F pdr

2

0

p2 r

20

p

2 r

Here the net force on dipole due to the wire will beattractive.DISTRIBUTED DIPOLE: Consider a half ringwith a charge +q uniformly distributed and anotherequal negative charge –q placed at its centre. Here–q is point charge while +q is distributed on thering. Such a system is called distributed dipole.

=

+q

R–qR

dq

d–d

The net dipole moment is pnet 2qR

If

/ 2

net

0

p 2 dp cos ; 2qR

sin / 2

If the arrangement is a complete circle,

2

netp 0.

FORCE BETWEEN TWO SHORTDIPOLES Consider two short dipoles seperatedby a distance r. There are two possibilities.

a) If the dipoles are parallel to each other.

2E 1E

2P1Pr

F 1 24

0

1 3p p

4 rAs the force is positive, it is repulsive. Similarly if

1 2p || p

the force is attractive.b) If the dipoles are on the same axis

r

2E 1E2P1P

F

1 24

0

1 6p p

4 rAs the force is negative, it is attractive.Quadrapole: We have discussed about elecricdipole with two equal and unlike point chargesseparated by a small distance. But in some cases


16 PINEGROVE

the two charges are not concentrated at its ends.(Like in water molecule) consider a situation asshown in the figure. Here three charges –2q, q andq are arranged as shown. It can be visualised asthe combination of two dipoles each of dipolemoment p = qd at an angle between them. Thearrangment of two electric dipoles are calledquadrapole. As dipole moment is a vector theresultant dipole moment of the system is

|p 2p cos / 2.

q

q

–2q q

q

–q

–q=

Few other quadrapoles are also as shown in thefollowing figures.

+q

–2q +q

+q

+q

–2q

+q

–q+q

–q–q–q 2q

Electric filed at the axis of a circularuniformly charged ring

dE cos

dE sin

dE sin

dq

Qa

dE cosdE

dE2 2a xy

x

Intensity of electric field at a point P that lies on theaxis of the ring at a distance x from its centre is

32 2 20

1

4

qxE

x R

where 2 2cos

x

a x

Where R is the radius of the ring. From the aboveexpression E = 0 at the centre of the ring.

E will be maximum when 0dE

dx .

Differentating E w.r.t x and putting it equal to zero

we get 2

Rx and max 2

0

2 1

43 3

qE

R

Electric field due to a Charged SphericalConductor (Spherical Shell )

‘q’ amount of charge be uniformly distributed overa spherical shell of radius ‘R’

Surface charge density, 24

q

R

When point ‘P’ lies outside the shell :

20

1

4

qE

r This is the same expression as obtained for

electric field at a point due to a point charge. Hencea charged spherical shell behave as a point chargeconcentrated at the centre of it.

2

2 20

1 .4

4 4

R qE

r r

; 2

20

.RE

r

When point ‘P’ lies on the shell :

0

E

When Point ‘P’ lies inside the shell: E = 0

2

1E

r

Distance from the centre

E r = R1

Note : The field inside the cavity is always zero this isknown as elctro static shielding

Electric filed due to a Uniformlycharged non – conducting sphere

Electric field intensity due to a uniformlycharged non-conducting sphere of charge Q,of radius R at a distance r from the centre ofthe sphereq is the amount of charge be uniformly distributedover a solid sphere of radius R.

= Volume charge density 343

q

R

When point ‘P’ lies inside sphere :

30

1

4

QrE

R for r R

0

.

3

rE

When point ‘P’ lies on the sphere:

20

1

4

qE

R ;

0

.

3

RE

17 PINEGROVE


When point ‘P’ lies outside the sphere:

20

1

4

qE

r ; 3

20

.

3

RE

r

E r

E

Rd

2

1E

r

Electric Field due to a charged Disc:Electric field due to a uniformly charged disc withsurface charge density of radius at a distance xfrom the centre of the disc is

2 20

12

xE

x R

If Q is the total charge on the disc, then

2 2 20

21

4

Q xE

R x R

Electric Potential: Work done to bring a unitpositive charge from infinite distance to a point inthe electric field is called electric potential at thatpoint .

it is given by W

Vq

It represents the electrical condition or state of the

body and it is similar to temperature. +vely charged body is considered to be at higherpotential and -vely charged body is considered tobe at lower potential. Electric potential at a point is a relative value butnot an absolute value.

Potential at a point due to a point charge 0

1

4

Q

r Potential due to a group of charges is the algebraic

sum of their individual potentials.

i.e. 1 2 3 . . . . . .V V V V Two charges +Q and -Q are separated by a distance

d, the potential on the perpendicular bisector of theline joining the charges is zero. When a charged particle is accelerated from restthrough a p.d. ' V ' , work done,

21 2

2

VqW Vq mv or v

m

The work done in moving a charge of q coulombbetween two points separated by p.d. 2 1V V is

2 1q V V . The work done in moving a charge from one point

to another point on an equipotential surface is zero. A hollow sphere of radius R is given a charge Q thepotential at a distance x from the centre is

0

1.

4

Qx R

R x

R

The potential at a distance when x>R is 0

1.

4

Q

x .

R

x

A sphere is charged to a potential. The potential atany point inside the sphere is same as that of thesurface. Inside a hollow conducting spherical shell,E=0, 0V . Relation among E, V and d in a uniform electric

field is V

Ed

(or) dV

Edx

Electric field is always in the direction of decreasing

potential .The component of electric field in any direction isequal to the negative of potential gradient in thatdirection.

V V VE i j k

x y z

An equipotential surface has a constant value ofpotential at all points on the surface .For single charge q

v2

v2

v2

v1

v1

v1

q

E

E

E

E

E

E

Electric field at every point is normal to theequipotential surface passing through that point

No work is required to move a test charge onunequipotential surface.

Zero Potential Point Two unlike charges Q1

and -Q2 are seperated by a distance ‘d’. The net

potential is zero at two points on the line joiningthem, one (x) in between them and the other (y)


18 PINEGROVE

outside them

1 2Q Q

x d x and

1 2Q Q

y d y

Potential due to a dipole: An electric dipoleconsists of two equal and opposite charges seperatedby a very small distance. If 'q' is the charge and 2athe length of the dipole then electric dipole momentwill be given by p = (2a)q.

a aB

+q

P

rN

–qO

M

A

Let AB be a dipole whose centre is at 'O' and 'P'be the point where the potential due to dipole is tobe determined. Let ,r be the position co-ordinates of 'P' w.r.t the dipole as shown in figure.Let BN & AM be the perpendiculars drawn on toOP and the line produced along PO. Fromgeometry ON acos OM . Hence thedistance ,BP from +q charge is r acos [because PB = PN as AB is very small incomparsion with r].For similar reason

AP r a cos AP PM .Hence potential at P due to charge +q situated at B

is 10

1

4 cos

qV

r a .

Similarly potential at P due to charge -q at A is

20

1

4 cos

qV

r a

.

Hence the total potential at P isV= V 1+V2

0 0

q qV

4 r a cos 4 r a cos

0

q 1 1V

4 r a cos r a cos

2 2 2

0

q 2a cosV

4 r a cos

But r >> a 2 2 2 2r a cos r

2

0

cos

4

pV

r.

Hence potential varies inversely as the square ofthe distance from the dipole.SPECIAL CASSES

1) On the axial line : For a point on the axial line

00 2axial 0V p / 4 r volts for a dipole.

2) Point on the equitorial line : For a point on theequitorial line 090 . equitorialV 0 Volts .

Equitorial line is a line where the potential is zero atany point.

Equipotential surfaces : Equipotential surfacein an electric field is a surface on which the potentialis same at every point. In other words, the locus ofall points which have the same electric potential iscalled equipotential surface.An equipotential surface may be the surface of amaterial body or a surface drawn in an electric field.The important properties of equipotentialsurfaces are as given below.

a) As the potential difference between any two pointson the equipotential surface is zero, no work isdone in taking a charge from one point to another.

b) The electric field is always perpendicular to anequipotential surface. In other words electric fieldor lines of force are perpendicular to theequipotential surface.

c) No two equipotential surfaces intersect. If theyintersect like that, at the point of intersection fieldwill have two different directions or at the samepoint there will be two different potentials which isimpossible.

d) The spacing between equipotential surfaces enablesto identify regions of strong and weak fields

dVE

dr. So 1

Edr

(if dV is constant).

e) At any point on the equipotential surface componentof electric field parallel to the surface is zero.In uniform field , the lines of force are straight andparallel and equipotential surfaces are planesperpendicular to the lines of force as shown in figure

19 PINEGROVE


equipotentialsurface

The equipotential surfaces are a family of concentricspheres for a uniformly charged sphere or for a pointcharge as shown in figure

equipotentialsurface

Equipotential surfaces in electrostatics are similarto wave fronts in optics. The wave fronts in opticsare the locus of all points which are in the samephase. Light rays are normal to the wave fronts.On the other hand the equipotential surfaces areperpendicular to the lines of force.

Note : 1) In case of non-uniform electric field, the fieldlines are not straight, and in that case equipotentialsurfaces are curved but still perpendicular to thefield.

2) Electric potential and potential energy are alwaysdefined relative to a reference. In general we takezero reference at infinity. The potential at a point Pin an electric field is V if potential at infinity is takenas zero. If potential at infinity is V0, the potential atP is (V–V0).

3) The potential difference is a property of two pointsand not of the charge q0 being moved.

ELECTRIC POTENTIAL DUE TO A LINEARCHARGE DISTRIBUTION

Consider a thin infinitely long line charge having auniform linear charge density placed along 1YY .Let P is a point at distance ‘r’ from the line chargethen manitude of electri field at point P is given by

02E

r

Y

Y

++++++++++++

Or dS

PGaussiansurface

E

l

We know that .V r E dr Here

02E

r

and .E dr Edr

So 02

V r Edr drr

0

log2 eV r r C

Where C is constant of integration and V(r) giveselectric potential at a distance ‘r’ from the linearcharge distribution

ELECTRIC POTENTIAL DUE TO INFINITEPLANE SHEET OF CHARGE

(NON CONDUCTING)Consider an infinite thin plane sheet of positivivecharge having a uniform surface charge density onboth sides of the sheet. by symmetry , it follws thatthe electric filed is perpendicular to the plane sheetof charge and directed in out ward direction.

The electric field intensity is 02

E

Electrostatic potential due to an infinite plane sheetof charge at a perpendicular distance r from the

sheet given by .V r E dr Edr

0 02 2V r dr r C

where C is constant ofintegration similarly the elec-tric pontential due to an infinite plane conductingplate at a perpendicular distance r from the plate is

given by .V r E dr Edr

0 0

V r dr r c

where C is constant of intergrationELECRTIC POTENTIAL DUE TO A CHARED

SPERICAL SHELL (OR CONDUCTINGSPHERE):

Shell

GaussianSurface

Charged spherical

Pq

R r

O

+++

++

++ +

++

++

E

dS


20 PINEGROVE

Consider a thin spherical shell of radius R and hav-ing charge+q on the spherical shell.

Case (i): When point P lies outside the spherical shell.The electric field at the point is

20

1

4

qE

r (for r > R)

The potential .V r E dr Edr 2

0 0

1 1

4 4

q qdr C

r r Where C is constant of integration

If r , 0V and 0C

0

1

4

qV r r R

r Case (ii) : When point P lies on the surface of spheri-

cal shell then r = Relectrostatice potential at P on the surface is

0

1

4

qV

R Case (iii) : For points inside the charged spherical

shell (r < R), the electric field E = 0

So we can write 0dV

dr

V is constant and is equal to that on the surface

So, 0

1

4

qV

R for r RThe varitaion of V with distance ‘r’ from centre isas shown in the graph.

V

r r = R

v 1/r0

1

4S

qV

R

ELECTRIC POTENTIAL DUE TO AUNIFORMLY CHARGED

NON-CONDUCTING SOLID SPHERE:Consider a charged sphere of radius R with totalcharge q uniformly distributed on it.

Case (i) : For points Outside the sphere (r > R)The electric field at any point is

20

1,

4

qE

r (for r > R)

The potential at any point outside the shell is

.V r E dr Edr

20 0

1 1

4 4

q qdr C

r r Where C is constant of integration

If , 0r V and C=0

0

1

4

qV r

r (r > R)

Case (ii) : When point P lies on the surface of spheri-cal shell then r = RThe electrostatic potential at P on the surface is

0

1

4

qV

R Case (iii) : FOr points inside the sphere (r < R)

The electric field is 30

1

4

qrE

R +

+

+

+

++

+

+

++

++

+

r

R

++

++

++

+

+

E

.dV E dr Edr 3

0

1

4s

v r r

v R R

qrdV E dr dr

R 2

30

1

4 2

r

s

R

q rV V

R

2 2

30 0

1 1

4 4 2 2

q q r RV

R R

2

20

1 3

4 2 2

q rV

R R

At the centre r = 0 then

Potential at centre 0 0

1 3 3 1

4 2 2 4C

q qV

R R The variation of V with distance ‘r’ from centre isas shown in the graph.

vvs

vs

r < R r < R

v 1/rr = R r

vs

21 PINEGROVE


Potential of a charged ring: A charge q isdistributed over the circumference of ring ( eitheruniformly or non-uniformly ) , then electric potential

at the centre of the ring is 1

.4 o

qV

R .

At distance ‘r’ from the centre of ring on its axis

would be 2 2

1.

4 o

qV

R r Electric potential of a uniformly charged disc

Consider a uniformly charged circular disc havingsurface charge density .

Potential a at point on its axial line at distance x from

the centre is 2 2

2 o

V R x x

At the centre of disc 0x 2 o

RV

For x R , 4 o

qV

x

Potential on the edge of the disc is o

RV

W.E-19: A charge Q is distributed over two con-centric hollow spheres of radii ‘r’ and R ( > r)such that the surface densities are equal. Findthe potential at the common centre.

Sol: If 1q and 2q are the charges on spheres of radii ‘r’’and R respectively, then in accordance with con-servation of charge

1 2q q Q ------(1)

And according to given problem 1 2 ,

i.e., 1 22 24 4

q q

r R or 2

12

2

q r

q R ------(2)

So from Eqs (1) and (2)

2

1 2 2

Qrq

r R and

2

2 2

QRq

r R -----(3)

Now as potential inside a conducting sphere is equalto that at its surface, so potential at the commoncentre,

1 21 2

0

1

4

q qV V V

r R

Substituting the value of 1q and 2q from Eq.(3)

2 2 2 20

1

4

Qr QrV

R r R r 2 2

0

1

4

Q R r

R r

W.E-20: If electric potential V at any point (x, y,z) all in metres in space is given by V = 4x2

volt. Calculate the electric field at the point(1m, 0m, 2m).

Sol : As electric field E is related to potential V throughthe relation

dVE

dr

2x

dV dE (4x ) 8x

dx dx

2y

dV dE (4x ) 0

dy dy

And, 2z

dV dE (4x ) 0

dz dz

So, x y zˆ ˆ ˆˆE i E j E k E 8xi

i.e., it has magnitude 8 V/m and is directed alongnegative x-axis.

W.E-21: A conducting spherical bubble of radiusr and thickness t (t >> r) is charged to apotential V. Now it collapses to form aspherical droplet. Find the potential of thedroplet.

Sol: Here charge and mass are conserved. If R is theradius of the resulting drop formed and is density

of soap solution, 3 24R 4 r t

3 2 1/3R (3r t)

Now potential of the bubble is 0

1 qV

4 r

or 0q 4 rV Now potential of resulting drop is

|

0

1 qV

4 R

1/3r

V3t

.

Potential Energy of System of Charges Two charges 1Q and 2Q are separated by a distance

'd'. The P.E. of the system of charges is 1 2

0

1.

4

QQU

d from U=W=Vq

dQ1 Q2


22 PINEGROVE

Three charges 1 2 3, ,Q Q Q are placed at the threevertices of an equilateral triangle of side 'a'. TheP.E. of the system of charges is

2 3 3 11 2

0

1

4

Q Q QQQQU

a a a or

1 2

0

1

4

Q QU

a

Q3

Q2Q1

aa

a

A charged particle of charge 2Q is held at rest at adistance 'd' from a stationary charge 1Q . When thecharge is released, the K.E. of the charge 2Q at

infinity is 1 2

0

1.

4

Q Q

d .

If two like charges are brought closer, P.E of thesystem increases. If two unlike chargtes are brought closer, P.E of thesystem decreses.For an attractive system U is always NEGATIVE.For a repulsive system U is always POSITIVE.

For a stable system U is MINIMUM.

i.e.dU

Fdx

= 0 (for stable system)

POTENTIAL ENERGY OF A SYSTEM OFTWO CHARGES IN AN EXTERNALFIELD: Consider two charges 1q and 2q located

at two points A and B having position vectgors 1r

and 2r respectively. Let 1V ang 2V be the potentialsdue to external sources at the two pointsrespectively.

The work done in bringing the charge 1q from infinity

to the point A is 1 1 1W qVIn bringing charge 2q , the work to be done notonly against the external field but also against the

filed due to 1q .

The work done in bringing the charge 2q from infinity

to the point B is 2 2 2W q V .

The workdone on 2q against the field due to 1q is

1 22

0 12

1

4

q qW

r where 12r is the distance between

1q and 2q .

The total work done in bringing the charge 2q

against the two fields from infinity to the point B is

1 22 2 2

0 12

1

4

q qW q V

r The total work done in assembling the configurationor the potential energy of the system is

1 21 1 2 2

0 12

1

4

q qW qV q V

r W.E-22: Charge q1 is fixed and another point

charge q2 is placed at a distance r0 from q1 ona frictionless horizontal surface. Find thevelocity of q2 as a function of seperation rbetween them (treat the changes as pointcharges and mass of q2 is m)

Sol : q1 q2

r0

According law of concervation of energy

1 1 2 2U K U K 21 2 1 2

0 0 0

q q q q1 1 10 mv

4 r 4 r 2

2 1 2

0 0

q q1 1 1mv

2 4 r r

; 1 2

0 0

q q 1 1v

2 m r r

W.E-23: A proton moves with a speed of 7.45 x 105

m/s directly towards a free proton originallyat rest. Find the distance of closest approachfor the two protons.

Given 90(1 / 4 ) 9 10 m / F; 27

Pm 1.67 10 kg

and e = 1.6 x 10-19 coulomb.Sol: As here the particle at rest is free to move, when

one particle approaches the other, due toelectrostatic repulsion other will also start movingand so the velocity of first particle will decreasewhile of other will increase and at closest approachboth will move with same velocity. So if v is thecommon velocity of each particle at closestapproach, by 'conservation of momentum'.

mu m m i.e., 1

u2

And by 'conservation of energy'

22 2 2

0

1 1 1 1 emu m m

2 2 2 4 r

So, 2

20

4er

4 mu

uas

2

And hence substituting the given data,19 2

9 1227 5 2

4 (1.6 10 )r 9 10 10 m

1.67 10 (7.45 10 )

23 PINEGROVE


W.E-24: A small ball of mass 2 x 10–3 kg having acharge of 1 C is suspended by a string oflength 0.8m. Another identical ball havingthe same charge is kept at the point ofsuspension. Determine the minimumhorizontal velocity which should be impactedto the lower ball so that it can make completerevolution :

Sol: To complete the circle at top most point T2 = 0F

V

q

FU

2 2

2o

q MVMg

4

22

o

qV g

4 M

...( 1 )

from law of conservation of energy

2 21 1mu mv mg 2

2 2 .... ( 2 )

from ( 1 ) and ( 2 );

2

o

qu 4g 5 86 m /s

4 m

W.E-25: If an electron enters into a space betweenthe plates of a parallel plate capacitor at anan angle with the plates and leaves at anangle to the plates, find the ratio of its kineticenergy while entering the capacitor to that whileleaving.

Sol: Let u be the velocity of electron while entering thefield and v be the velocity when it leaves the plates.Component of velocity parallel to the plates willremain unchanged.

Hence u cos u cos u cos

v cos

2

2 2

2

1mu

u cos21 v cos

mv2

W.E-26: Figure shows two concentric

conductiong shells of radii 1r and 2r carrying

uniformly distributed charages 1q and 2q .respectively. Find out an expression for thepotential of each shell.

r2

r1

+q1

+q2

Sol: The potential of each sphere consists of two points:One due to its own charge, andSecond due to the charge on the other sphere.Using the principle of superposition, we have

1 21 , ,r surface r insideV V V and

1 22 , ,r outside r surfaceV V V Hence,

1 21

0 1 0 2

1 1

4 4

q qV

r r

and 1 2

20 2 0 2

1 1

4 4

q qV

r r W.E-27: In the previous example, if the charge

1 0q q and the outer shell is earthed, thena) determine the charge on the outer shell, andb) find the potential of the inner shell.

Sol: a) We know that charge on facing surfaces is equal

and opposite. So, if charge on inner sphere is 0,q

then charge on inner surface of shell should be 0q .

Now, let charge on outer surface of shell be 2q .As the shell is earthed. So its potential should bezero. So,

00 22

2 2 2

0 0shell

k qkq kqV q

r r r

Hence, charge on outer surface of shell is zero. Final


24 PINEGROVE

charges appearing are shown in figb) Potential of inner sphere:

00 01

1 2 0 1 2

1 1

4

k qkq qV

r r r r

W.E-28: Consider two concentric spherical metalshells of radii ‘a’ and b > a. The outer shellhas charge Q, but the inner shell has nocharge, Now, the inner shell is grounded. Thismeans that the inner shell will come at zeropotential and that electric field lines leave theouter shell and end on the inner shell.a) Find the charge on the inner shell.b) Find the potential on outer sphere.

Sol: a) When an object is connected to earth

(grounded), its potential is reduced to zero. Let 'qbe the charge on A after it is earthed as shown in fig

–qQ + q

The charge 'q on A induces 'q on

inner surface of B and 'q on outer surface of B.In equilibrium, the charge distribution is as shownin figPotential of inner sphere = potential due to chargeon A+ potential due to charge on B = 0

' ' '

0 0 0

04 4 4A

q q Q qV

a b b

or ' aq Q

b

This implies that a charge /Q a b has been

transferred to the earth leaving negative charge onA.Final charge distribution will be as shown in fig..

++

+

+

+

+++

+

+

+

++

++

+

+

++ + +

+

+

Q b a

bQa

b

Qa

b

As b>a, so charge on the outer surface of outer

shell will be

0Q b a

b

.

b) Potential of outer surface BV potential due to

charge on A + potential due to charge on B.'

, ,0 0

1 1

4 4B a out b both surface

q QV V V

b b

2

0 0 0

1 1

4 4 4

aQ

Q b aQbb b b

W.E-29: Two circular loops of radii 0.05 and0.09m, respectively, are put such that their axescoincide and their centres are 0.12 m apart.Charge of 610 coulomb is spread uniformlyon each loop. Find the potential differencebetween the centres of loops.

q1

q2

R2R1

O2O1

r1r2

x

Sol: The potential at the centre of a ring will be due tocharge on both the rings and as every element of aring is at a constant distance from the centre, so

1 21 2 2

0 1 2

1

4

q qV

R R x

4 49

2 2

10 109 10

5 9 12

5 51 1

9 10 2.40 105 15

V similarly,

2 12 2 2

0 2 1

1

4

q qV

R R x

25 PINEGROVE


or 3 5

2

1 1 1989 10 10

9 13 117V

5

2 1.69 10V V So, 5

1 2 2.40 1.69 10 71V V kV W.E-30: A circular ring of radius R with uniform

positive charge density per unit length islocated in the y - z plane with its centre at theorigin O. A particle of mass ‘m’ and positivecharge ‘q’ is projected from the point

3 ,0,0p R on the negative x-axis directly

towards O, with initial speed v. Find thesmallest (non-zero) value of the speed such thatthe particle does not return to P?

qP

R

O

Q

3 ,0,0R

r

Sol: As the electric field at the centre of a ring is zero,

the particle will not come back due to repulsion if itcrosses the centre fig.

2

0 0

1 1 1

2 4 4

qQ qQmv

r R

But here, 2Q R and 223 2r R R R

So,2

0

1 1 2 11

2 4 2

R qmv

R

or 02

qv

m

So, min

02

qV

m

C.U.Q

CHARGE & CONSERVATION OF CHARGE1. Two identical metallic spheres A and B of

exactly equal masses are given equal positiveand negative charges respectively. Then1) mass of A > Mass of B2) mass of A < Mass of B3) mass of A = Mass of B4) mass of A Mass of B

2. Two spheres of equal mass A and B are given+q and -q charge respectively then1) mass of A increases 2) mass of B increases

3) mass of A remains constant4) mass of B decreases

3 A soap bubble is given a negative charge, thenits radius.1) Decreases 2) Increases3) Remanins unchanged4) Nothing can be predicted as information isinsufficient

COULOMB’S LAW4. Two charges are placed at a distance apart. If

a glass slab is placed between them, forcebetween them will1) be zero 2) increase3) decrease 4) remains the same

5. A negatively charged particle is situated on astraight line joining two other stationaryparticles each having charge +q. The directionof motion of the negatively charged particlewill depend on1) the magnitude of charge2) the position at which it is situated3) both magnitude of charge and its position4) the magnitude of +q

6. Four charges are arranged at the corners of asquare ABCD as shown in the figure. The forceon the positive charge kept at the centre ‘O’is

A B

CD

+Q +2Q

2Q

O

+Q

1) zero

2) along the diagonal AC

3) along the diagonal BD

4) perpendicular to side AB7. Two identical +ve charges are at the ends of a

straight line AB. Another identical +ve chargeis placed at ‘C’ such that AB=BC. A, B and Cbeing on the same line. Now the force on ‘A’1) increases 2) decreases3) remains same 4) we cannot say

8. Two identical pendulums A and B aresuspended from the same point. Both are givenpositive charge, with A having more chargethan B. They diverge and reach equilibriumwith the suspension of A and B makingangles 1 and 2 with the vertical respectively..

1) 1 2 2) 1 2 3) 1 2 4) The tension in A is greater than that in B

9. Two metal spheres of same mass aresuspended from a common point by a lightinsulating string. The length of each string issame. The spheres are given electric charges


26 PINEGROVE

+q on one end and +4q on the other. Which ofthe following diagram best shows the resultingpositions of spheres?

1)+q +4q

+q +4q2)

3)4)

+q+4q

+4q

+q

10. Two point charges q and 2q are placed ata certain distance apart. Where should a thirdpoint charge be placed so that it is inequilibrium?1) on the line joining the two charges on the right of 2q2) on the line joining the two charges on theleft of q3) between q and 2q4) at any point on the right bisector of the l i n ejoining q and 2q .

ELECTRIC FIELD11. Figure shows the electric lines of force

emerging from a charged body. If the electricfield at ‘A’ and ‘B’ are AE and BE respectivelyand if the displacement between ‘A’ and ‘B’ is‘r’ then

AB

1) A BE E 2) A BE E3) B

A

EE

r 4) 2

BA

EE

r

12. Figure shows lines of force for a system of twopoint charges. The possible choice for thecharges is

q1 q2

1) 1 24 , 1.0q C q C 2) 1 21 , 4q C q C 3) 1 22 , 4q C q C 4) 1 23 , 2q C q C

13. Drawings I and II show two samples of electricfield lines

I II

1) The electric fields in both I and II are produced.by negative charge located somewhere on the leftand positive charges located somewhere on the right2) In both I and II the electric field is the sameevery where3) In both cases the field becomes stronger onmoving from left to right4) The electric field in I is the same everywhere,but in II the electric field becomes stronger onmoving from left to right

14. An electron is projected with certain velocityinto an electric field in a direction opposite tothe field. Then it is1) accelerated 2) retarded3) neither accelerated nor retarded4) either accelerated or retarded

15. The acceleration of a charged particle in auniform electric field is1) proportional to its charge only2) inversely proportional to its mass only3) proportional to its specific charge4) inversely proportional to specific charge

16. An electron and proton are placed in anelectric field. The forces acting on them are

1F and 2F and their accelerations are 1a and

2a respectively then

1) 1 2F F 2)

1 2 0F F 3) 1 2a a 4) 1 2a a

17. The bob of a pendulum is positively charged.Another identical charge is placed at the pointof suspension of the pendulum. The timeperiod of pendulum1) increases 2) decreases3) becomes zero 4) remains same.

18. Intensity of electric field inside a uniformlycharged hollow sphere is1) zero 2) non zero constant3) change with r4) inversely proportional to r

19. A positive charge q0 placed at a point P near acharged body experiences a force of repulsion

27 PINEGROVE


of magnitude F, the electric field E of thecharged body at P is

1) 0

F

q 2) 0

F

q 3)

0

F

q 4) F

20. A cube of side b has charge q at each of itsvertices. The electric field at the centre of thecube will be (KARNATAKA CET 2000)

1) zero 2) 2

32q

b3) 22

q

b4) 2

q

b

21. An electron and proton are sent into an electricfield. The ratio of force experienced by themis1) 1 : 1 2) 1 : 18403) 1840 : 1 4) 1 : 9.11

22. An electron enters an electric field with itsvelocity in the direction of the electric lines offorce. Then1) the path of the electron will be a circle2) the path of the electron will be a parabola3) the velocity of the electron will decrease4) the velocity of the electron will increase

23. A charged bead is capable of sliding freelythrough a string held vertically in tension. Anelectric field is applied parallel to the string sothat the bead stays at rest of the middle of thestring. If the electric field is switched offmomentarily and switched on1) the bead moves downwards and stops as soonas the field is switched on2) the bead moved downwards when the field isswitched off and moves upwards when the field isswitched on3) the bead moves downwards with constantacceleration till it reaches the bottom of the string4) the bead moves downwards with constantvelocity till it reaches the bottom of the string

24. An electron is moving with constant velocityalong x-axis. If a uniform electric field isapplied along y-axis, then its path in the x-yplane will be1) a straight line 2) a circle3) a parabola 4) an ellipse

25. An electron of mass eM , initially at rest ,moves through a certain distance in a uniform

electric field in time 1t . proton of mass pM

also initially at rest, takes time 2t to movethrough an equal distance in this uniform

electric field. Neglecting the effect of gravity

the ratio 2 1/t t is nearly equal to

1) 1 2) /p eM M 3) /e pM M 4) 1836

26. Dimensions of 0 are

1) 1 3 4 2M L T A 2) 0 3 3 3M L T A 3) 1 3 3M L T A 4) 1 3 2M L TA

27. Two point charges q and -2q are placed somedistance d apart. If the electric field at thelocatiion of q is E, that at the location of -2q is(1987)

1) 2

E 2) –2E 3) 2

E4) – 4E

28.dV

Edr

, here negative sign signified that

1) E is opposite to V 2) E is negative3) E increases when V decreases4) E is directed in the direction of decreasing V

29. An electron moves with a velocity v

in an

electric field E . If the angle between v

and

E is neither 0 nor , then path followed bythe electron is1) straight line 2) circle3) ellipse 4) parabola

30. A charged particle is free to move in an electricfield1) It will always move perpendicular to the line offorce2) It will always move along the line of force in thedirection of the field.3) It will always move along the line of forceopposite to the direction of the field.4) It will always move along the line of force in thedirection of the field or opposite to the direction ofthe field depending on the nature of the charge

31. Two parallel plates carry opposite chargessuch that the electric field in the space betweenthem is in upward direction. An electron isshot in the space and parallel to the plates.Its deflection from the original direction willbe1) Upwards 2) Downwards3) Circular 4) does not deflect


28 PINEGROVE

ELECTRIC POTENTIAL ANDPOTENTIAL ENERGY

32. Potential at the point of a pointed conductor is1) maximum 2) minimum3) zero 4) same as at any other point

33. An equipotential line and a line of force are1)perpendicular to each other2)parallel to each other3) in any direction 4) at an angle of 045

34. When a positively charged conductor is placednear an earth connected conductor, itspotential1) always increases 2) always decreases3) may increase or decrease 4) remains the same

35. If a unit charge is taken from one point toanother over an equipotential surface, then1) work is done on the charge2) work is done by the charge3) work on the charge is constant4) no work is done

36. Electric potential at some point in space is zero.Then at that point1) electric intensity is necessarily zero2) electric intensity is necessarily non zero.3) electric intensity may or may not be zero4) electric intensity is necessarily infinite.

37. When an electron approaches a proton, theirelectro static potential energy1) decreases 2) increases3) remains unchanged 4) all the above

38. An electron and a proton move through apotential difference of 200V. Then1) electron gains more energy2) proton gains more energy3) both gain same energy4) none of them gain energy

39. Two charges +q and –q are kept apart. Thenat any point on the right bisector of line joiningthe two charges.1) the electric field strength is zero2) the electric potential is zero3) both electric potential and electric field strengthare zero4) both electric potential and electric field strengthare non - zero

40. When ‘n’ small drops are made to combine toform a big drop, then the big drop’s1) Potential increases to n1/3 times original potential

and the charge density decreases to n1/3 timesoriginal charge2) Potential increases to n2/3 times original potentialand charge density increases to n1/3 times originalcharge density3) Potential and charge density decrease to n1/

3 times original values4) Potential and charge density increases to ‘n’times original values

41. A hollow metal sphere of radius 5cm is chargedsuch that the potential on its surface is 10V.The potential at the centre of the sphere is1) 0 V 2) 10 V3) same as at point 5cm away from the surface4) same as at point 25cm from the surface

42. The work done (in Joule) in carrying a chargeof ‘x’ coulomb between two points having apotential difference of ‘y’ volt is

1) x

y 2) 2x

y3)

y

x4) xy

43. Two charges q and -q are kept apart. Then atany point on the perpendicular bisector of linejoining the two charges. (2008E)1) the electric field strength is zero2) the electric potential is zero3) both electric potential and electric field strengthare zero4) both electric potential and electric field strengthare non-zero

44. Electric potential at the centre of a chargedhollow spherical conductor is1) zero2) twice as that on the surface3) half of that on the surface4) same as that on the surface

45. Which of the following pair is related as in workand force1) electric potential and electric intensity2) momentum and force3) impulse and force4) resistance and voltage

46. The equipotential surfaces corresponding tosingle positve charge are concentric sphericalshells with the charge at its origin. The spacingbetween the surfaces for the same change inpotential1) is uniform throughout the field

29 PINEGROVE


2) is getting closer as r 3) is getting closer as 0r 4) can be varied as one wishes to

47. Four identical charges each of charge q areplaced at the corners of a square. Then at thecentre of the square the resultant electricintensity E and the net electric potential Vare1) 0, 0E V 2) 0, 0E V 3) 0, 0E V 4) 0, 0E V

48. Two positive charges q and qare placed atthe diagonally opposite corners of a square andtwo negative charges -q and -q are placed atthe other two corners of the square. Then atthe centre of the square the resultant electricintensity E and the net electric potential V are1) 0, 0E V 2) 0, 0E V 3) 0, 0E V 4) 0, 0E V

49. Two copper spheres of the same radii, onehollow and the other solid, are charged to thesame potential, then1) hollow sphere holds more charge2) solid sphere holds more charge3) both hold equal charge4) we can’t say

50. Equipotential surfaces are shown in figure aand b. The field in

V02V03V0

x0 x0

9V0 4V0 V0

r02r0

3r0

FIGURE (B)FIGURE (A)

1) a is uniform only 2) b is uniform only3) a and b is uniform 4)both are nonuniform

51. Due to the motion of a charge, its magnitude1) changes2) does not changes3) increases (or) decreases depends on its speed4) can not be predicted

52. Induction preceeds attraction because1) an uncharged body can attract an unchargedbody due to induction of opposite charge on it2) a charged body can attract an uncharged bodydue to induction of same charge on it.3) a charged body can attract an uncharged bodydue to induction of opposite charge on it.

4) a charged body can attract another chargedbody due to induction of same charge on it.

53. The coulomb electrostatic force is defined for1) two spherical charges at rest2) two spherical charges in motion3) two point charges in motion4) two point charges at rest

54. The Electric field is given by 0

FE

q

, here the

test charge ‘q0’ should bea) Infinitesimally small and positiveb) Infinitesimally small and negative1) only a 2) only ‘b’3) a (or) b 4) neither ‘a’ or ‘b’

55. The p.d B cV V between two points from Cto B1) does not depend on the path2) depends on the path3) depends on test charge4) independent of electric field

56. Match List-I with List-IIList-I List-IIa) proton and e) gains same velocity electron in an elctric field forsame timeb) proton and f) gains same KE in anpositron electric field for same time.c) Deutron and g) experience same

- particle force in electric fieldd) electron and h) gains same KEpositron when accelerated

by same potentialdifference.

1) , , ,a h b g c e d f 2) , , ,a h b g c f d e 3) , , ,a g b h c e d f 4) , , ,a e b f c g d h

57. Match List-I with List-IIList-I List-IIa) Electric potential e) inversly proportionalinside a charged to square of the

conducting sphere distance 2( )rb) Electric potential f) directly proportional


30 PINEGROVE

charged sphere outside the conductingto distance

r from the centre

c) Electric field g) constantinside the nonconductingcharged sphere

d) Electric field h) inverslycharged sphere outside a

conductingproportional to

distance ( )r

1) , , ,a f b e c g d h 2) , , ,a e b f c h d g 3) , , ,a h b g c e d f 4) , , ,a g b h c f d e

58. Match the followingList-I List-IIa) Fluid flow d) Temperature

differenceb) Heat flow e) Pressure

differencec) Charge flow f)Potential difference1) , ,a e b d c f 2) , ,a d b e c f 3) , ,a f b e c d 4) , ,a e b f c d

59. Match List-I with List-IIList-I List-IIa) Two like charges e) the force betweenare brought nearer them decreases.b) Two unlike f) potential energycharge of some of the systembrought nearer increasesc) When a third g) mutual forces arecharge of same not affected nature is placedequidistance fromtwo like chargesd) When a dielectric h) potential energymedium is introduced of the system between two charges decreases

1) , , ,a h b f c g d e 2) , , ,a f b h c g d e 3) , , ,a h b f c e d g 4) , , ,a g b e c f d h

60. Match the following :a) Electric field e) Constant outside a conducting charged sphereb) Electric potential out f) directly propor side the conducting national to charged sphere distance from

centrec) Electric field inside g) inversely propor

a non-conducting tional to the charged sphere distance

d) Electric potential in h) inversely side a charged proportional to conducting sphere the square of the

distance

1) , , ,a h b g c e d f 2) , , ,a e b f c h d g 3) , , ,a h b g c f d e 4) , , ,a g b h c f d e

DIPOLE61. The angle between electric dipolemoment

p and the electric field E when the dipole isin stable equilibrium

1) 0 2) / 4 3) / 2 4) 62. ‘Debye’ is the unit of

1) electric flux 2) electric dipolemoment3) electric potential 4) electric field intensity

63. The electric field at a point at a distance r froman electric dipole is proportional to

1) 1

r2) 2

1

r3) 3

1

r4) 2r

64. An electric dipole placed with its axis in thedirection of a uniform electric fieldexperiences1) a force but not torque2) a torque but no force3) a force as well as a torque4) neither a force nor a torque

31 PINEGROVE


65. An electric dipole is placed in a non uniformelectric field increasing along the +ve directionof X - axis. In which direction does the dipole

Y

Y

q

X q

X

1) move along + ve direction of X - axis, rotateclockwise2) move along - ve direction of X - axis, rotateclockwise3) move along + ve direction of X - axis, rotateanti clockwise4) move along - ve direction of X - axis, rotateanti clockwise

66. An electric dipole placed in a nonuniformelectric field experiences1) a force but no torque2) a torque but no force3) a force as well as a torque4) neither a force nor a torque

67. If aE be the electric field intensity due to a

short dipole at a point on the axis and rE be

that on the perpendicular bisector at the samedistance from the dipole, then

1) a rE E 2) 2a rE E3) 2r aE E 4) 2a rE E

68. The electric potential due to an extremelyshort dipole at a distance r from it isproportional to

1) 1

r2) 2

1

r3) 3

1

r4) 4

1

r69. An electric dipole when placed in a uniform

electric field will have minimum potentialenergy, if the angle between dipole momentand electric field is1) zero 2) / 2 3) 4) 3 / 2

70. The angle between the electric dipole momentand the electric field strength due to it, on the

equatorial line is1) 00 2) 900 3) 1800 4) 600

71. A metallic shell has a point charge q kept insideits cavity. Which one of the following diagramscorrectly represents the electric lines of forces?

1) 2)

4)3)

ASSERTION & REASONIn each of the following questions, a statementof Assertion (A) is given followed by acorresponding statement of Reason (R) justbelow it. Mark the correct answer.1) Both ‘A’ and ‘R’ are true and ‘R’ is thecorrect explanation of ‘A’2) Both ‘A’ and ‘R’ are true and ‘R’ is not thecorrect explanation of ‘A’3) ‘A’ is true and ‘R’ is false4) ‘A’ is false and ‘R’ is true

72. Assertion(A) : Force between two pointcharges at rest is not changed by the presenceof third point charge between them.Reason(R): Force depends on the magnitudeof the first two charges and seperation betweenthem

73. Assertion (A): Electric potential at any pointon the equatorial line of an electric dipole iszeroReason (R): Electric potential is scalar

74. Assertion (A) : Electrons always move from aregion of lower potential to a region of highepotentialReason (R) : Electrons carry a negativecharge

75. Assertion(A): A metallic shield in form of ahollow shell may be built to block an electricfield.Reason (R): In a hollow spherical shield, theelectric field inside it is zero at every point.

76. Assertion (A): For practical purpose, the earthis used as a reference for zero potential in


32 PINEGROVE

electrical circuits.Reason (R): The electrical potential of asphere of radius R with charge Q uniformly

distributed on the surface is given by 04

Q

R77. Assertion(A): Coulomb force between charges

is central forceReason (R): Coulomb force depends onmedium between charges

78. Assertion(A): Electric and gravitational fieldsare acting along same line. When proton and - particle are projected up veritically alongthat line, the time of flights is less for proton.Reason (R): In the given electric fieldacceleration of a charged particle is directlyproportional to specific charge

79. Assertion(A): When a proton with certainenergy moves from low potential to highpotential then its KE decreases.Reason (R): The direction of electric field isopposite to the potential gradient and workdone against it is negative.

80. Assertion(A): In bringing an electron towardsa proton electrostatic potential energy of thesystem increases.Reason (R): Potential due to proton is positive

81. Assertion(A): The surface of a conductor isan equipotential surfaceReason (R): Conductor allows the flow ofcharge

82. Assertion (A) : A charge ' 1q ' exerts some force

on a second charge ' 2q ' . If a third charge '

3q ' is brought near , the force exerted by 1q

on 2q does not changeReason (R): The elecrtostatic force betweentwo charges is independent of presence of thirdcharge

83. Assertion (A) : A point charge 'q' is rotatedalong a circle around another point charge Q.The work done by electric field on the rotatingcharge in half revolution is zero.Reason (R) : No work is done to move a chargeon an equipotential line or surface.

84. Assertion: (A): Work done by electric forceis path independent.Reason: (R): Electric force is conservative

85. Assertion (A): In bringing an electron towardsa proton electrostatic potential energy of the

system increases.Reason (R): Potential due to proton is positive.

86. Assertion(A): Two particles of same chargeprojected with different velocity normal toelectric field experience same forceReason (R): A charged particle experiencesforce, independent of velocity in electric field

87. Assertion(A): The coulomb force is thedominating force in the universeReason (R): The coulomb force is strongerthan the gravitational force.

88. Assertion(A): A circle is drawn with a point

positive charge q at its centre. The work

done in taking a unit positive charge oncearound it is zeroReason (R): Displacement of unit positivecharge is zero

89. Assertion(A): Electric potential at any pointon the equatorial line of electric dipole is zero.Reason (R): Electric potential is scalar

90. Assertion(A): The potential at any point dueto a group of ' 'N point charges is simplyarrived at by the principle of superpositionReason (R): The potential energy of a systemof two charges is a scalar quantity

91. Assertion (A): The electrostatic potentialenergy is independent of the manner in whichthe cofiguration is achievedReason (R): Electrostatic field is conservativefield

STATEMENT QUESTIONS92. Statement-1:- For a charged particle moving

from point P to point Q, the net work done byan electrostatic field on the particle isindependent of the path connecting point P topoint Q.Statement-2:- The net work done by aconservative force on an ojecte moving alonga closed loop is zero1) Statement-1 is true, statement-2 is true,Statement-2 is the correct explanation of statement-1.2) Statement-1 is true, statement-2 is true,Statement-2 is not the correct explanation ofstatement-1.3) Statement-1 is false, Statement-2 is true.4) Statement-1 is true, Statement-2 is false

93. A dielectric slab of thickness d is inserted in aparallel plate capacitor whose negative plate

33 PINEGROVE


is at x = 3d. The slab is equidistant from theplates. The capacitor is given some charge. As‘x’ goes from 0 to 3d:1) the magnitude of the electric field remains thesame2) the direction of the electric field remains the same3) the electric potential increases continuously4) the electric potential dicreases at first, thenincreases and again dicreases

94. Choose the wrong statement1) Work done in moving a charge on equipotentialsurface is zero.2) Electric lines of force are always normal to anequipotential surface3) When two like charges are brought nearer, thenelectrostatic potential energy of the system getsdecreased.4) Electric lines of force diverge from positivecharge and converge towards negative charge.

95. Out of the following statementsA. Three charge system can not have zeromutual potential energyB. The mutual potential energy of a system ofcharges is only due to positive charges1) A is wrong and B is correct2) A is correct and B is wrong3) Both A and B are correct4) Both A and B are wrong

96. Statement A: Electrical potential may exist ata point where the electrical field is zeroStatement B : Electrical Field may exist at apoint where the electrical potential is zero.Statement C : The electric potential inside acharge conducting sphere is constant.1) A, B are true 2) B,C are true3) A,C are true 4) A,B,C are true

97. Statement A: If an electron travels along thedirection of electric field it gets acceleratedStatement B: If a proton travels along thedirection of electric field it gets retarded1) Both A & B are true 2) A is true, B is false3) A is false, B is true 4) Both A & B are false

98. A : Charge cannot exist without mass but masscan exist without charge.B : Charge is invariant but mass is variant withvelocityC : Charge is conserved but mass alone may

not be conserved.1) A, B, C are true 2) A, B, C are not true3) A, B are only true 4) A, B are false, C is true

99. A particle of mass m and charge q is fastenedto one end of a string fixed at point O. Thewhole system lies on a frictionless horizontalplane. Initially, the mass is at rest at A. Auniform electric field in the direction shown isthen switched on. Then

A

BO

E

l

600

1) the speed of the particle when it reaches B is

2qE

m

2) the speed of the particle when it reaches B is

qE

m

3) the tension in the string when particles reaches

at B is 2

Eq.

4) the tension in the string when the particle reachesat B is qE.

100. A conducting sphere A of radius a, with chargeQ, is placed concentrically inside a conductingshell B of radius b. B is earthed. C is thecommon centre of the A and B

B

QA

C

a

p) The field at a distance r from C, where

a r b , is 2

Qk

rq) The potential at a distance r from C, where

a r b , is Q

kr

r) The potential difference between A and B is


34 PINEGROVE

1 1kQ

a b

s) The potential at a distance r from C, where

a r b , is 1 1

kQr b

Choose the correct answer1. p and r are true 2. q is true3. p,r,s are true 4. p,q,r,s are true

101. A block of mass m is attached to a spring offorce constant k. Charge on the block is q. Ahorizontal electric field E is acting in thedirection as shown. Block is released with thespring in unstretched position

k

smooth

q, mE

a) block will execute SHM

b) Time period of oscillation is 2m

k

c) amplitude of oscillation is qE

kd) Block will oscillate but not simpleharmonicallyChoose the correct answer1) a and b are true 2) d is true3) a,b,c are true 4) a,b,c,d are true

102. A charge is moved against repulsion. Thenthere isA) decreasing its kinetic energyB) increasing its potential energyC) increasing both the energiesD) decreasing both the energies.1) A, B, C, D are true 2) A, B, C are true3) A, B are true 4) A only true

103. Which of the following statements are correct?a) The electrostatic force does not dependon medium in which the charges are placedb) The electrostatic force between twocharges does not exist in vacuumc) The gravitational force between massescan be usually neglected in comparision withelectrostatic forced) Any excess charge given to a conductor,not always resides on the outer surfaceof theconductor.1) both a & c 2) only ' 'c 3) both c & d 4) all

104. The property of the electric line of forcea) The tangent to the line of force at any pointis parallel to the directio of ' 'E at the pointb) No two lines of force intersect each other1) both a & b 2) only a 3) only b 4) a or b

105. Which of the following statements are correct.a) Electric lines of force are just imaginarylinesb) Electric lines of force will be parallel to thesurface of conductorc) If the lines of force are crowded, them fieldis strongd) Electric lines of force are closed loops1) both a & c 2) both b & d3) only a 4) all

106. Statement(A): Negative charges always movefrom a higher potential to lower potential pointStatement (B): Electric potential is vector.1) A is true but B is false2) B is true but A is false3) Both A and B false4) Both A and R are true

107. Statement (A): A solid conducting sphereholds more charge than a hollow conductingsphere of same radiusStatement (B) : Two spheres A and B areconnected by a conducting wire. No chargewill flow from A to B, when their radii are Rand 2R and charges on them are 2q and qrespectively1) A is true, B is false2) A is false B is true3) Both A and B are true4) Both A and B are false

108. A ring with a uniform charge Q and radius R,is placed in the yz plane with its centre at theorigina) The field at the origin is zero

b) The potential at the origin is Q

kR

c) The filed at the point (x, 0, 0) is 2

Qk

x

d) The field at the point (x, 0, 0) is 2 2

Qk

R xChoose the correct answer1) a and b are true 2) c is true3) a,b,c are true 4) a,b,c,d are true

35 PINEGROVE


109. A positively charged thin metal ring of radiusR is fixed in the xy plane, with its centre at theorigin O. A negatively charged particle P isreleased from rest at the point (0, 0, z0), wherez0>0. Then the motion of P isa) Periodic, for all value of z0 satisfying

00 z b) Simple harmonic, for all values of z0satisfying 00 z R c) Approximately simple harmonic, providedz0<<Rd) Such that P crosses O and continues to movealong the negative z-axis towards zChoose the correct answer1) a and b are true 2) c is true3) a,c,d are true 4) a,b,c,d are true

110. A circular ring carries a uniformly distributedpositive charge. The electric field (E) andpotential (V) varies with distance (r) from thecentre of the ring along its axis as

E

r

E

r

r

V

r

V

a) b)

c) d)

Choose the correct answer1) b and c are true 2) a is true3) a,b,c are true 4) a,b,c,d are true

111. Two concentric shells of radii R and 2R havegiven charges q and – 2q as shown in figure.In a region r < R

2q

q

R r

2R

a) E = 0 b) E 0 c) V = 0 d) V 0Choose the correct answer1) a and c are true 2) c is true3) a,d,c are true 4) a,b,c,d are true

C.U.Q - KEY1) 2 2) 2 3) 2 4) 3 5) 2 6) 37) 1 8) 3 9)1 10) 2 11) 1 12) 113) 4 14) 1 15) 3 16) 2 17) 4 18) 119) 2 20) 1 21) 1 22) 3 23) 4 24) 325) 2 26) 1 27) 3 28) 4 29) 4 30) 431) 2 32) 4 33) 1 34) 2 35) 4 36) 337) 1 38) 3 39) 2 40) 2 41) 2 42) 443) 2 44) 4 45) 1 46) 3 47) 3 48) 249) 3 50) 1 51) 2 52) 3 53) 2 54) 155) 1 56) 1 57) 1 58) 1 59) 2 60) 361) 1 62) 1 63) 3 64) 4 65) 1 66) 367) 2 68) 2 69) 1 70) 3 71) 3 72) 173) 1 74) 1 75) 1 76) 2 77) 2 78) 179) 2 80) 4 81) 2 82) 2 83) 1 84) 185) 4 86) 1 87) 4 88) 2 89) 2 90) 291) 1 92) 1 93) 2 94) 3 95) 4 96) 497) 4 98) 1 99) 2 100) 3 101) 3 102) 3103) 2 104) 1 105) 1 106) 3 107) 4 108) 1109) 1 110) 1 111) 1

EXERCISE - I ( C.W) ELECTRIC CHARGES AND

CONSERVATION OF CHARGE1. One million electrons are added to a glass rod.

The total charge on the rod is1) 1310 C 2) 131.6 10 C 3) 121.6 10 C 4) 1210 C

2. A body has a charge of 209.6 10 coulomb. It is1) possible2) not possible3) may (or) may not possible4) Data not sufficient

COULOMB’S LAW3. A force of 4N is acting between two charges in

air. If the space between them is completelyfilled with glass 8r , then the new force willbe1) 2N 2) 5N 3) 0.2N 4) 0.5N

4. There are two charges c1 and c2 keptat certain seperation . The ratio of electrostatic forces acting on them will be in the ratioof1) 1 : 2 2) 2 : 1 3) 1 : 1 4) 1 : 4


36 PINEGROVE

5. Two identical metal spheres possess +60Cand –20C of charges. They are brought incontact and then separated by 10 cm.Theforce between them is1) 1336 10 N 2) 1436 10 N3) 1236 10 N 4) 123.6 10 N

6. A charge q is placed at the centre of the linejoining two equal charges Q. The system ofthree charges will be in equilibrium if q is equalto

1) 2

Q 2) 4

Q 3) 4

Q 4) 2

Q

7. Three charges -q, +q and -q are placed at thecorners of an equilateral triangle of side ‘a’.The resultant electric force on a charge +qplaced at the centroid O of the triangle is

1) 2

20

3

4

q

a 2) 2

204

q

a 3) 2

202

q

a 4) 2

20

3

2

q

a8. A charge of 2 C is placed at x=0 and a

charge of 32 C at x=60 cm. A third charge –Q be placed on the x-axis such that itexperiences no force. The distance of thepoint from 2 C is(in cm)1) -20 2) 20 3) 15 4) 10

ELECTRIC FIELD9. Two charges of 50 C and 100 C aree

separated by a distance of 0.6m. Theintensity of electric field at a point midwaybetween them is

1) 650 10 Vm 2) 65 10 V

m3) 610 10 V

m 4) 610 10 Vm

10. Two point charges Q and -3Q are placed some

distnace apart. If the electic field at thelocation of Q is E

, the field at the location of-3Q is

1) E

2) E 3) 3

E

4) 3

E

11. A mass m carrying a charge q is suspendedfrom a string and placed in a uniformhorizontal electric field of intensity E. Theangle made by the string with the vertical inthe equilibrium position is

1) 1tan

mg

Eq 2)

1tanm

Eq

3) 1tanEq

m 4)

1tanEq

mg

12. A proton of mass ‘m’ charge ‘e’ is releasedfrom rest in a uniform electric field of strength‘E’. The time taken by it to travel a distance

‘d’ in the field is

1) 2de

mE2)

2dm

Ee3)

2dE

me4)

2Ee

dm13. An infinite number of charges each of

magnitude q are placed on x - axis at distancesof 1,2, 4, 8, ... meter from the origin. Theintensity of the electric field at origin is

1) 03

q

2) 06

q

3) 02

q

4) 04

q

14. A uniformly charged thin spherical shell of

radius R carries uniform surface chargedensity of per unit area. It is made of twohemispherical shells, held together by pressingthem with force F.F is proportional to

1)2 2

o

1R 2)

2

o

1R 3)

2

o

1

R

4)

2

2o

1

R


15. The p.d. between two plates separated by adistance of 1 mm is 100 V. The force on anelectron placed in between the plates is1) 510 N 2) 241.6 10 N3) 141.6 10 N 4) 191.6 10 N

16. An infinite number of charges each equal to'q' are placed along the X-axis at x = 1, x = 2,x = 4, x = 8 ..... The potential at the point x = 0due to this set of charges is

1) 4 o

Q

2) 2

4 o

Q

3) 3

4 o

Q

4) o

Q

17. A, B, C are three points on a circle of radius 1

cm. These points form the corners of anequilateral triangle. A charge 2C is placed atthe centre of the circle. The work done incarrying a charge of 0.1 C from A to B is1) Zero 2) 1118 10 J 3) 111.8 10 J 4) 1154 10 J

18. Charges +q, -4q and +2q are arranged at thecorners of an equilateral triangle of side 0.15m.If q=1 C, their mutual potential energy is1) 0.4J 2) 0.5J 3) 0.6J 4) 0. 8J

19. An electron of mass ‘M’ kg and charge ‘e’coulomb travels from rest through a potentialdifference of ‘V’ volt. The final velocity of theelectron is (in m/s)

1) 2eV

M2)

2MV

e3)

2eV

M 4)

2MV

e20. A charge ‘Q’ is placed at each corner of a cube

of side ‘a’. The potential at the centre of thecube is (2008 M)

37 PINEGROVE


1) 0

8Q

a 2) 0

4

4

Q

a 3) 0

4

3

Q

a 4) 0

2Q

a21. A uniform electric field pointing in positive x-

direction exists in a region let A be the orgin Bbe the point on the x-axis at x = +1 cm and Cbe the point on the Y axis at y = +1cm. Thenthe potentials at the points A, B and C satisfy1) V

A < V

B 2) V

A > V

B3) V

A < V

C4) V

A > V

C.

22. The electric field at the origin is along the +vex-axis. A small circle is drawn with the centreat the origin cutting the axes at the points A,B, C and D having coordinates (a, 0), (0, a),(-a, 0), (0, -a) respectively. Out of points onthe periphery of the circle, the potential isminimum at

B (0,a)

A (a, 0)E

D (0, a)C ( a, 0)1) A 2)B

3)C 4)D

DIPOLE23. An electric dipole is along a uniform electric

field. If it is deflected by 600, work done by anagent is 192 10 J. Then the work done by anagent if it is deflected by 300 further is1) 192.5 10 J 2) 192 10 J3) 194 10 J 4) 162 10 J

24. The dipole moment of the given system is2q

q q

ll

1) 3ql along perpendicular bisector of q - q line 2) 2 ql along perpendicular bisector of q - q line 3) ql 2 along perpendicular bisector of q - q line 4) 0

EXERCISE - I ( C.W ) KEY1)2 2) 2 3) 4 4) 3 5) 1 6) 27) 3 8) 1 9)2 10) 3 11) 4 12) 213) 1 14) 1 15) 3 16) 2 17) 1 18) 319) 3 20) 3 21) 2 22)1 23) 2 24) 1

EXERCISE - I ( C.W ) HINTS1. Q ne n is integer 2.Q ne n is integer

3. | FF

K 4.

1 22

0

14

q qF

r

5. 2

1 2

20

14 4

q qF

d

6.2 2

0 0

1 10

4 4

2

QQ qQ

l l

7.1 2

20

14

q qF

r 8. 2

1

1

dx

qq

9. 1 22 2

0 01 2

1 1

4 4

q qE

x x 10. 3

0

1

4

QE r

r

11. tanqE mg 12. 21

2

qEs t

m

13. 2 2 20

1 1 1

4 1 2 4

qE

14. Pressure

2

o2

and Force =2

2

o

R2

15. F Eq =

Vq

d

16.0

1 1 1 1

4 1 2 4 8

QV

17. Equipotential surface

18. 2 3 1 31 2

0 1 2 3

1

4

q q q qq qU

r r r

19. 21

2mv eV 20.

0

1

4

qV

d21. Along the field direction potential decreases.

22. V E dr 23. 1 1 cosW pE and

2 1 2cos cosW pE 24.

1 2 3Rp lq p and P ql EXERCISE - I(H.W)

COULOMB’S LAW1. A charge Q is divided into two parts 1q and 2q

such that they experience maximum force ofrepulsion when separated by certain distance.The ratio of Q, 1q and 2q is1) 1 : 1 : 2 2) 1 : 2 : 2 3) 2 : 2 : 1 4) 2 : 1 : 1

2. Two charges each 1 c are at 2 3P i j k m

and Q i j k m . Then the force betweenthem is _____1) 100N 2) 10N 3) 410 dyne 4) 100 dyne

3. Two charges of 200 C and 200 C areeplaced at the corners B and C of an equilateraltriangle ABC of side 0.1 m. The force on a


38 PINEGROVE

charge of 5 C placed A is1) 1800 N 2)1200 3N 3) 600 3N 4) 900N

4. Two equally charged pith balls 3 cm apart repeleach other with a force of 54 10 newton. Thecharge on each ball is

1) 92 10 C 2) 92 10 C 3) 9210

3C 4) 92

103

CELECTRIC FIELD

5. An electron 319.1 10mass kg is sent into anelectric field of intensity

69.1 10 newton/coulomb. The accelerationproduced is

1) 1821.6 10 m

s 2) 6

21.6 10 ms

3) 18

21.6 10 ms

4) 621.6 10 m

s

6. The electric field at (30, 30) cm due to a chargeof -8 nC at the origin in NC-1 is

1) 400 i j 2) 400 i j3) 200 2 i j 4) 200 2 i j

7. Two charges of 10 C and -90 C areeseparated by a distance of 24 cm. Electrostaticfield strength from the smaller charge is zeroat a distance of1) 12 cm 2) 24 cm 3) 36 cm 4) 48 cm

8. Two electric charges of 910 C and 910 Care placed at the corners A and B of anequilateral triangle ABC side 5cm.The electricintensity at C is1)1800N/C 2)3600 N/C 3)900N/C 4)2700 N/C


9. If 204 10 eV is required to move a charge of 0.25coulomb between two points, the potentialdifference between these two points is

1) 256 volt2) 1

256 volt

3) 19256 10 volt 4) 250 volt10. Two electric charges of 9 C and -3 C aree

placed 0.16m apart in air. There are two pointsA and B on the line joining the two charges atdistances of (i) 0.04m from -3 C and inbetween the charges and (ii) 0.08m from -3 C and out side the two charges. Thepotentials at A and B are1) 0V, 5V 2) 0V, 0V 3) 5V, 0V 4) 5V, 10V

11. Four charges 3 , 1 , 5C C C and 7 Care arranged on the circumference of a circleof radius 0.5 m. The potential at the centre is1) Zero 2) 418 10V3) 418 10V 4) 3288 10V

12. A positive point charge ‘q’ is carried from apoint ‘B’ to a point ‘A’ in the electric field of apoint charge +Q. If the permittivity of freespace is 0 , the work done in the process isgiven by

+Q A B

a

b

1) 0

1 1

4

qQ

a b 2)

0

1 1

4

qQ

a b

3) 2 20

1 1

4

qQ

a b 4) 2 2

0

1 1

4

qQ

a b

13. An electric cell does 5 joules of work in carrying10 Coulomb’s of charge around a closedcircuit. The emf of the cell is1) 2V 2) 0.5V 3) 4V 4) 1V

14. Two positive charges 12 C and 10 C areeinitially separated by 10cm. The work done inbringing the two charges 4cm closer is1) 7.2J 2) 3.6J 3) 8.4J 4) 12.4J

15. An insulated charged conducting sphere ofradius 5 cms has a potential of 10V at thesurface. What is the potential at centre?1) 10V 2) zero3) same as that at 5 cms from the surface4) same as that at 25 cms from the surface

16. A positive charge 'Q' is fixed at a point.Anegatively charged particle of mass 'm' andcharge 'q' is revolving in a circular path ofradius 'r 1' with 'Q' as the centre. The work tobe done to change the radius of the circularpath from r 1 to r2 in Joules is

1) 0 2)

21o r

1

r

1

4

Qq

3)

21oo r

1

r

1

4

Qq

4

14)

12o r

1

r

1

4

Qq

17. Figure bellow shows a square array of chargedparticles, with distance d between adjacenetparticle. What is the electric potential at pointP at the centre of the square if the electricpotential is zero at infinity ?

2q4q q

q

q

q P

2qq

1) Zero 2)d4

2q-

03)

d44q-

0 4) .d4

q

018. The radii of two charged metal spheres are 5cm

and 10cm both having the same charge 60mC.

39 PINEGROVE


If they are connected by a wire1) A charge of 20mC flows through the wire fromlarger to smaller sphere2) A charge of 20mC flows through the wire fromsmaller to larger sphere3) A charge of 40mC flows through the wire fromsmaller to larger sphere4) No charge flows through the wire because bothspheres have same charge.

19. The electric potential at a point (x, 0, 0) is

given by V = 2 3

1000 1500 500

x x x then the

electric field at x = 1 m is (in volt/m)

1) ˆ5500i 2) ˆ5500i 3) ˆ5500i 4) zero

DIPOLE20. An electric dipole of moment p is placed in

the position of stable equilibrium in uniformelectric field of intensity E. It is rotatedthrough an angle from the intial position.The potential energy of electric dipole in theposition is1) cospE 2) sinpE 3) 1 cospE 4) cospE

21. An electric dipole of moment p

is placednormal to the lines of force of electric intensity

E

, then the work done in deflecting it through

an angle of 0180 is1) pE 2) 2pE 3) 2pE 4) zero

EXERCISE - I (H.W) KEY 1) 4 2) 4 3) 4 4) 2 5) 1 6) 3 7) 18) 2 9) 1 10) 2 11) 1 12) 1 13) 2 1 4 )4 15) 1 16) 2 17) 3 18) 2 19)2 20) 421) 4

EXERCISE - I (H.W) HINTS

1. 1 2F q q 2. 1 2

20

14

q qF

r

3.1 2

1 2 20

14

q qF F

r ; 1 2rF F F because angle

between then is 0120

4.2

20

14

qF

d 5. eE

am

6. 30

1

4

QE r

r

7. 2

1

1

dx

qq

8. 2

0

14

QE

a

9. W q V 10.1 2

0 1 2

1

4

q qV

r r

11.0

1

4

QV

r 12.

1 2

0 1 2

1 1

4

q qW

r r

13.W

emfq

14.1 2

0 1 2

1 1

4

q qW

r r

15.0

1

4

QV

R 16.

1 2

0 1 2

1 1

4

q qW

r r

17.0

1

4

QV

r 18. V=constant andQ R

19.dV

Edx

20. .U p E 21. 1 1 cosW pE

EXERCISE - II (C.W)COULOMB’S LAW

1. Two charges when kept at a distance of 1mapart in vacuum hava some force of repulsion.If the force of repulsion between these twocharges be same, when placed in an oil ofdielectric constant 4, the distance ofseparation is1) 0.25m 2) 0.4m 3) 0.5m 4) 0.6m

2. The excess (equal in number) number ofelectrons that must be placed on each of twosmall spheres spaced 3 cm apart with force ofrepulsion between the spheres to be 1910 Nis1) 25 2) 225 3) 625 4) 1250

3. Two small conducting spheres each of mass49 10 kg are suspended from the same point

by non conducting strings of length 100 cm.They are given equal and similar charges untilthe strings are equally inclined at 045 each tothe vertical. The charge on each sphere is .....coulomb1) 61.4 10 2) 61.6 10 3) 62 10 4) 61.96 10

4. Two point charges of magnitude 4 C and -9 C are 0.5m apart. The electric intensity iszero at a distance ‘x’ m from ‘ A’ and ‘y’ mfrom ‘B’. ‘x’ and ‘y’ are respectively

4mc 9mcp

A B0.5m

1) 0.5m, 1.0m 2) 1.0m, 1.5m3) 2.0m, 1.5m 4) 1.5m, 2.0m


40 PINEGROVE

5. A charge +q is fixed to each of three cornersof a square. On the empty corner a charge Qis placed such that there is no net electrostaticforce acting on the diagonally opposite charge.Then

1) 2Q q 2) 2 2Q q 3) 2Q q 4) 4Q q

6. Electrical force between two point charges is200N. If we increase 10% charge on one ofthe charges and decrease 10% charge on theother, then electrical force between them forthe same distance becomes1) 198 N 2) 100 N 3) 200 N 4) 99 N

7. N fundamental charges each of charge ‘q’ areto be distributed as two point charges seperatedby a fixed distance, then the maximum tominimum force bears a ratio (N is even andgreater than 2)

1)

2

2

N4

1N 2) 1N

N4 2

3) 1N4

N2

4) 1N

N2 2

8. A particle A having a charge of 62 10 C and

a mass of 100g is placed at the bottom of asmooth inclined plane of inclination 300. Thedistance of another particle of same mass andcharge, be placed on the incline so that it mayremain in equilibrium is1) 27 cm 2) 16 cm 3) 30 cm 4) 45 cm

9. Two identical particles of charge q each areconnected by a massless spring of forceconstant k. They are placed over a smoothhorizontal surface.They are released whenunstretched. If maximum extension of thespring is r, the value of k is : (neglectgravitational effect)

1) 0

1qk

r r 2) 2

20

1 1

4

qk

l r

3) 0

2 1qk

r r 4) 0

2qk

r rELECTRIC FIELD

10. In the figure shown, the electric field intensityat 1 , 6 , 9r m r m r m in 1Vm is

V

10V r (m)

0 2 8 10

1)-5, -1.67, +5 2) -5, 0, +5

3) 0,1.67,0 4) +5, 1.67, -5

11. Point charges of 93 10 C are situated ateach of three corners of a square whose sideis 15 cm. The magnitude and direction ofelectric field at the vacant corner of the square

is1) 2296 V/m along the diagonal2) 9622 V/m along the diagonal3) 22.0 V/m along the diagonal 4) zero

12. A large flat metal surface has uniform chargedensity . An electron of mass m and chargee leaves the surface at an angle at point Awith speed v , and return to it at point B. Themaximum value of AB is ____

1) 0vm

e

2) 2

0v m

e

3) 2

0

v e

m 4) 2

0

v e

m

13. ‘n’ charges Q, 4Q, 9Q, 16Q ..... are placed atdistances of 1, 2, 3 ..... metre from a point ‘0’on the same straight line. The electricintensity at ‘0’ is

1) 204

Q

n 2) 04

Q

n 3) Infinity 4) 04

nQ

14. Two point charges 1 2q C and 2 1q C aree

placed at distances b=1 cm and a=2 cm fromthe origin on the y and x axes as shown infigure. The electric field vector at point (a, b)will subtend an angle with the x - axis givenby

y

q1 P (a, b)

q2O a

b

x

1) tan 1 2) tan 2 3) tan 3 4) tan 4 15. A non-conducting ring of radius 0.5 m carries

of total charge of 1.11x10-10c distributed non-uniformly on its circumference producing anelectric field E everywhere in space. The value

of the integral

l 0

l

E.dl

(l=0 being

centre of the ring) in volts is1) +2 2) -1 3) -2 4) zero


16. Three charges +q, -q and -q are kept at thevertices of an equilaterial triangle of 10cm side.The potential at the mid point in between -q, -q, if q = 5 C is1) 56.4 10V 2) 412.8 10V 3) 46.4 10V 4) 512.8 10V

17. Two charges each ‘Q’ are released when thedistance between is ‘d’. Then the velocity ofeach charge of mass ‘m’ each when the distancebetween them is ‘2d’ is

1) 08

Q

dm 2) 04

Q

dm 3) 04

Q

dm 4) 02

Q

dm

41 PINEGROVE


18. An oil drop carrying charge ‘Q’ is held inequilibrium by a potential difference of 600Vbetween the horizontal plates. In order to holdanother drop of twice the radius in equilibriuma potential drop of 1600V had to be maintained.The charge on the second drop is

1) 2

Q2) 2Q 3)

3

2

Q4) 3Q

19. A body of mass one gram and carrying a charge810 C passes through two points P and Q. The

electrostatic potential at Q is OV. The velocityof the body at Q is 10.2ms and at P is

10.028ms . The potential at P is1) 150V 2) 300V 3) 600V 4) 900V

20. Three charges each 20 C are placed at thecorners of an equilateral triangle of side 0.4m.The potential energy of the system is1) 618 10 J 2) 9J 3) 69 10 J 4) 27J

21. An electric field is expressed as E 2i 3j .

The potential difference A BV V betweentwo points A and B whose positions vectors are

given by Ar i 2 j and

Br 2i j 3k is1) -1 V 2) 1 V 3) 2 V 4) 3 V

22. Figure shows three circular arcs, each of radiusR and total charge as indicated. The netelectric pontential at the centre of thecurvature is

1) o

Q

2 R 2) o

Q

4 R 3) o

2Q

R 4) o

Q

R23. Two identical conducting very large plates

1 2P and P having charges 4Q and 6Q areplaced very closed to each other at separationd. The plate area of either face of the plate isA. The potential difference between plates

1 2P and Pis

1) 1 2P Po

QdV V

A 2) 1 2P P

o

QdV V

A

3) 1 2P P

o

5QdV V

A 4) 1 2P P

o

5QdV V

A

DIPOLE

24. An electric dipole consists of two oppositecharges of magnitude 1 C separated by adistance of 2cm. The dipole is placed in anelectric filed 5 110 Vm . The maximum torque

that the field exert on the dipole is1) 310 Nm 2) 132 10 Nm3) 33 10 Nm 4) 34 10 Nm

25. An electric dipole is formed two particles fixedat the ends of a light rigid rad of length l. Themass of each particle is m and charges are -qand +q The system is suspended by atorsionless thread in an electric field ofintensity E such that the dipole axis is parallelto the field if it is slightly displaced, the periodof angular motion is

1)1 2

2

qE

ml 2)2ml

qE 3)2

2

ml

qE 4)

1

2 4

ml

qE26. Two point charges - q and +q are located at

points (0,0,-a) and (0,0,a) respectively. The

electric potential at point (0,0,z) is z a

+ Q

2Q

3Q

R

300

450

1) 204

qa

z 2) 04

q

a3) 2 2

0

2

4

qa

z a 4) 2 20

2

4

qa

z a 27. Two equal charges ‘q’ of opposite sign are

separated by a small distance ‘2a’. The electricintensity ‘E’ at a point on the perpendicularbisector of the line joining the two charges ata very large distance ‘r’ from the line is

+ 4Q + 4Q

P1 P2

d

1) 20

1

4

qa

r 2) 30

1 2

4

qa

r3) 2

0

1 2

4

qa

r 4) 30

1

4

qa

rEXERCISE - II (C.W) KEY

1) 3 2) 3 3) 1 4) 2 5) 2 6) 1 7) 38) 1 9) 2 10) 2 11) 1 12) 2 13) 4 14)115) 1 16) 4 17) 1 18) 4 19) 3 20) 4 21)122) 1 23) 2 24) 2 25) 3 26) 3 27) 2

EXERCISE - II (C.W) HINTS

1.1 t

kt

2.1 2

20

1

4 q qF

r and q = ne


42 PINEGROVE

3. F = w tan where 1 2

20

14

q qF

r4. Distance of null point

2

1

1

dx

Q

Q

+ve for like charges -ve for unlike charges

5. 2

220 0

1 12 0

4 4 2

q Qq

a a

6.1 2

20

14

q qF

r ; 11 1

110

100q q and 1

2 290

100q q

7.

2

max 2min 1 1

NF

F N 8.

0

21

4 2sin

qmg

r

9. CF kx 10. dV

Edr

11. 2

0

12 1/ 2

4

qE E E

r r = length of the side

12. Field near metal surface E=0

Force on electron = eE =0

e

Acceleration of electron a =0

e

m

It will act as projectile with max range

=2 2

0

u um

a e 13.

1 22 2 21 2

1. ....

4n

o n

Q Q QE

x x x

14.2

1

ETan

E

15. o

qV K ,V 0

R ;

t 0

o

t

E.dl V V

16. 1 2 3V V V V ; 1 2

0

1

4 / 2

qV V

a

3

0

1

4 32

qV

a 17. gain in K.E = loss in P.E 18.

3

1 1 2

2 2 1

.V R Q

V R Q

19.2 21

2 Q p P Qm v v q V V 20.

2 3 3 11 2

0 12 23 13

1

4

q q q qq qU

r r r

21.

2 1

B A

1 2

V V 2dx 3dy

22. 1 2 3V V V V 23. 1 2P P

o

QV V

A / d

24. max 2pE aqE 25. sin PE ; I ; sinI PE

I = moment of inertia =2

2

ml

Time period 2I

pE

26. The distance of point P from charge +q is 1r z a and from charge -q is 2r z a Potential at P is

2 20 1 2 0

1 1 2

4 4

q q qa

r r z a

27. Similar to B on equitorial line of a short bar magnet

EXERCISE - II (H.W)COULOMB’S LAW

1. Two equally charged identical metal spheresA and B repel each other with a force F.Another identical uncharged sphere C istouched to A and then placed midway betweenA and B. The net force on C is in the direction1) F towards A 2) F towards B3) 2F towards A 4) 2F towards B

2. Two unlike charges seperated by a distanceof 1m attract each other with a force of0.108N . If the charges are in the ratio 1:3,theweak charge is

1)2 C 2) 4 C 3) 6 C 4) 5 C3. Three charges each equal to 910 C are placed

at the corners of an equilateral triangle of side1m. The force on one of the charges is

1) 99 10 N 2) 99 3 10 N3) 927 10 N 4) 918 10 N

43 PINEGROVE


4. Two particles each of mass ' 'm and carryingcharge ' 'Q are seperated by some distance.Ifthey are in equilibrium under mutualgravitational and electro static forces, then

/Q m (in c/Kg) is of the order of

1) 510 2) 1010 3) 1510 4) 20105. There point charges + q, – q and + q are

placed at the vertices P, Q and R of anequilateral triangle as shown. If

2

20

1 qF

4 r

, where 'r' is the side of the

triangle, the force on charge at 'P' due tocharges at Q and R is

P X

Y

R

1) F along positive x–direction

2) F along negative x–direction

3) 2 F along positive x–direction

4) 2 F along negative x–direction6. Three point charges +q, +q and –q are placed

at the corners of an equilateral triangle ofside 'a'. Another charge +Q is kept at thecentroid. Force exerted on Q is:

1) 2o

1 2qQ

4 a 2) 2o

1 6qQ

4 a

3) 2o

1 8qQ

4 a 4) 2o

1 14qQ

4 a7. Three charges 1 2 3q , q and q are placed

as shown in fig. The X-component of the forceon 1q is proportional to Y

q3

q1

q1Xb

a1) 32

2 2

qqcos

b a 2) 32

2 2

qqsin

b a

3) 2 22 2

q qcos

b a 4) 2 2

2 2

q qsin

b a

ELECTRIC FIELD8. The breakdown electric intensity for air is

63 10 V/m. The maximum charge that canbe held by a sphere of radius 1 mm is1) 0.33 C 2) 0.33 nC 3) 3.3 C 4) 3.3 C

9. There is a uniform electric field of strength310 /V m along y-axis. A body of mass 1 g and

charge 610 C is projected into the field fromorigin along the positive x-axis with a velocity

10 m/s. Its speed in m/s after 10s is (neglectgravitation)1) 10 2) 5 2 3) 10 2 4) 20

10. The point charges 1 , 1C C and 1C areplaced at the vertices A, B and C of anequilateral triangle of side 1m. Then(A) The force acting on the charge at A is

99 10 N(B) The electric field strength at A is

9 19 10 NC1) A is correct but B is wrong2) B is correct but A is wrong3) Both A and B are wrong4) Both A and B are correct

11. A pendulum bob of mass m carrying a chargeq is at rest in a uniform horizontal electric fieldof intensity E. The tension in the thread is

1) 2 2T Eq mg 2) 2

2ET mg

q

3)

2 2E m

Tq g

4) T mg Eq ELECTRIC POTENTIAL AND

POTENTIAL ENERGY12. Four charges 8 8 810 ; 2 10 ; 3 10 and 82 10

coulomb are placed at the four corners of asquare of side 1m the potential at the centreof the square is1) zero 2)360 volt 3) 180 volt 4) 360 2 volt

13. Two metal spheres of radii 1R and 2R areecharged to the same potential. The ratio of thecharge on the two spheres is

1) 1 2) 1

23) 1 2R R 4)

1

2

R

R14. Two concentric, thin metallic spherical shells

of radii 1R and 2 1 2R R R bear charges 1Q and

2Q respectively. Then the potential at radius

‘r’ between 1R and 2R will be 0

1

4 times

1) 1 2Q Q

r

2)

1 2

1

Q Q

R r 3)

1 2

1 2

Q Q

R R 4)

1 2

2 2

Q Q

R R

15. An electric charge 310 C is placed at theorigin (0, 0) of X-Y coordinate system. Two

points A and B are situated at 2, 2 and(2, 0) respecitvely. The potential differencebetween the points A and B will be:1) 9 V 2) zero 3) 2 V 4) 4.5 V


44 PINEGROVE

16. A charge 2 C at the origin, 1 C at 7cmand 1 C at 7cm are placed on X axis. Themutual potential energy of the system is1) 0.051J 2) 0.045J 3) 0.045J 4) 0.064J

17. Four equal charges Q are placed at the fourcorners of a square of side ' 'a each. Workdone in removing a charge Q from its centreto infinity is

1) zero 2) 2

0

24

Qa

3) 2

0

2Qa 4)

2

02Q

a 18. The electrostatic potential V at any point

(x,y,z) in space is given by 24V x1) The y - and z - components of the electrostaticfield at any point are not zero2) The x - component of electric field intensity at

any point is given by 8xi

3) The x - component of electric field intensity at a

point (2, 0,2) is 8i

4) The y - and z - components of the field areconstant in magnitude.

DIPOLE19. The self potential energy of hydrogen chloride

whose dipole moment is 303.44 10 C - m andseparation between hydrogen and chlorineatoms is 101.01 10 m is1) 191.036 10 J 2) 53.2 10 J3) 74.5 10 J 4) 61.65 10 J

EXERCISE - II (H.W) KEY 1)1 2) 1 3) 2 4) 2 5) 2 6) 2 7) 2 8) 2

9)3 10) 4 11) 1 12) 4 13) 4 14) 2 15) 2 16) 417) 3 18) 2 19) 1

EXERCISE - II (H.W) HINTS

1.1 2

20

1

4 q qF

r 2. 1 2

20

1

4 q qF

r

3.2

1 2 20

1

4 qF F

r ; 2

20

13.

4R

qF

r

4. 2 2

2 20

14e g

q GmF and F

r r

5. 1 2F F and angle between them is 0120

6. 20

12

4

qF

r

where 3

ar

8. 2

1.

4 o

QE

d 9. v u at where

Eqa

m

10.1 2

2

1.

4 o

q qF

r ; 2

1.

4 o

qE

r1 2F F F ; 1 2E E E

12.0

1

4

QV

r 13.

0

1

4

QV

R

14. Potential is constant within the sphere and isadditive.

15. 1 2

1 1

4 o

qV

r r 16.

1 2

0

1.

4

q qPE

r17. Workdone = Electrostatic potential

energy at the centre of the square

18.dV

Edx

19. 202 3.41 10

2

pp qa q

a

219

0

11.036 10

4 2

qPE J

a EXERCISE - III

ELECTROSTATIC FORCE1. A ball of mass m = 0.5 kg is suspended by a

thread and a charge q = 0.1 C is supplied.When a ball with diameter 5cm and a likecharge of same magnitude is brought close tothe first ball, but below it, the tension decreasesto 1/3 of its initial value. The distance betweencentres of the balls is1) 20.12 10 m 2) 40.51 10 m3) 50.2 10 m 4) 20.52 10 m

2. Five point charges each +q, are placed on fivevertices of a regular hexagon of side L, Themagnitude of the force on a point charge ofvalue – q placed at the centre of the hexagon(in newton) is

1)Zero 2)2

20

3

4q

L 3)

2

204

q

L 4)

2

204 3

q

L3. Two small objects X and Y are permanently

separated by a distance 1 cm. Object X has acharge of + 1.0 C and object Y has a chargeof - 1.0 C . A certain number of electrons

45 PINEGROVE


are removed from X and put onto Y to makethe electrostatic force between the two objectsan attractive force whose magnitude is 360 N.Number of electrons removed is

1) 138.4 10 2) 126.25 10 3) 114.2 10 4) 103.5 104. Two identical positive charges are fixed on the

y-axis, at equal distance from the origin O, Apartical with a negative charge starts on thenegative x-axis at a large distance from O,moves along the x-axis passed through O andmoves far away from O. Its acceleration a istaken as positive along its direction of motion.The particle's acceleration a is plotted againstits x-co-ordinate. Which of the following bestrepresents the plot?

a

xo

a

xOO x

a

1) 2)

3) 4)

a

xO

5. Two equal negative charges –q each are fixedat points (0, –a) and (0,a) on y-axis. A positivecharge Q is released from rest at the point (2a,0) on the x-axis. The charge Q will1) execute simple harmonic motion about the origin2) move to the origin and remain at rest3) move to infinity4) execute oscillatory but not simple harmonicmotion

6. In a liquid medium of dielectric constant K andof specific gravity 2, two identically chargedspheres are suspended from a fixed point bythreads of equal lengths. The angle betweenthem is 90º. In another medium of unknowndielectric constant K1, and specific gravity 4,the angle between them becomes 120º. Ifdensity of material of spheres is 8 gm/cc thenK1 is :

1) K2

2) 3

K3)

3K

24)

K

37. The force of attraction between two charges

separated by certain distance in air is F1. Ifthe space between the charges is completelyfilled with dielectric of constant 4 the forcebecomes F2. If half of the distance between the

charges is filled with same dielectric the forcebetween the charges is F3. Then F1 : F2 : F3 is1) 16 : 9 : 4 2) 9 : 36 : 163) 4 : 1 : 2 4) 36 : 9 : 16

8. Two small spheres of masses, 1M and 2M areesuspended by weightless insulating threads oflengths 1L and 2L . the sphere carry charges

1Q and 2Q respectively. The spheres areesuspended such that they are in level withanother and the threads are inclined to thevertical at angles of 1 and 2 as shown below,,which one of the following conditions isessential , if 1 2 .

1

L1

M1 Q1

2

L2

M2 Q2

1) 1 2M M but 1 2Q Q 2) 1 2M M3) 1 2Q Q 4) 1 2L L

ELECTRIC FIELD9. If the electric field between the plates of a

cathode ray oscilloscope be 41.2 10 /N C , thedeflection that an electron will experience if itenters at right angles to the field with kineticenergy 2000 eV is (The deflection assembly is1.5cm long.)1) 0.34 cm2) 3.4 cm3) 0.034 mm 4) 0.34 mm

10. A electric field of 4 11.5 10 NC existsbetween two parallel plates of length 2 cm. Anelectron enters the region between the platesat right angles to the field with a kinetic energyof 2000 .kE eV The deflection that theelectron experiences at the deflecting platesis1) 0.34 mm 2) 0.57 mm 3) 7.5 mm 4) 0.75 mm

11. A bob of a simple pendulum of mass 40gm witha positive charge 64 10 C is oscillating with atime period 1T .An electric field of intensity

43.6 10 N/C is applied vertically upwards.Now

the time period is 2T the value of 2

1

TT is (g =

10m/s2)1)0.16 2) 0.64 3)1.25 4)0.8


46 PINEGROVE

12. A particle of mass m and charge q is placedat rest in a uniform electric field E and thenreleased. The kinetic energy attained by theparticle after moving a distance y is1) 2qEy 2) 2qE y 3) qEy 4) 2q Ey

13. Four equipotential curves in an electric fieldare shown in the figure. A,B,C are three pointsin the field.If electric intensity at A,B,C are

, ,A B CE E E then

C B A

120V 90V 60V 60V

1) A B CE E E 2) A B CE E E 3) A B CE E E 4) A B CE E E

14. A particle of mass 1Kg and carrying 0.01C isat rest on an inclined plane of angle 300 with

horizontal when an electric field of 1490

3NC

applied parllel to horizontal .The coefficient offriction is

1) 0.5 2) 1

33)

32

4)3

715. Electric field on the axis of a small electric

dipole at a distance r is 1E

and 2E

at a distanceof 2r on a line of perpendicular bisector. Then

1) 2 1 / 8E E

2)

2 1 /16E E

3 )2 1 / 4E E

4)

2 1 / 8E E

16. A particle having charge that on an electronand mass 1.6 x 10–30 kg is projected with aninitial speed 'u' to the horizontal from the lowerplate of a parallel plate capacitor as shown.The plates are sufficiently long and haveseparation 2cm. Then the maximum value ofvelocity of particle not to hit the upper plate.(E=103 V/m upwards).

uE = 10 V/m3

45°1)2 x 106 m/s 2) 4 x 106 m/s3) 6 x 106 m/s 4) 3x 106 m/s

17. An electric field is acting vertically upwards.A small body of mass 1 gm and charge -1Cis projected with a velocity 10 m/s at an angle450 with horizontal. Its horizontal range is 2mthen the intensity of electric field is :(g = 10 m/s2)1) 20,000 N/C 2) 10,000 N/C3) 40,000 N/C 4) 90,000 N/C

18. A thin copper ring of radius ‘a’ is charged withq units of electricity. An electron is placed atthe centre of the copper ring. If the electronis displaced a little, it will have frequency.

1) 30

1

2 4

eq

ma 2) 30

1

2 4

q

ema 3)

04

eq

ma 4) 304

q

ema 19. A thin fixed ring of radius 1 metre has a

positive charge 51 10 C uniformly distributedover it. A particle of mass 0.9gm and having anegative charge of 61 10 C is placed on theaxis at a distance of 1 cm from the centre ofthe ring. Assuming that the oscillations havesmall amplitude, the time period of oscillationsis1) 0.23s 2) 0.39s 3) 0.49 s 4) 0.63s

20. A sphere carrying charge 0.01 C is kept atrest without falling down, touching a wall byapplying an electric field 100 N/C.If thecoeffcient of friction between the sphere andthe wall is 0.2 , the weight of the sphere is1) 4N 2) 2 N 3) 20 N 4) 0.2 N

21. A particle of mass 1kg and carrying positivecharge 0.01 C is sliding down an inclined planeof angle 030 with the horizontal. An electricfield E is applied to stop the particle. If thecoefficient of friction between the particle and

the surface of the plane is 1

2 3, E must be

E cos

E sin

mg sin mg

E

= 30 0

1)1260 V/m 2)245 V/m

3) 140 3 V/m 4)490

3V/m

22. Two identical point charges are placed at aseparation of l . P is a point on the line joiningthe charges, at a distance x from any onecharge. The field at P is E. E is plotted againstx for values of x from close to zero to slightlyless than l . Which of the following bestrepresents the resulting curve?

47 PINEGROVE


1) 2)

3) 4)

E

O X lX l

E

E

Xl

O

O O X l

E

23. A particle of charge q and mass m moves ina circular orbit of radius r about a fixed charge

Q . The relation between the radius of theorbit r and the time period T is

1) 3

2016

Qqr T

m 2) 3 2

3016

Qqr T

m 3)

2 33

016

Qqr T

m 4) 2 3

016

Qqr T

m 24. A thin semicircular ring of radius ‘r’ has a

positive charge distributed uniformly over it.The net field E at the centre ‘O’ is (AIEEE

2010) i

iO

1) 2 202

qj

r 2) 2 204

qj

r 3) 2 2

04

qj

r 4) 2 202

qj

r ELECTRIC POTENTIAL

25. Two thin rings each having a radius R areplaced at distance d apart with their axescoinciding.The charges on the two rings are+q, -q. The potential difference between therings

1) 20

.4 .

Q Rd 2) 2 2

0

1 12

QR R d

3) 2 20

1 14

QR R d

4)0

26. Two metal sphres A and B have their capacitiesin th ratio 3:4. They are put in contact witheach other and an amount of charge

67 10 C is given to the combination. Next, thetwo spheres are separated and kept wide theapart so that one has no electrical infuence onthe other. The potential due to the smallersphere at a distance of 50m from it is1) 540V 2) 270V 3) 1180V 4) zero

27. A solid conducting sphere having a charge Qis surrounded by an uncharged concentricconducting spherical shell. The potentialdifference between the surface of solid sphereand the shell is V. The shell is now given acharge –3Q. The new potential differencebetween the same surfaces will be 1) –

2V 2) 4V 3) V 4) 2V28. A spherical charged conductor has surface

charge density . The intensity of electricfield and potential on its surface are E and V.Now radius of sphere is halved keeping thecharge density as constant. The new electricfield on the surface and potential at the centreof the sphere are1) 4E, V 2) E, V/2 3) E, V 4) 2E, 4V

29. Two spherical conductors A and B of radii 1mm and 2mm are seperated by a distance of 5cm and are uniformly charged. If the spheresare connected by a conducting wire then in theequilibrium condition the ratio of electric fieldsat surfaces of A and B is1) 4: 1 2) 1 : 2 3) 2 : 1 4) 1 : 4

30. A charge +q is fixed at each of the pointsx=x0,x=3x0, x=5x0 . . . . . on the x - axis anda charge-q is fixed at each of the points

0 0 02 , 4 , 6 .......x x x x x x . Here 0x is apositive constant. Take the electric potentialat a point due to a charge Q at a distance r

from it to be 04

Q

r . Then the potential at

the origin due to the above system of chargesis

1) 0 2) 0 08 log 2e

q

x 3) 4)

0 0

log 2

4eq

x 31. A non – conducting ring of radius 0.5 m carries

a total charge of 1.11 x 10 –10 C distributed non– uniformly in its circumference producing anelectric field E everywhere in space. The value

of the integral 0l

lEdl

( l = 0 being centre

of the ring) in volt is :1) +2 2) –1 3) –2 4) zero

32. Some equipotential surfaces are shown infigure. The electric field strength is

Y10V 20V 30V

x cm15cm10cm5cm

300

001) 100 V/m along x-axis

2) 100 V/m along y-axis

3) 400 V/m at an angle 1200 with x-axis

4) 400

3V/m an angle 1200 with x-axis


48 PINEGROVE

33. A field of 100Vm–1 is directed at 300 to positvex - axis. Find VBA if OA = 2m and OB = 4m

B

AO300

1) 100 3 2 V 2) 100 2 3 V

3) 100 2 3 V 4) 200 2 3 V34. Here is a special parallelogram with adjacent

side lengths 2a and a and the one of thepossible angles between them as 60°. Twocharges are to be kept across a diagonal only.The ratio of the minimum potential energy ofthe system to the maximum potential energyis1) 3 : 7 2) 3 : 7 3) 1 : 2 4) 1 : 4

35. Two concentric spherical conducting shells ofradii R and 2R carry charges Q and 2Qrespectively. Change in electric potential on theouter shell when both are connected by a

conducting wire is : 0

1

4k

1) zero 2) 3

2

kQ

R3)

kQ

R4)

2kQ

R36. The longer side of a rectangle is twice the length

of its shorter side. A charge q is kept at onevertex. The maximum electric potential due tothat charge at any other vertex is V, then theminimum electric potential at any other vertexwill be1) 2V 2) 3 V 3) / 5V 4) 5 V

37. There are three uncharged identical metallicspheres 1,2 and 3 each of radius r and areplaced at the vertices of an equilateral triangleof side d. A charged metallic sphere havingcharge q of same radius r is touched to sphere1, after some time it is taken to the location ofsphere 2 and is touched to it, then it is takenfar away from spheres 1,2 and 3. After thatthe sphere 3 is grounded, the charge on sphere3 is (neglect electrostatic induction byassuming d>> 2r)

1) Zero 2) 3

4

qr

d

3)

2

qr

d

4)

4

qr

d

38.

EP (a , b, 0)

(0 , 0, 0)Q (2a , 0, 0)

R (a , b, 0)A point charge q moves from point P to point Salong the path PQRS in a unifrom electric field

E

pointing parallel to the positive directionof the x-axis. The coordinates of the points P,Q, R and S are (a,b,0), (2a,0,0), (a, –b, 0) and(0,0,0) respectively. The work done by the fieldin the above process is given by the expression1) qaE 2) –qaE

3) 2 2( )q a b E 4) 2 23qE a b39. The potential at a point x (measured in m)

due to some charges situated on the x-axis is

given by 2

20

4V x

x

volt. The electric field

E at x = 4 m is given by

1) 5

3 V

m and in the positive x - direction

2) 10

9 V

m and in the negative x - direction

3) 10

9 V

m and in the positive x-direction

4) 5

3 V

m and in the negative x-direction

40. Two points charges q1 and q2 (=q1/2) are placedat points A(0, 1) and B (1, 0) as shown in thefigure. The electric field vector at point P(1, 1)makes an angle q with the x–axis, then theangle q is

q1

q2

Bx

O

AP(1,1)

y

1)1 1

tan2

2) 1 1

tan4

3) 1tan 1 4) 1tan 0

49 PINEGROVE


41. Figure shows three spherical and equipotentialsurfaces 1,2 and 3 round a point charge q. Thepotential difference V1–V2 = V2 – V3. If t 1 andt2 be the distance between them. Then

3

2

1

t1

t2

q

1) t1=t

2 2) t

1>t

2

3) t1<t

2 4) 1 2t t

42. A half ring of radius ‘ r ’ has a linear chargedensity .The potential at the centre of thehalf ring is

1) 04

2) 2

04 r

3) 04 r

4) 2

04 r

43. The distance between plates of a parallel platecapacitor is 5d. The positively charged plateis at x=0 and negativily charged plates is atx=5d. Two slabs one of conducotor and theother of a dielectric of same thickness d areinserted between the plates as shown in figre.Potential (V) versus distance x graph will be

+q q

xO d 2d 3d 4d 5d

CO

ND

UC

TO

R

DIE

LEC

TR

IC

V

X5d4d3d2ddO

1)

V

X5d4d3d2ddO

2)

V

X5d4d3d2ddO

3) 4)

V

X5d4d3d2ddO

44. A dielectric slab of thickness d is inserted in aparallel plate capacitor whose negative plateis at x=0 and positive plate is at x=3d. Theslab is equidistant from the plates. Thecapacitor is given some charge. As one goesfrom 0 to 3d,1) the magnitude of the electric field remains thesame2) the direction of the electric field remains the same3) the electric potential increases continuously'4) the electric potential increases at first, thendecreases and again increases.

45. A solid sphere of radius R is chargeduniformly. The electrostatic potential V isplotted as a function of distance r from thecentre of the sphere. Which of the followingbest represents the resulting curve ?

VR

ro

VR

o

VR

ro

3) 4)

1) 2)

VR

ro

POTENTIAL ENERGY

46. Along the X-axis, three charges 2

q,-q and

2

q

are placed at x = 0, x =a and x =2arespectively . The resultant electric potentialat x=a+r(if a ,<<r) is ( 0 is the permittivity offree space

1) 204

qa

r 2) 2

304

qa

r 3)

2

30

( / 4)

4

q a

r 4) 04

q

r 47. An electron travelling from infinity with

velocity ‘v’ into an electric field due to twostationary electrons separated by a distanceof 2m. If it comes to rest when it reaches themid point of the line joining the stationaryelectrons.The initial velocity ‘ v ‘ of theelectron is1) 16m/s 2) 32m/s3) 16 2 /m s 4)32 2 /m s

48. Work performed when a point charge82 10 C is transformed from infinity to a point

at a distance of 1cm from the surface of theball with a radius of 1cm and a surface chargedensity = 8 210 /C cm1) 41.1 10 J 2) 411 10 J3) 40.11 10 J 4) 4113 10 J

49. A conducting sphere A of radius a, with acharge Q, is placed concentrically inside aconducting shell B of radius b. B is earthed. Cis the common centre of A and B.


50 PINEGROVE

B

A

Ca

b

Q

1) The field at a distance r from C, where a r b ,

is 2

Qk

r2) The potential at a distance r from C, where

a r b , is 2

Qk

r3) The potential difference between A and B is4) The potential at a distance r from C, wherea r b , is

50. Given figure shows an arrangement of six fixedcharged particle. The net electrostatic force Facting on charge +q at the origin due to othercharges is

300300

a a

+qq

OB CAq +q +qq

2a 2a

a

1) 2

20

6

4

q

a 2) zero

3) 2

20

7

2

q

a 4) 2

20

33

24

q

a

51. 2q and 3q are two charges separated by adistance 12 cm on x-axis. A third charge q isplaced at 5 cm on y-axis as shown in figure.Find the change in potential energy of thesystem if 3q is moved from initial position to apoint on X-axis in circular path:

2q 3q (12,0)(5,0)

( 5)0,

( ) 0,

q

1) 2

4o

q2)

26

4 91 o

q3)

218

4 91 o

q4)

23

4o

q

52. An electron travelling from infinity withvelocity 'V' into an electric field due to twostationary electrons seperated by a distanceof 2m. If it comes to rest when it reaches the

midpoint of the line joining the stationaryelectrons, the initial velocity 'V' of theelectron is (in m/sec)1) 16 2) 32 3) 16 2 4) 32 2

53. A particle of mass m and charge q is projectedvertically upwards. A uniform electric field E

is acted vertically downwards. The mostappropiate graph between potential energyU (gravitational plus electrostatic) and heighth(<< radius of earth) is : (assume U to be zeroon surface of earth)

U U

hh

U U

hh

1) 2)

3) 4)

54. Three charged particles are initially inposition 1. They are free to move and theycome in position 2 after some time. Let U1and U2 be the electrostatic potential energiesin position 1 and 2. Then :1) U

1>U

22) U

2>U

13) U

1=U

2 4) 2 1U U

55. Two identical thin rings, each of radius R,are coaxially placed a distance r apart. If Q1and Q2 are respectively the charges uniformlyspread on the two rings, the work done inmoving a charge q from the centre of one ringto that of the orther is

1) zero 2) 1 2 02 1 / 24q Q Q R

3) 1 2 02 / 4q Q Q R

4) 1 2 0/ 2 1 24q Q Q R

56. The electric potential at a point (x, 0, 0) is

given by V = 2 3

1000 1500 500

x x x then the

electric field at x = 1 m is (in volt/m)

1) ˆ5500i 2) ˆ5500i 3) ˆ5500i 4) zero

57. Six charges are placed at the vertices of aregular hexagon as shown in the figure. Theelectric field on the line passing through pointO and perpendicular to the plane of the figureat a distance of x (>> a) from O is

51 PINEGROVE


Q Q

Q Q

Q Q

a

1) 30

Qa

x 2) 30

2Qa

x 3) 30

3Qa

x 4) zero

DIPOLE58. A small electric dipole is placed at origin with

its dipole moment directed along positive x -axis. The direction of electric field at point 2,2 2,0 is

1) along z - axis 2) along y - axis3) along negative y -axis 4) along negative z-axis

59. Two electric dipoles each of dipolemoment306.2 10p C m are placed with their axis

along the same line and their centres at adistanced= 810 cm . The force of attractionbetween dipoles is

1) 162.1 10 N 2) 122.1 10 N3) 102.1 10 N 4) 82.1 10 N

60. Two charges 193.2 10 C and 193.2 10 C placed 02.4A apart form an electric dipole. Itis placed in a uniform electric field of intensity

54 10 /V m the work done to rotate theelectric dipole from the equilibrium position by1800 is

1) 233 10 J 2) 236 10 J3) 2312 10 J 4) Zero

61. Two opposite and equal charges4 x10–8 coulomb when placed 2 x 10–2 cm away,from a dipole. If this dipole is placed in anexternal electric field 2x 10–2 newton/coulomb, the value of maximumtorque and the work done in rotating it through1800 will be

1) 432x10 Nm and 432x10 J2) 464x10 Nm and 464x10 J3) 464x10 Nm and 432x10 J

4) 432x10 J and 464 10x Nm62. An electric dipole is made up of two particles

having charges 1 c , mass 1 kg and other

with charge 1 c and mass 1 kg separatedby distance 1m. It is in equilibrium in a uniformelectric field of 20 x 103 V/m. If the dipole isdeflected through angle 20, time taken by it tocome again in equilibrium is1) 2.5 s 2) 2.5 s 3) 5 s 4) 4

63. A point particle of mass M is attached to oneend of a massless rigid non-conducting rod oflength L. Another point particle of the samemass is attached to the other end of the rod.The two particle carry charges + q and – qrespectively. This arrangement is held in aregion of a uniform electric field E such thatthe rod makes a small angle (say of about5

0) with the field direction (see figure). The

expression for the minimum time needed forthe rod to become parallel to the field after itis set free.

A q

q

E

B

O

1)2 2

mLt

qE

2)

mLt

2 qE

3) 2mL

t2 qE

4)

3mLt

2 2qE

EXERCISE - III - KEY

1) 4 2) 3 3) 2 4) 2 5) 4 6) 47) 4 8) 2 9) 4 10) 4 11) 3 12) 313) 3 14) 4 15) 2 16) 1 17) 3 18) 119) 4 20) 4 21) 3 22) 4 23) 2 24) 425) 2 26) 1 27) 3 28) 2 29) 3 30) 431) 1 32) 3 33) 1 34) 1 35) 1 36) 337) 2 38) 2 39) 3 40) 1 41) 3 42) 143) 2 44) 2 45) 3 46) 2 47) 2 48) 249) 1 50) 2 51) 3 52) 2 53) 1 54) 155) 2 56) 2 57) 1 58) 2 59) 4 60) 261) 4 62) 1 63) 1


52 PINEGROVE

EXERCISE - III - HINTS

1.1 2

2

1

4 o

q qT mg

r

2 ,3&5. 0

1

4 r

F 1 22

q q

d6. F mgTan7.

0

1

4 r

F 1 22

q q

d8. There are three forces acting on each sphere are

(i) tension (ii) weight(w) (iii) electrostatic force ofrepulsion for sphereIn equailibirum, from figure

1 1 1tan /F M g From sphere 2, in equilibirum from figure

2 2 2tan /F M g for 1 2F For 1 2 only for

1 2

1 2

F F

M g M g

But, 1 2F F and then 1 2M M9. Deflection

2

4

eExy

K where K is kinetic energy..

10. 2

.4

eEy K K E

k

11. T =2

eff

lg

12. K.E =FS K.E = qEy

13.dV

Edx

14. N =mg sin + qE sin

mg sin = cosN qE 15. 3

2axis

kpE

r and 32

bicectorkp

Er

16. Maximum height

2 2sin

2

u

EQg

m

17. Range 2 sin 2u

EQg

m

18. 3/2 32 20 0

1

4 4

qx qxE

aa x 2

2 30

1

4

d x qexm

dt a So motion is S.H.M.

23

0

1

4

qe

ma

19. 304

Qq xF kx

R and 2 mT

k20. mg qE21. sin cos sin cosmg qE qE 23.

22

0

1

4

QqF mr

r ;

2

T

24. 2

0

1 sin / 2

4 / 2

qE

r

2 20

sin / 2

2

qE j

r

25. 1 2 20

1 14Q

VR R d

2 2 2

0

1 14

QV

R R d

1 2V V V 26.

11

1 2

rq q

r r

; 1

10

1

4

qV

r27. Pd between the two spheres is independent of

charge on outer shell.

28.0

E and

0

RV

29. .

QV K

R .

2 2V Q

KR

.Q

KR

1

. .2

Q QK K

R d 2d R

When the two conducting spheres are connectedby a conducting wire, charge will flow from onesphere (having higher potential) to other (havinglower potential) till both acquire the same potential.

There fore, V

Er

1 2

2 1

22 :1

1

E r

E r

30.0 0 0 0 0

1 1 1 1

4 2 3 4

qV

x x x x

=0 0

1 1 11

4 2 3 4

q

x

0 0

log 24

q

x 31. .V E dr

53 PINEGROVE


32.dV

Edx

33. .V E r 34. long and short diagonal lengths are

2 2 2 cosp q pq 35.

0

1 3

4 2

QV

R

36. If the charges is kept at ‘A’ then maximum andminimum potentials at D and C respectively

37. commen potential38. w = Fs ; W =q.E S

39.dx

dvE

41. 1 21 2

1 1V V kq

r r

1 2 1 22 1

V V r rr r

kq

; but 2 1r r t 1 2t r r if P.D is constant then 2 1r r t 42. potential due to small element ‘ p ’ at the centre

.dl

dv Kr

; 0

0 0

14

14 4

v dv k dl dlr r

rr

43.dx

dvE E inside the conductor is zero.

44. The direction of E is constant.

45.0

1

4

QV

r

46. Force of interaction

3 30

2 1 1

4

pq

d l d l

47.0

2 1 214

1

2

q qmv

r48 . Potential at a distance 2cm from its centre

= 2

'0 0

4

4 4

Q r

r r

2

'0 0

1

2 100

r

r

since r=1 cm and 'r =2 cm

PD b/w the two points is equal to 0200

work done =VQ= 0200

X 82 10 = 411 10 J

49. field concept

50. concept of force

51. f iU U U and 1 2

0

1

4

q qU

r

52. 0iu , 2

0

12 .

4 / 2f

eu

d

212

PE KE mv calculate ‘ v ‘

53. conceptual.54. Particles moves in a direction where potential energy

of the system decreased.

55.1 2

10 04 4 2

Q QV

R R

and 1 2

200 44 2

Q QV

RR

1 2 2 1W q V V 56.

dVE

dx

57. concept of field58. Use vector representation59. Force of interaction

3 30

2 1 1

4

pq

d l d l

60. 2 1 2 2 2W PE PE PE qdE 61. pESin 62. 2

IT

pE

63. 2I

TpE

********


54 PINEGROVE

Electric flux:It is the measure of total number of electric lines offorce crossing normally the given area. The total flux passes through the given surface is

given by .E A E

n

cosEA where is the angle made by the normal with theelectric field.For a closed body outward flux is taken to bepositive while inward flux is taken to be negative.

A1 A2

A3

E

a) Flux through A1 : negative

b) Flux through A2 : positive

c) Flux through A3 : 0

GAUSS LAWi) According to this law, the total flux linked with a

closed surface called Gaussian surface is 0(1/ )times the net charge enclosed by the closed surface.

ii) Alternatively, Gauss law can be stated as the surfaceintegral of electric field E over a closed surface is

equal to 1/ 0 times the charge (q) enclosed by thatclosed surface.

i.e., 0

.q

E dA

q is the total charge enclosed by the Gaussiansurface.

iii) Coulomb’s law can be derived from Gauss law.iv) The electric field E

is the resultant field due to all

charges, both those inside and those outside theGaussian surface.

v) The electric field due to a charge outside theGaussian surface contributes zero net flux throughthe surface, Because as many lines due to that chargeenter the surface as leave it.

+Q -Q

E

-+

a) Flux from surface 10

QS

b) Flux from surface 20

QS

c) Flux from S3 = flux from surface S4 = 0Applications of Gauss Law :

i) If a dipole is enclosed by a surface

-Q +Q 0encQ 0 ii) The net charge Qenc is the algebraic sum of the

enclosed positive and negative charges. If Qenc ispositive then the net flux is outwards. If Qenc isnegative then the net flux is inwards.

1 2 30

1( )Q Q Q

iii) If a closed body (not enclosing any charge) is placedin an electric field (either uniform or non - uniform)total flux linked with it will be zero

(A) = 0T

Sphere

Y

Xa

a

(B) = = in out rEa2

= 0

E

Z

iv) If charge is kept at the centre of cube

Q

0

1( )total Q ;

0

1( )

6face Q v) If charge is kept at the centre of a face, first we

should enclose the charge by assuming a Gaussiansurface (an identical imaginary cube)

GAUSS LAW

55 PINEGROVE


(a) (b)

Q

Total flux emerges from the system (Two cubes)

0total

Q Flux from given cube (i.e., from left side 5 faces

only) 02cube

Q vi) If a charge is kept at the corner of a cube

(a) (b)

For enclosing the charge seven more cubes arerequired so total flux from the 8 cube system is

0T

Q .

Flux from given cube 08cube

Q .

Flux from one face opposite to the charge, of thegiven cube

0

0

/8

3 24face

Q Q (Because only three faces

are seen).Electric field at a point due to a line ofcharge: A thin straight wire over which ‘q’amount of charge be uniformly distributed. bethe linear charge density i.e, charge present per unitlength of the wire.

02

qE

rl 02E

r

+

+

+

+

dS dE

ds

This implies electric field at a point due to a linecharge is inversely proportional to the distance ofthe point from the line charge.Electric field intensity at a point due to a thininfinite charged sheet [Non conducting plate]‘q’ amount of charge be uniformly distributed over

the sheet. Charge present per unit surface area ofthe sheet be . i.e surface charge density

dS

Eds

P

02

qE

A ;

02

qE where

A

E is independent of the distance of the point from

the charged sheet.Electric field intensity at a point due to a thickinfinite charged sheet [Conducting plate] :‘q’ amount of charge be uniformly distributed overthe sheet. Charge present per unit surface area ofthe sheet be .

P

dS

E

ds

ds

dSE

0 0

qE

A

Electric field at a point due to a thick charged sheetis twice that produced by the thin charged sheet ofsame charge density.

APPLICATIONS OF GAUSS LAW Electric field due to long uniformly charged

cylinder:

L

R

r

Consider a long cylinder of radius R which is uni-formly charged on its surface with charge density .We know that at the interior points of a metal bodyelectric field strength is zero. Let us find the electricfield at a point and at a distance r from the axis ofthe cylinder. Consider a cylindrical Gaussian sur-face of radius r and length L as shown in the figure.From Gauss's law, we can write

en0

1E.ds (q )

Here enclosedq 2 RL Here electric flux through the circular faces is zero.So, from Gauss law

0

2 RLE.ds

or 0

2 RLE2 rL


56 PINEGROVE

0

RE

r

The variation of E with distance r from the axis isas shown in the graph.

E

r = R r

E 1

r

Electric field due to uniformly charged non-conducting cylinder: Consider a long cylinder ofradius R charged with volume charge density uniformly. Let us find electric field at a distance rfrom the axis of the cylinder. Consider a cylindricalGaussian surface of length L and radius r as shown,

R

+

+

+

+

r

L

encl

0

qE .ds

; where 2enclq R L

Here electric flux through the circular faces is zero.Case (i): If r > R, then from Gauss's law

2

0

R LE.ds

2

0

R LE2 rL

or

2

0

RE

2 r

out1

Er

Case (ii): If r = R, then 0

RE

2

Case (iii): If r < R, 2enclq r L

from Gauss law encl

0

qE.ds

2

0

r LE2 rL

(or) 0

rE

2

inE r

In vector form 0

rE

2

The variation of E with distance r from the axis isas shown in the graph.

EE r

E 1

r

r = R rELECTRIC INTENSITY AND ELECTRICPOTENTIAL DUE TO INFINITE PLANE

SHEET OF CHARGE (NONCONDUCTING):If E is the magnitude of electric field at point P, thenelectric flux crossing through the gaussian surfaceis given by

EE

Area = A

GaussianSuracePlane sheet

of charge

= E x area of the end face (circular caps) of thecylinderor = E x 2A ..........(i)According to Gauss's theorem, we have

0

q

Here, the charge enclosed by the gaussian surface,q A

0

A..........(ii)

From equations (i) and (ii), we have

0

AE 2A

or 0

ˆ2

E n

Where n is unit vector normal to the plane andaway from it.Thus, we find that the magnitude of the electric fieldat a point due to an infinite plane sheet of charge isindependent of its distance from the sheet of charge.Electric intensity due to two thin parallelcharged sheets:Two charged sheets A and B having uniform chargedensities A

and B respectively..

In region I :

E 0

1

2 A B

57 PINEGROVE


In region II:

AE

I.

II.

III.

A

A

B

AE

BE

r

B

BE

0

1

2II A BE In region III :

0

1

2III A BE Electric field due to two oppositelycharged parallel thin sheets :

EA

EB

EA

EB

EB

EA

III III

0

10

2IE

0 0

1[ ]

2IIE

0

1( ) 0

2IIIE Electric field due to a charged Spherical shell‘q’ amount of charge be uniformly distributed overa spherical shell of radius ‘R’

Surface charge density, 24

q

R

When point ‘P’ lies outside the shell (r>R):

20

1

4

qE

r This is the same expression as obtained for

electric field at a point due to a point charge. Hencea charged spherical shell behaves as a point chargeconcentrated at the centre of it.

2

2 20

1 .4

4 4

R qE

r R

2

20

.RE

r

When point ‘P’ lies on the shell (r =R) :

0

E

When Point ‘P’ lies inside the shell (r <R):E = 0 The electric intensity at any point due to a chargedconducting solid sphere is same as that of a chargedconducting sperical shell.Electric Potential (V) due to a UniformlyCharged sphe-rical c onducting shell (Hollowsphere)

qP2

r

rP1

P3

R

When point 3P lies outside the sphere r R ,

the electric potential, 0

1

4V q

r

When point 2P lies on the surface r R ,

0

1

4

qV

R When point 1P lies inside the surface r R ,

0

1

4

qV

R Note: The electric potential at any point inside the

sphere is same and is equal to that on the surface.

E

O

E = 0in

r = R

2

1

rE out

(A)

OR

OR

VS

O

V 0inr = R

(B)

V out 1

r

Note: The electric potential at any point due to a chargedconducting sphere is same as that of a chargedconducting spherical shell


58 PINEGROVE

W.E-1: A particle that carries a charge ‘–q’ isplaced at rest in uniform electric field 10 N/C. It experiences a force and moves. In acertain time ‘t’, it is observed to acquire avelocity 10 10i j m/s. The given electric fieldintersects a surface of area 1m2 in the x–zplane. Find the Electric flux through thesurface.

Sol: Force on charge F qE

particle moves opposite to E with V

unit vector in the direction of V is i j

2 2

unit vector in the direction of E is i j

2 2

i j

E 102 2

ie. A 1 j

Electric flux 2E A 5 2 Nm / C

W.E-2: The electric field in a region is given by

0ˆx

E E iL

. Find the charge contained inside

a cubical volume bounded by the surface x =0, x = L, y =0 , y = L, z = 0 and z = L.

Sol: At x = 0, E = 0 and at x = l, 0ˆE E i

The direction of the field is along the x-axis, so itwill cross the yz-face of the cube. The flux of thisfield

y

z

x

X = L

E0

left face right face ; 2 20 00 E L E L

By Gauss's law, 0

q 20 0 0q E L

W.E-3: A square surface of side lm in the planeof the paper. A uniform electric field E (V/m)also in the plane of the paper, is limited onlyto the lower half of the square surface, theelectric flus (in SI units) associated with thesurface is.

E

Sol: A, electric flux, .E E ds 0cos cos90 0E ds E ds

Thus, the lines are parallel to the surface.

W.E- 4 : A hollow cylinder has a chartge ‘q’ cou-lomb within it. If is the electric flux in unitof V-m, associated with the curved surface B,the electirc flux linked with the plane surfaceA in unit of V-m, will be

C A

B

Sol: We have, 0

total A B c

q '

0

2q '

0

1

2

q

W.E-5: The adjacent diagram shows a charge +Qheld on an insulating suppot S and enclosedby a hollow spherical conductor, O representsthe centre of the spherical conductor and P isa point such that OP=x and SP=r. The elec-tric field at point, P will be

x

r P

O

S

Charge + Q

Sol: According to Gauss’s’ theorem,

0

. inQE ds ;

2

0

.4Q

E x or 204

QE

xW.E-6: The electrostatic potential inside a charged

spherical ball is given by 2ar b , where,‘r’ is the distance from the centre, a and b areconstants. Then the charge density the ball is

Sol: Here, 2ar b ; As 2ar b

0

.q

E ds ; 2 3

00

2 .4 8q

ar r q a r

343

q

r

; 06a

59 PINEGROVE


SOLID ANGLESolid angle is the three dimensional anglesubtended by the lateral surface of a cone atits vertex

solid angle Let us calculate the solid angle subtended by asurface X at a point O. Join all the points of theperiphery of the surface X to the point O by straightlines as shown. It gives a cone with vertex at O.

r2

r1 S1

S2

xO

Fig. (b)

By taking centre at O, we draw several sphericalsections on this cone of different radii as shown.Let the area of spherical section which is of radiusr1 be s1 and the area of section of radius r2 be s2.The ratios of area of any surface intersectedby cone to the square of radius of that sphereis a constant and it gives actually the solid angleFrom the figure, solid angle subtended by surface

X at the point O is given by 1 22 21 2

s s

r r.

Note: SI unit of solid angle is steradian and it is adimensionless quantity. one steradian is the solid angle subtended atthe centre of the sphere by the surface of thesphere having area equal to square of theradius of the sphere.The surface subtending solid angle need not benormal to the axis of the cone. For exampleconsider a surface X of area ds as shown. Theaxis of cone formed by the surface at O is notnormal to the surface. In this cone solid angle subtended at point O can be given as

2

dscos

r

xr

ds

O

Here is the angle between ds and axis of thecone.

RELATION BETWEEN SEMI- VERTEXANGLE OF A CONE AND SOLID ANGLE

SUBSTENDEDConsider a spherical surface of radius R. Let X bea surface on that sphere which substends a semivertex angle ( in radian) at the centre of the sphere.Now consider an elemental strip of this section ofradius r = R sin and angular width d as shown.Then surface area of this strip is given by ds 2 Rsin Rd .The total area of spherical section can be obtainedby integrating this elemental area from0 to .Total area of spherical section is

2

0

S ds 2 R sin d

20

2 R cos 22 R 1 cosr

d

O

R

Rd

If is solid angle subtended by this section at thecentre O, then its area is given by 2S R (asdiscussed earlier) So, we can write

2 2R 2 R (1 cos ) and 2 1 cos Note : The solid angle substended by a hemispherical

surface at its centre is given by 02 1 cos90 2 steradians

If = 1800 in the previous case, we get the solidangle substended by a closed surface 02 1 cos180 4 steradians

The total solid angle substanded by a closed surfaceis always 4 steradians, irrespective of the size andshape of the closed surface.

W.E-7: A point charge q is placed at a distance dfrom the centre of a circular disc of radius R.Find electric flux through the disc due to thatcharge


60 PINEGROVE

q R

d Sol : We know that total flux originated from a point

charge q in all directions is 0

q

. This flux is originated

in a solid angle 4 . In the given case solid anglesubtended by the cone subtended by the disc atthe point charge is 2 1 cos So, the flux of q which is passing through the surfaceof the disc is

0 0

q q1 cos

4 2

From the figure,

2 2

dcos

d R so

2 20

q d1

2 d R

W.E-8: Two point charges +Q1 and -Q2 are placed

at A and B respectively. A line of force emanatesfrom Q1 at an angle with the line joining AAand B. At what angle will it terminate at B?

Q1 Q2

A B

Sol: We know that number of lines of force emerge isproportional to magnitude of the charge. The fieldlines emanating from Q1, spread out equally in alldirections. The number of field lines or flux through

cone of half angle is 1Q2 1 cos

4 . Similarly

the number of lines of force terminating on -Q2 at

an angle is 2Q2 1 cos

4 . The total lines of

force emanating from Q1 is equal to the total linesof force terminating on Q2

1 2Q Q2 1 cos 2 1 cos

4 4

or 1 2Q Q1 cos 1 cos

2 2 ; 2 2

1 2Q sin / 2 Q sin / 2

1

2

Qsin / 2 sin / 2

Q

1 1

2

Q2 sin sin / 2

Q

CAVITY IN THE CONDUCTORWe have discussed that there will be no electricfield inside a charged conductor and all the chargeresides on its outer surface only. Suppose thatcharged conductor has a cavity or cavities and there

are no charges within the cavity or cavities, eventhen charge resides on the outer surface of theconductor. There will be no charge on the walls ofthe cavity or cavities. This can be verfied very easilyusing Gauss's law by enclosing the cavity with aGaussian surface.

cavity

Fig.E.ds 0 For the dotted surface.

q = 0 inside cavity..Consider a conductor with spherical cavity insideit. There is no charge on the conductor. Now apoint charge +q is kept at the centre of the cavity.Due to this charge, a charge -q is induced on theinner surface of cavity . The total flux originated by+q will terminate on the cavity walls and no fieldlines enter into the conductor body

-q

P+q

Fig. (a)

-q+q

-q

Fig. (b)

We can consider a Gaussian surface around thecavity and prove that induced charge on the cavitywalls is -q. The reason is electric field E is zeroinside the material of the conductor. The Totalenclosed charge within the Gaussian surface is zero.Here the conductor is initially uncharged. Fromconservation of charge, we can say that on the outersurface of the conductor a charge +q will beinduced. At any point inside the material ofconductor, say at P, the electric field produced by+q in the cavity is cancelled by the field producedby charges induced on the walls of cavity and onthe outer surface of the conductor. If the point chargeis not at the centre of the spherical cavity, even theninduced charges on the cavity walls and on the outersurface of the conductor are -q and +q only.But the distribution of induced charges will changein such a way that at any point P in the material ofthe conductor resultant electric field is zero.Suppose the conductor has charge q0 on it initially.This charge resides on the outer surface of the

61 PINEGROVE


conductor. If point charge q is kept inside the cavity,induced charges on the walls of cavity and on theouter surface of the conductor are the same asbefore. i.e., – q and + q. But the total charge on theouter surface of the conductor is (q0 +q) now.If the charge inside the cavity is displaced, theinduced charge distribution on inner surface of thebody changes such that at any point inside thematerial of the conductor resultant field is zero. Inthis case the charge distribution on outer surface ofthe conductor does not change and only the chargedistribution on the cavity walls will change.Now the charge inside the cavity is fixed. If anothercharge is brought towards the conductor fromoutside., it will not affect the charge distributioninside the cavity and only the distribution of chargeon the outer surface will be affected.

MECHANICAL FORCE ON THE CHARGEDCONDUCTOR

We know that like charges repel each other. So,when a conductor is charged, the charge on anypoint of the conductor is repelled by the charge onits remaining part. It means surface of a chargedconductor experiences mechanical force.

1E

2E

P2

P1

1E

2E

ds

Consider a charged conductor as shown. Let dsbe the surface area of a small element on theconductor. The electric field at point P1 near theconductor surface can be considered as thesuperposition of fields 1E and 2E . Here 1E is thefield produced by that elemental surface and 2E isthe field due to the remaing surface of the conductor.

1 2E E E But we know that

0

E at P1 which is just outside

the conductor and is zero at P2 which is just insidethe conductor

So at P1 we have 1 20

E E

and at P2 we have E1 – E2 = 0

1 20

E E2

Now the force experienced by small surface ds dueto the charge on the rest of the surface is

2

2 20

dsF dq E ds E

2

and

22

00

Force F 1E

Area ds 2 2

ELECTRIC PRESSURE ON A CHARGED

SURFACEFrom the above derivation we observed that a smallsurface of a charged conductor will experience a

force by the remaining surface. The force per unit

area of the surface is 2

20

0

1E or

2 2

This is known as electric pressure on the chargedmetal surface.

2e 0

1P E

2 Suppose a charged body is in an external electricfield. Let us find out the electric pressure on thesurface of that charged body.Consider a surface uniformly charged with chargedensity . On that surface 'ds' is the surface areaof a small element. The charge on that element isdq ds

ds

E

The given surface is in an external electric fieldrepresented by the field lines as shown.Let E be the intensity of electric field on theelemental surface. Here angle betweenEandds

is . In this caseEhas two comperments.Component parallel to the surface is

11E E sin and component normal to the surface is

E E cos Here force due to E11 on the surface is tangentialwhich tries to stretch the surface. Where as the force

due to E applies outward pressure on the surface.Now outward force on the elemental surface is

dF (dq)E dsE So, the outwards electric pressure on the surface is

e edF

P E P E cosds


62 PINEGROVE

W.E-9: A thin spherical shell radius of r has acharge Q uniformly distributed on it. At thecentre of the shell, a negative point charge -qis placed. If the shell is cut into two identicalhemispheres, still equilibrium is maintained.Then find the relation between Q and q?

R

-q

+Q

Sol : Here the outward electric pressure at every pointon the shell due to its own charge is

22

1 20 0

1 QP

2 2 4 r

; 2

1 2 40

QP

32 r

Due to -q, the electric field on the surface of theshell is

20

1 qE

4 r

This electric field pulls every point of the shell ininward direction. The inward pressure on thesurface of the shell due to the negative charge is

2P E ; 2 20

Q 1 q

44 r r

2 40

Qq

16 r

For equlibrium of the hemispherical shells 2 1P P

or 2

2 4 2 40 0

Qq Q

16 r 32 r ;

Qq

2

WE.10 : If r and T are radius and surface tensionof a spherical soap bubble respectively thenfind the charge needed to double the radiusof bubble

Sol: For smaller bubble 3

1 0 14T 4

P P and V rr 3

For larger bubble

23

2 0 20

4T 4P P and V R

R 2 3 where 2

q

4 R

for air in the bubble, 1 1 2 2P V P V

2

3 30 0 2 4

0

4T 4T qP r P R

r R 16 R 2

2

3 3 2 20 2

0

qP R r 4T R r 0

32 R

But R = 2r

2

3 20 2

0

qP 7r 4T 3r 0

32 2r

2

3 202

0

q7P r 12Tr

64 r

2 2 30 0q 64 r 7P r 12T

12

0 0q 8 r r 7P r 12T

CONCEPTUAL QUESTIONS

ELECTRIC FLUX AND GAUSS LAW1. A cubical Gaussian surface encloses electric

flux of 30 C per unit permittivity of a charge.The electric flux through each face of the cubeper unit permittivity is1) 30 C 2) 15 C 3) 10 C 4) 5 C

2. As one penetrates uniformly chargedconducting sphere, what happens to the electricfield strength1) decreases inversely as the square of the distance2) decreases inversely as the distance3) becomes zero4) increases inversely as the square of distance

3. Mark the correct option1) Gauss law is valid only for unsymmetrical chargedistributions2) Gauss law is valid only for charge placed invacuum3) The electric field calculated by Gauss law is thefield due to the charges outside the Gaussiansurface.4) The flux of the electric field through a closedsurface due to all the charges is equal to the fluxdue to the charges enclosed by the surface

4. If the flux of the electric field through a closedsurface is zero1) The electric field must be zero every where onthe surface2) The electric field must not be zero everywhereon the surface3) The charge inside the surface must be zero4) The charge in the vicinity of the surface must bezero

5. An infinite plane sheet of a metal is chargedto charge density 2/C m in a medium ofdielectric constant K. Intensity of electric fieldnear the metallic surface will be

1) o

EK

2) 2 o

E 3) 2 o

EK

4) 2 o

KE

6. The electric flux from a cube of edge l is .Its value if edge of cube is made 2l and chargeenclosed is halved is1) / 2 2) 2 3) 4 4)

63 PINEGROVE


7. If the electric flux entering and leaving anenclosed surface respectively is 1 and 2 , theelectric charge inside the surface will be

1) 1 2 / o 2) 1 2 / o 3) 1 2 o 4) 2 1 o

8. Electric flux at a point in an electric field is1) positive 2) negative3) zero 4) positive or negtive

9. Electric flux over a surface in an electric fieldmay be1) positive 2) negative3) zero 4) positive, negative, zero

10. A charge Q is placed at the mouth of a conicalflask. The flux of the electric field through theflask is

1)zero 2) 0/Q 3)02

Q

4)02

Q

11. A charge Q is placed at the mouth of a conical

flask. At the centre of the circular crossectionflux of the electric field through it is

1) zero 2) 0/Q 3)02

Q

4)02

Q

12. Electric field intensity at a point due to an

infinite sheet of charge having surface chargedensity is E. If sheet were conductingelectric intensity would be1) E/2 2) E 3) 2 E 4) 4 E

13. Two thin infinite parallel sheets (nonconducting) have uniform surface densities ofcharge + and – . Electric field in the spacebetween the two sheets is1) 0/ 2) 0/ 2 3) 02 / 4) zero

14. In the above question, if the sheets were thickand conducting, value of E in the space betweenthe two sheets would be1) 02 / 2) 0/ 3) zero 4) 04 /

15. In the above problem the value of E in thespace outside the sheets is.

1) 0/ 2) 0/ 2 3) zero 4) 02 / 16. The Gaussian surface for calculating the

electric field due to a charge distribution is1) any closed surface around the charge distribution2) any surface near the charge distribution3) a spherical surface4) a closed surface at a every point of which electricfield has a normal component which is zero or afixed value

17. The electric flux over a sphere of radius 1m is . If radius of the sphere were doubled withoutchanging the charge enclosed, electric fluxwould become1) 2 2) / 2 3) / 4 4)

18. A charge q is placed at the centre of a cube.What is the electric flux associated with oneof the faces of cube

1)0

q 2) 0q

3)

0

6q 4)06

q 19. A charge Q is placed at the corner of a cube.

The electric flux through all the faces of thecube is

1)0

Q 2)06

Q 3) 08

Q 4)03

Q 20. A point charge +q is placed at mid point of a

cube of side ‘L’. The electric flux emergingfrom the cube is

1) 0

q 2)26qL 3) 2

06q

L 4)zero

21. A charge q is enclosed as shown below, theelectric flux is

q q q(ii) (iii) i

1) maximum in (i) 2) maximum in (ii)3) maximum in (iii) 4) equal in all

22. An ellipsoidal cavity is carved with in a perfectconductor. A positive charge q is placed at thecentre of the cavity. The points A and B areon the cavity surface as shown in the figurethena) Electric field near A in the cavity = Electric fieldnear B in the cavityb) Charge density at A = Charge density at Bc) Potential at A = Potential at Bd) Total electric flux through the surface of the cavity

is q/ 0 .

q B

A1) a,b,c,d are correct2) a,b,c are correct3) only a and b are correct4) only c and d are correct

23. A metallic shell has a point charge ' 'q keptinside its cavity. Which one of the followingdiagrams correctly represents electric lines offorces


64 PINEGROVE

1) 2)

3) 4)

24. Two infinitely long thin styraight wires havinguniform linear charge densities and 2 arearranged parallel to each other at a distance rapart . The intesity of the electric field at apoint midway between them is

1)

0

2r 2)

0r

3) 02 r 4)

0

32 r

25. Find the total flux due to charge q associatedwith the given hemispherical surface

q

a)

qc)

q

e)

q

b)

q

d)

qf)

1) 0 0 0

, 0, , 0, 02

q q qa b c d e f

2) 0 0 0

0, , 0, , 02

q q qa b c d e f

3) 0 0 0 0

, , 0, , 02

q q q qa b c d e f

4) 0 0 0 0

0, , 0, ,2

q q q qa b c d e f

26. A : A metallic sheild in the form of a hollowshell may be built to block an electric field.R : In a hollow spherical shield, the electricfield inside it is zero at every point.1) Both ‘A’ and ‘R’ are true and ‘R’ is the correctexplanation of ‘A’.2) Both ‘A’ and ‘R’ are true and ‘R’ is not thecorrect explanation of ‘A’.3) ‘A’ is true and ‘R’ is false4) ‘A’ is false and ‘R’ is true

27. A thin spherical shell of radius R has chargeQ spread uniformly over its surface. Whichof the following graphs most closely represents

the electric field E r produced by the shell

in the range 0 r , where r is the distancefrom the centre of the shell?

O R r

E (r)

O R r

E (r)

O R r

E (r)

O R r

E (r)

1) 2)

3) 4)

28. Three positive charges of equal value q areplaced at vertices of an equilateral triangle.The resulting lines of force should be sketchedas in (3)

1)

3)

2)

4)

29. An uncharged metal sphere is placed betweentwo equal and oppositely Charged metal plates.The nature of line of force will be

1) 2)

3)

4)

ASSERTION& REASON1) Both (A) and (R) are true and (R) is thecorrect explanation of (A)2) Both (A) and (R) are true and (R) is not thecorrect explanation of (A)3) (A) is true but (R) is false4) (A) is false but (R) is true

30. Assertion: A device used to measure E

islocated at some distance from a fixed pointcharge. In this situation, the device measures

65 PINEGROVE


0E as the magnitude of electric field intensity..Now an uncharged conducting sphere with avery small hole is lowered by an insulatingthread so that it surrounds the point charge.Now, the reading of the device becomes zero.Reason: Electrostatic shielding is thephenomenon in which inside of hollowconductor is shielded for outside electric field

31. Assertion: E in outside vicinity of aconductor depends only on the local chargedensity Reason: E

in outside vicinity of a conductor

is given by 0

32. Assertion: Four point charges 1 2 3, ,q q q and

4q are as shown in Fig. The flux over the shown

Gaussian surface depends only on charges 1q

and 2q

Reason: Electric field at all points on

Gaussian surface depends ony on charges 1q

and 2q Gaussian surface

q1

q2

q3

q4

33. Assertion: A point charge q is placed near anarbitray shaped solid conductor as shown infigure. The potential difference between thepoints A and B within the conductor remainsame irrespective of the magnitude of chargeq.Reason: The electric field inside a solidconductor is zero under electrostaticconditions.

A

B

34. Assertion: Two point charges +Q and -Q arefixed at point A(+a,0,0) and point B(-a,0,0)respectively. Then the magnitude of electricflux due to electric field of either point chargethrough infinite y-z plane (that is x=0 plane) isless than magnitude of net electric flux due to

electric field of both charges through that plane(x = 0 plane).Reason: The magnitude of net electric fluxthrough a surface due to a system of pointcharges is equal to sum of magnitude of electricflux through that surface due to each of thepoint charge of the system.

35. Assertion: In a region whereuniform electric field exists, the net chargewithin volume of any size is zero.Reason: The electric flux through any closedsurface in region of uniform electric field iszero.

36. Assertion: A point charge q is placed at centreof spherical cavity inside a spherical conductoras shown. Another point charge Q is placedoutside the conductor as shown in Fig. Now asthe point charge Q is pushed away fromconductor, the potential difference (VA – VB)between two points A and B within the cavityof sphere remains constantReason: The electric field due to charges onouter surface of conductor and outside of theconductor is zero at all points inside theconductor.

BAq Q

37. Assertion: The electrostatic force on a chargedparticle located on a equipotential surface iszeroReason: Componant of E along equipotentialsurface is zero.

38. Assertion: We cannot produce electric field ina neutral conductor.Reason: Neutral conductor cannot produceelectric field.

39. Two parallel metallic plates have surface

charge densities 1 and 2 as shown in figure.Match the following:

1 2


66 PINEGROVE

Column I Column II

a) If 1 2 0 (p) Electric field in region III is towards right

b) If 1 2 0 (q) Electric field in regionI is zero

c) If 1 2 0 (r) Electric field in regionI is towards right

1) a-p, b-q, c-r 2) a-q, b-p, c-r3) a-r b-q, c-r 4) a-p, b-r, c-r

40. Two spherical shells are as shown in figure.Suppose r is the distance of a point from theircommon centre. Then,

q1

q2

R2

R1

Column I Column IIa) Electric field for (p) is constant for

1r R 2q and vary for 1qb) Electric potential for (q) is zero for

1r R 2q and vary for 1qc) Electron potential for (r) is constant for both

1 2R r R 1 2q and q

d) Electric field for (s) is zero

1 2R r R 1) a-r, b-s, c-p, d-q 2) a-s, b-r, c-p, d-q3) a-s, b-s, c-p ,d-q 4) a-r, b-q, c-p ,d-q

C. U. Q - KEY1) 4 2) 3 3) 4 4) 3 5) 1 6) 17) 4 8) 3 9) 4 10) 3 11) 3 12) 313) 1 14) 1 15) 3 16) 4 17) 4 18) 419) 3 20) 1 21) 4 22) 4 23) 1 24) 225) 1 26) 1 27) 1 28) 3 29)2 30) 431) 2 32) 3 33) 1 34) 1 35) 2 36) 137) 4 38) 2 39) 2 40) 3

EXERCISE- I (C.W)

ELECTRIC FLUX AND GAUSS LAW1. A charged spherical conductor has a surface

charge density of 0.7 2/C m . When its chargeis increased by 0.44C, the charge densitychanges by 0.14 2/C m . The radius of thesphere is1) 5 cm 2) 1 0 m 3) 0.5 m 4) 5 m

2. The electric field in a region of space is given

by 15 2E i j NC . The electric flux due to

this field through an area 22m lying in the YZplane in S.I. units is

1) 10 2) 20 3) 10 2 4) 2 293. Number of electric lines of force emerging

from 1C of positive charge in vacuum is

1) 128.85 10 2) 99 103) 91/ 4 9 10 4) 111.13 10

4. A charge of 5 C is placed at the centre of aspherical gaussian surface of radius 5 cm. The

electric flux through the surface is0

1

times of

1) 0.1 N-m2/C 2) 0.5 N-m2/C3) 1 N-m2/C 4) 5 N-m2/C

5. In a region where intensity of electric field is15 NC , 40 lines of electric force are crossing

110 NC will be1) 20 2) 80 3) 100 4) 200

6. A half ring of radius R has a charge of perunit length. The potential at the centre of the

half ring is (1

4o

k )

1)kR

2)k

R

3)kR

4) k

7. An electron is placed at the centre of aConducting sphere of radius 0.2 metre havinga charge 25 10 coulomb. The force on theelectron is1) zero 2) 911 10 N3) 922.5 10 N 4) 92.5 10 N

8. Eightcharges,1 C ,-7 C ,-4 C , 10 C ,2 C ,-

5 C ,-3 C and 6 C are situated at the eightcorners of a cube of side 20 cm. A spheric alsurface of radius 80 cm encloses this cube. Thecenter of sphere coincides with center of thecube. Then the outgoing flux from the sphericalsurface(in units Vm) is1) 336 10 2) 3684 103) zero 4) 372 10

67 PINEGROVE


EXERCISE- I (C.W) KEY1) 3 2) 1 3) 4 4) 4 5) 2 6) 47) 1 8) 3

EXERCISE- I (C.W) HINTS

1.QA

2. . 5 2E E ds E s

4.0

E

q 5. q 6.

kqq r and V

r

7. 0,insideE F Eq EXERCISE- I (H.W)

1. Caluclate the net flux emerging from givenenclosed surface - Nm2 C–1

2C

3C5C

1) 4.5 × 1013 2) 45 × 1012

3) zero 4) 1.12 × 1012

2. A charge Q is situated at the centre of a cube.The electric flux through one of the faces ofthe cube is

1) 0Q / 2) 0Q / 2 3) 0Q / 4 4) 0Q /63. The magnitude of the electric field on the

surface of a sphere of radius r having a uniformsurface charge density is1) 0/ 2) 0/ 2 3) 0/ r 4) 0/ 2 r

4. If the electric flux entering and leaving anenclosed surface respectively is 1 2and , theelectric charge inside the surface will be

1) 2 1 0 2) 1 2 0/

3) 2 1 0/ 4) 1 2 0

5. A charge q is placed at the centre of the openend of cylinderical vessel. Find the flux of theelectric field through the surface of the vessel.

1) 0

q

2 2) 0

q3) 0

q

3 4) zero

EXERCISE- I (H.W) KEY1) 1 2) 4 3) 1 4) 1 5) 1

EXERCISE- I (H.W) HINTS

1.11

120

2 5 3. 4.5 10

8.85 10

qE ds

2.

0 0

.6

q qE ds through on of the face

3.2

0

4q

E r ; 20 04

qE

r

4. 2 1

0

.E ds

5.0

.q

E ds EXERCISE- II (C.W)

ELECTRIC FLUX AND GAUSS LAW1. The inward and outward electric flux for a

closed surface in units of N-m2/C arerespectively 38 10 and 34 10 . Then the totalcharge inside the surface in S.I units is (where

0 =permitivity constant )

1) 34 10 2) – 34 10 3) –2R RE

4 )

304 10

2. A cylinder of radius R and length L is placedin the uniform electric field E parallel to thecylinder axis. The total flux from the two flatsurfaces of the cylinder is given by

1) 22R E 2)2 R

E 3)

2 R R

E4) zero

3. A cube is arranged such that its length , breadth, height are along X,Y and Z directions . Oneof its corners is situated at the origin . Lengthof each side of the cube is 25cm . Thecomponents of electric field are

400 2 /xE N C , 0yE and 0zE respectively. The flux coming out of the cubeat one end will be1) 225 2 /Nm C 2) 25 2 /Nm C3) 2250 2 /Nm C 4) 225 /Nm C

4. If a hemispherical body is placed in a uniformelectric field E then the flux linked with thecurved surface is

B

n

E

1) 22 ER 2) 2ER

3) 24 ER 4) 26 ER


68 PINEGROVE

5. A thin conducting ring of radius R is given acharge +Q . The electric field at the centre Oof the ring due to the charge on the part AKBof the ring is E. The electric field at the centredue to the charge on the part ACDB of the

ring is

C

A

D

BO

K

1)3 E along OK 2)3E along KO

3)E along OK 4) E along KO6. In a unifrorm electric field find the total flux

associated with the given surfaces (R is radius)E

a)

Eb)

E

c)

1) 0, 0, 0a b c 2) 20, , 0R Ea b c 3) 22 , , 0R Ea RE b c 4) 2 , 0, 0a R E b c

7. Surface charge density of soap bubble ofradius ' 'r and surface tension T is . If P isexcess pressure, the value of is

1)

32

0

4TP

r 2)

1

2

0

42 TP

r

3) 4T

r 4)

0

122

4T

Pr

8. An infinitely long thin straight wire has

uniform linear charge density of 1/3 coul.m-1. Then the magnitude of the electricintensity at a point 18 cm away is

1) 1 10.33 10 NC 2) 1 13 10 NC 3) 1 10.66 10 NC 4) 1 11.32 10 NC

9. Consider two concentric spherical surface 1S

with radius a and 2S with radius 2a, bothcentred on the origin. There is a charge +q atthe origin, and no other charges.Compare the

flux 1 through 1S with the flux 2 through 2S

1) 1 24 2) 1 22

3) 1 2 4) 1 2 / 2 10. The electric field on two sides of a large

charged plate is shown in fig. The chargedensity on the plate in S.I. units is given by

( o is the permittivity of free space in S.I.units)

1) o2 2) o4 3) o10 4) zero

EXERCISE- II (C.W) KEY1) 4 2) 4 3) 1 4) 2 5) 3 6) 17) 2 8) 1 9) 3 10) 2

EXERCISE- II (C.W) HINTS

1.0

E

q 3. 1 1 2 2 3 3. . .E A E A E A 4. .E A 5. 0 0E

9. Flux through both will be same as net chargeenclosed by both is same

10. Electric field due to plate= o2

EXERCISE- II (H.W)1. The number of electric lines of force originating

from a charge of 1C is1) 1.129×1011 2) zero3) 1.129×10–11 4) 1.129×1010

2. A cube of side l is placed in a uniform field E,where E = Ei . The net electric flux throughthe cube is1) Zero 2) l2E 3) 4l2E 4) 6l2E

3. A point charge +q is placed at the centre of acube of side L. The electric flux emerging fromthe cube is

1) 0

q2) Zero 3)

2

0

6qL4) 2

0

q

6L4 A long thin flat sheet has a uniform surface

charge density . The magnitude of theelectric field at a distance ‘ r ‘ from it is givenby

1) 0/ 2) 0/ 2 3) 0/ r 4) 0/ 2 r

5 A charge of 8.85C is placed at the centre of aspherical Guassian surface of radius 5 cm. Theelectric flux through the surface is1) 1012 V/m 2) 10–12 V/m3) 108 V/m 4) 1010 V/m

69 PINEGROVE


6. The inward and outward electric flux for aclosed surface in units ofN–m2/C arerespectively 8 × 103 and 4 × 103. Then the totalcharge inside the surface in S.I. units is (where

o = permitivity in free space )1) 4 × 103 2) – 4 × 103

3)

3

0

4 104) 3

o4 10

7. The total flux linked with unit negative chargeput in air is

1) 0

1out wards 2) 0

1inwards

3) 0

1

4outwards 4) 0

1

4inwards

EXERCISE- II (H.W) KEY

1) 4 2) 4 3) 1 4) 2 5) 4 6) 4 7) 1

EXERCISE- II (H.W) HINTS

1. No of lines 11

120

11.129 10

8.85 10

q

2. In ward equat to outward flux so net flux is zero

3. Total flux 0

q

4.02

E

5.12

12

8.8510

8.85 10E

6. Net flux 34 10 3

0

4 10q

3

04 10q 7. Due to -ve charge flux is inward

EXERCISE- IIIELECTRIC FLUX AND GAUSS LAW

1. A solid metallic sphere has a charge +3Q.Concentric with this sphere is a conductingspherical shell having charge +Q. The radiusof the sphere is a and that of the spherical shellis b, (b>a). What is the electric field at adistance R (a<R<b) from the centre.

1) 02

Q

R 2) 0

3

2

Q

R 3) 0

2

3

4

Q

R 4) 0

2

4

2

Q

R2. Two parallel plane sheets 1 and 2 carry uniform

charge densities 1 2and as shown in fig.electric field in the region marked II is

1 2

+++++++++++

+++++++++++

IIIIII

Sheet 1 Sheet 2

1 2

1) 1 2

02

2)

1 2

02

3) 1 2

02

4) 1 2

02

3. Electric charge is uniformly distributed alonga long straight wire of radius 1 mm. The chargeper cm length of the wires is Q coulomb.Another cylindrical surface of radius 50 cm andlength 1 m symmetrically encloses the wire asshown in the figure. The total electric flux

passing through the cylindrical surface is

+++

50cm

1m

+++

1) 0

Q

2) 0

100Q

3) 0

10Q

4) 0

100Q

4. A charge Q is distributed uniformly on a ring

of radius r. A sphere of equal radius r isconstructed with its centre at the periphery ofthe ring as shown in figure. Find the flux ofthe electric field through the surface of thesphere.

A

01

0

B

Ring

Sphere1)

03

Q

2) 0

q

3)

02

q

4) zero

5. 21 3 4, ,q andq q q are point charges located at

points as shown in the figure as S is a sphericalGaussian surface of radius R. Which of thefollowing is true according to the Gauss’s law

R

q4S

q2q1

q3


70 PINEGROVE

1) 1 2 321 3

0

.2

q q qE E E d A

2) 1 2 3421 3

0

.q q q

E E E E d A

3) 1 2 3 4421 3

0

.q q q q

E E E E d A

4) None of the above

6. Shown below is a distribution of charges. Theflux of electric field due to these chargesthrough the surface S is

+q

+q

+q

1) 03 /q 2) 02 /q 3) 0/q 4) Zero

7. A thin spherical conducting shell of radius Rhas a charge q. Another charge Q is placed atthe centre of the shell. The electrostaticpotential at a point P at a distance R/2 fromthe centre of the shell is

1) 0

2

4

Q

R 2) 0 0

2 2

4 4

Q q

R R

3) 0 0

2

4 4

Q q

R R 4)

0

2

4

q Q

R

8. A charge ' 'q is distributed over two concentrichollow conducting spheres of radii a and b (b >a) such that their surface charge densites areequal. The potential at their common centre is

1) Zero 2) 202 24

a bq

a b

3) 0

1 1

4

q

a b 4) 0

2 24

q a b

a b

9. Two concentric sphere of radii 1 2a and a carryy

charges 1 2q and q respectively. If the surface

charge density is same for both spheres,the electric potential at the common centre willbe

1) 1

0 2

a

a

2)

2

0 1

a

a

3) 1 20

a a 4) 1 2

0

a a

10. Assume three concentric conducting sphereswhere charge 1 2q and q have been placed oninner and outer sphere where as middle spherehas been earthed. Find the charge on the outersurface of middle spherical conductor

r1

r2

r3

+q1 +q2

O1) 2

23

rq

r 2) 1q

3) 2q 4)

2

13

rq

r11. Three concentric metallic spheres A, B and C

have radii a,b and c (a < b < c ) and surfacecharge densities on them are , and respectively. The values of AV and BV will be

cb

a

A

B

C

–1)

0 0

,a

a b c b cb

2) 2

,a

a b cc

3) 0 0,

aa b c b c

b

4) 2 2

0

,a b

cc c

0

a b c

12. A charged ball hangs from silk thread whichmakes an angle ' ' with large chargedconducting sheet ' 'P as shown. The surface

charge density of the sheet is proportionalto

+

1) cos 2) cot3) sin 4) tan

COMPREHENSIONThe electric field in a region is given by

E x i . Here is is a constant of properdimensions.

71 PINEGROVE


Y

ZD H

GC

X

B F

EA

13. Find the total flux passing through a cubebounded by surfaces

, 2 , 0, , 0, .x l x l y y l z z l 1) 3l 2) 32 l 3) 33 l 4) 34 l

14. The charge contained inside the above cube is

1) 032 l 2) 0

3l3) 0

34 l 4) 033 l

15. Two point charges q and q are seperatedby a distance 2a. Find the flux of the electricfield vector acrossthe circle of radius r isshown.

–q+qr

a a

1) 2 20

12

q a

a r

2) 2 20

1q a

a r

3) 2 2

0

21

q a

a r

4) Zero

16. A long string with a charge of per unit lengthpasses through an imaginary cube of edge a.The maximum flux of the electric field throughthe cube will be

1) 0/a 2) 02 /a 3) 2

06 /a 4) 03 /a 17. A rod with linear charge density is bent in

the shape of circular ring. The electric potentialat the centre of the circular ring is

1)04

2)

02

3)

0

4)

0

2

18. The electric field components in the figure are1/ 2, 0, 0x y zE x E E where = 800 N /

m2. If a = 0.1 m is the side of cube then thecharge within the cube is

z a

ax

a

y

a

1) 129.27 10 C 2) 126 10 C3) 122.5 10 C 4) Zero

19. Three very large plates are given charges asshown in the figure. If the cross-sectional areaof each plate is the same, the finalcharge distribution on plate C is

Q –5Q 10Q

A B C

Outer

Innera) +5Q on the inner surface , +5Q on the outersurfaceb) +6Q on the inner surface, +4Q on the outersurfacec) +7Q on the inner surface, +3Q on the outersurfaced) +8Q on the inner surface, +2Q on the outersurface

20. An electric dipole of dipole moment P is keptat a distance r from an infinite long chargedwire of linear charge density as shown. Theforce acting on the dipole is

1) 202

P

r

2) 2

0

P

r

3) 2

0

2P

r

4) 2

04

P

r

21. A point charge q is a distance r from the centreO of an uncharged spherical conducting layer,whose inner and outer radii equal to a and brespectively. The potential at the point

0

qO if r a is times

4

1) 1 1 1

r a b 2)

1 1 1

a r b

3) 1 1 1

b c r 4)

1 1 1

a b r

22. One-fourth of a sphere of radius R is removedas shown in fig. An electric field E existsparallel to x-y plane. Find the flux through theremaining curved part.1) 2R E 2) 22 R E


72 PINEGROVE

3) 2R E / 2 4) 22 R EEXERCISE - III - KEY

1) 3 2) 4 3) 2 4) 1 5) 2 6) 27) 3 8) 4 9) 4 10) 1 11) 1 12) 413) 1 14) 2 15) 4 16) 4 17) 2 18) 119)3 20) 1 21) 3 22) 3

EXERCISE- III - HINTS

2.1 2netE E E

1 2

0 02 2

1 2

02netE

3. The total flux passing through cylindrical surface

4. From the geometry of the figure. 1OA OO and

1 1O A O O . Thus, 1OAO is equilateral triangle.

Hence 01 60AOO or 0120AOB .

The are 1AO B of the ring subtends an angle 0120at the centre O. Thus, third of the ring is inside thesphere.

The charge enclosed by the sphere 3

Q . From

Gauss’s law, the flux of the electric field through the

surface of the sphere is 03

Q

.

5. 1 2 3 40

.q

E E E E ds 1 2 3

3

q q q 6. The

1 0

2q q qds

0 0

/ 100Q cm Q 8. 1 2

2 24 4

q q

a b

21

22

q a

q b , 1 2q q q

2

1 1 2

bq q q

a

2

1 21

bq q

a

;

2 2

1 2

a bq q

a

2

1 2 2

qaq

a b ;

2

2 2 2

qbq

a b

potential at commoncentre

2 2 2 20 04 4

q a qbV

a b a b 2 2 2 2

04

q a bV

a b a b

2 204

q a bV

a b

10.1 1 2

2 2 3

0q q q q

r r r

2

23

rq q

r

11.2 2 2

0

1 4 4 4

4A

a b cV

a b c

0AV a b c

and

2 2 2

0

1 4 4 4

4B

a b cV

b b c

2

0B

aV b c

b

17.

QA

18. Magnitude of E at the left face

12

LE a at right

face 12

2RE a 2

R LE E a and 0q 22. 1 2E. A A

*******

73 PINEGROVE


Electric Capacity: The ratio of charge to potential of a conductor is

called its capacity. Q

CV

Unit : farad (F)Parallel Plate Capacitor: If two plates eachof area A are seperated by a distance 'd' then its

capacity 0ACd

(air as medium),

0k AC

d

(dielectric medium)

When a dielectric medium is introduced betweenthe plates of a parallel plate capacitor, its capacityincreases to 'k' times the original capacity.

When a dielectric slab of thickness 't' is introducedbetween the plates of a parallel plate capacitor,

+ + + ++ +

- - - -- -

air

d t

0 0

11

A Anew capacity

td td t

kk

GAUSS METHODLet us consider a case of parallel plate capacitor inwhich a medium of dielectric constant K is partiallyfilled as shown in figure.Then the field is uniform in air as well as in mediumbut they will have different values. let 't' be thethickness of the medium whose relative permittivityis K. The remaining space of (d - t) thickness beoccupied by air.

x yq

airx1 y1

d

K

Fig. (b)

Imagine a Gaussian surface enclosing the plate asshown.

+ + + + + ++ + + + +

x y

x1 y1

Fig. (c)If E0 is the field in air, then from Gauss law

0 0

0 0

q qE ds E A or

00

qE

A .... (a)

Similarly by considering a Gaussian surface throughthe medium, then by Gauss law,

0 0

q qE.ds EA

K K

where E is a field in the medium

0

qE

A K .... (b)

The P.D. between the two plates of the capacitor.

0V E d t E.t

0 0

q qV= d t t

A A K

0

q td t

A K

or 0

q qC

qV d t t / KA

0AC

td t

K

When a metal slab of thickness 't' is introducedbetween the plates of a parallel plate capacitor, new

capacity oA

d t

.

( for metal k = ) The method for the calculation of capacitance

requires integration of the electric field between two

ELECTRIC CAPACITANCE


74 PINEGROVE

conductors or the plates which are separated witha potential difference Vab

i.e. a

abb

V E.dr or V V E.dr from this

ab

qC

V

When a thin metal sheet 0t is introduced

between the plates of a parallel plate capacitor, thencapacity remains unchanged.

A dielectric slab of thickness 't' is introducedbetween the plates, to restore the original capacity,if the distance between the plates is

increased by x, then 1

1x tk

. Two dielectric slabs of equal thickness are

introduced between the plates of a capacitor asshown in figure, then new capacity

1 22

CK K .

A/2

dK2K1

A/2

If the two dielectrics are of different face areas 1A

and 2A but of same thickness, then capacity,,

01 1 2 2C K A K A

d

If two dielectric slabs of constants 1k and 2k are

introduced as shown in figure, new capacity

1 2

1 2

2.

k kC

k k

2

d

2

d

K1

K2

If number of dielectric slabs of same cross sectionalarea ‘A’ and of thicknesses 1 2 3, , ,........nt t t t and

constants 1 2, ....... nk k k are introduced between theplates, effective capacity

0

11 2

1

..... ...... nn

n

AC

ttd t t t

k k

In the above case if the dielectric media are

completely filled between the plates, effective

capacity 0

1

1

..... n

n

AC

tt

k k

The capacity of a parallel plate capacitor is

independent of the charge on it, potential differencebetween the plates and the nature of plate material.

In a capacitor, the energy is stored in the electricfield between the two plates.

Capacity of a spherical conductor = 04 r , wherer is the radius of the sphere.

If we imagine earth to be a uniform solid sphere

then capacity of earth is 04 R Where R = Radius of the earth = 36400 10mNote : For the earth, 6R 6.4 10 m

The capacity of earth is

60 9

1C 4 R 6.4 10 711 F

9 10

W.E-1: A metal slab of thickness, equal to half the

distance between the plates is introduced be-tween the plates of a parallel plate capacitoras shown. Find its capacity.

air

d/2

q

d

Sol: When capacitor is partially filled with dielectric

capacity 0

(1 )1

AC

d tk

For metal slab of thickness t = d/2,

0 AC

d t

(K for metal slab)

0 0A A

2 .d dd2

75 PINEGROVE


W.E-2: Two conductors carrying equal andopposite charges produce a non uniformelectric field along X - axis given by

2

0

QE (1 Bx )

Awhere A and B are

constants. Separation between the conductorsalong X-axis is X. Find the capacitance ofthe capacitor formed.

Sol: Potential difference between the conductors is

given by X

0

V V V Edx X

2

00

QV (1 Bx )dx

A or

X3

0 O

Q BxV x

A 3

3

0

Q BXX

A 3

Capacity 0

2

AQC

V BXX 1

3

W.E-3: Find the capacitance of a system of two

identical metal balls of radius a if the dis-tance between their centres is equal to b, withb>>a. The system is located in a uniform di-electric with permittivity K.

Sol: Let q and -q be the charges on two balls. Then

1 ballV V V V 2 ballV V V V

The potential difference between the balls

b

r

a a

q -q

1 2V V 2V b a

a

2 E dr

b a

20 0a

1 q 2q 1 12 dr

4 K 4 K a b ar

1 2

0

q qC

V V b 2a2q4 K a b a

02 K a(b a)

(b 2a)

For b >> a, we can write 0C 2 K a .

W.E-4: Capacitor has square plates each of side‘l’ making an angle ' ' with each other asshown. Then for small value of , the capaci-tance ‘C’ is given by

l

l

d

Sol: At one side, distance between plates d,At another side,distance d sin d l lMean distance between the plates

d dd

2 2

l l

Capacity 2

0 0AC

d d2

ll

12 2

0 01 1d 2d d 2d

l ll l

Spherical condenser

V = 0

10

4p q

q qV V

a b

0

1

4

b aq

ab

qC

V

b

-Q

+Q

Pa

q

(a) 04ab

Cb a

, if inner sphere is charged

and outer sphere is earthed.

ba

(b) 2

04b

Cb a

,

If inner sphere is earthed and outer sphere ischarged.


76 PINEGROVE

Cylindrical Capacitor: A cylindrical capacitorconsists of two coaxial cylinders and its capacitanceis given by

l

O

02

log e

lc

b

a

Where l is the length of each of cylinder and a andb are the radii of the inner and outer cylinders.

Force between the plates of a capacitorConsider a parallel plate capacitor with plate areaA. Let Q and –Q be the charges on the plates ofcapacitor. Let F be the force of attraction betweenthe plates. Let E be the field between the capacitorplates.The expression for the force can be derivedby energy method. Let the distance between theplates be x.So electric field energy between the plates is

20

1U E (Ax)

2

20

dU 1E A

dx 2

Xdx

F

By definition 20

dU 1F E A

dx 2

(Conservative force)

So the force of attraction between the plates is F =

20

1E A

2

Note : For an isolated charged capacitor 2

0

QF

2 A .

This force does not depend on the separationbetween the plates, and so the constant amount offorce is needed to change the separation.

Note : For a capacitor having constant potential

difference across the plates the force2 2

0

C VF

2 A

2 2 20

20

A V

2 Ad

2

0 2

1 VF A

2 d

In this case force depends on the separationbetween the plates. Thus to change the separationvariable force is needed.

CAPACITORS IN SERIESIn series combination, the capacitors are firstarranged in a series order such that the secondplate of first capacitor is connected to the firstplate of second capacitor, the second plate ofsecond capacitor is connected to first plate ofthird capacitor and so on. And finally the firstplate of first capacitor and second plate of lastcapacitor are connected to opposite terminalsof battery.Let us consider three capacitors of capacities C1,C2 and C3 connected in series across a source ofpotential difference 'V' as shown in figure.

V1

+q -q

C1

V2

+q -q

C2

V3

+q -q

C3

V

At the moment, the system is connected to thesource, left plate of first condenser acquires positivecharge due to conduction. This inturn will producenegative charge of equal magnitude, on the left faceof second plate of first condenser due to induction.The process continues for the remaining twocondensers. Hence the charge acquired by all thethree capacitors will be same.As the capacitors are different, the potentialsdeveloped across them will be different.

1 1 2 2 3 3q C V C V C V 1 2 3

1 2 3

q q qV ,V ,V

C C C

But 1 2 3V V V V

1 2 3

1 1 1V q

C C C

.... (1)

If a single capacitor when connected across the

77 PINEGROVE


same source draws the same charge, thatcapacitance is said to be the equivalent capacitanceof the three capacitors. If CS is the equivalentcapacitance.

Sq

CV

S

qV

C -----(2)

Substituting (2) in (1)

S 1 2 3

q q q q

C C C C

1 2 3s

1 1 1 1

C C C C

In generalS n

1 1

C C

The resultant capacity of series combination issmaller than the least capacity of the capacitors ofthe combination.

In series, ratio of charges on three capacitors is 1 :1 :1.

The ratio of potential differences across threecapacitors is

1 2 31 2 3 1 2 3

Q Q Q 1 1 1V : V : V : : : :

C C C C C C P.D across first capacitor is

11

1 2 3

1C

V V1 1 1C C C

similary we can find V2 and V3.

W.E-5: The equivalent capacity between A and Bin the given circuit is

8 F

8 F

A

12 F 12 FB

Sol: Here 12 F and 12 F are short circuited. Hencethey are not charged. Take only 8 F and 8 F parallel combination.C 8 8 16 F

W.E-6: When the space between the plates of aparallel plate condenser is completely filledwith two slabs of dielectric constants K1 andK2 and each slab having area A and thickness

equal to d

2 as shown in the figure

K1

K2

Fig. The equivalent circuit is as shown

A

A

K1

K2

2

d

2

d

a) Capacity of the upper half 1 o1

2K AC

d

b) Capacity of the lower half 2 o

22K A

Cd

c) C1 and C2 may be supposed to be conencted in

series.d) Effective capacity

1 2

1 2

C CC

C C 0 1 2 1 2

01 2 1 2

A 2K K 2K KC

d K K K K

Here C0 is the capacity of the condenser with airmedium.

e) Effective dielectric constant K= 1 2

1 2

2K K

K K

Capacitors in parallelCapacitors are said to be connected in parallelif the two plates of any capacitor are connectedone to positive terminal and the other tonegative terminal of the source, then theconnection is said to be parallel connection.

C1 C2 C3

Vq1 q2 q3

Fig.

Let us consider three capacitors of capacities C1,C2 and C3 connected in parallel across a source‘V’ as shownThe moment capacitors are connected, charge isdrawn from the voltage source and this charge isdrawn along three branches and thus gets shared.


78 PINEGROVE

As all capacitors are connected in parallel, thepotential across any of the capacitors is same . Herecharge gets shared depending upon theircapacitances for maintaining same potential.

31 2

1 2 3

qq qV

C C C 1 2 3 1 2 3q q q C V C V C V 1 2 3q V C C C

1 2 3q

C C CV ... (1)

If a single capacitor when connected to the samesource draws a charge q then that capacitor is saidto be the effective or equivalent capacitor for thethree parallel capacitors.If the effective capactiance is Cp,

Pq

CV

... (2)

from (1) and (2)

P 1 2 3C C C C In general P nC C s

The resultant capacity of parallel combination isgreater than the largest capacity of the capacitorsof the combination.

In parallel, ratio of P.D. on three capacitors is 1 : 1:1.

The ratio of charges on three capacitors is 1 2 3 1 2 3 1 2 3Q :Q :Q C V:C V:C V C :C :C

The charge on first capacitor is

11

1 2 3

CQ Q

C C C

similarly we can find Q2 and Q3. When n identical capacitors each of capacity Care first connected in series and next connected inparallel then the ratio of their effective capacities

s pC

C ; C nCn

2s

p

Cn :1

C

W.E-7 : In the net work three identical capacitorsare connected as shown. Each of them canwithstand to a maximum 100 V potentialdifference. What is the maximum voltage thatcan be applied across A and B so that nocapacitor gets spoiled.

A B

C

C

C

Sol: Let maxq be the max-charge supplied by the batterybetween A and B so that no capacitor gets spoiled.For each capacitor

max 0q CV C(100) 100C For the combination max equivalent maxq C V

max max2

100C C V V 150 V3

Among 150V, potential difference across parallelcombination is 50V and the potential differenceacross the other capacitor is 100V.

W.E-8: Calculate the capacitance of a parallel platecapacitor, with plate area A and distancebetween the plates d, when filled with a dielecticwhose permittivity varies as

0( ) 02d

x kx x ; 0( ) ( )

2d

x k d x x d

Sol:

X=0 X=d

dx

X

The given capacitor is equivalent to two capacitorsin series. Let C1 and C2 be their capacities. Then

1 2

1 1 l

C dC dC

Consider an element of width dx at a distance xfrom the left plate. Then 0

1

kx AdC

dx

for d0 x

2

and 0

2

k(d x) AdC

dx

for d

x d2

on substituting these two values we get

0

0

2 Kd1 l 2n

C dC KA 2

0

0

2 KdKAC n

2 2

W.E-9: When the space between the plates of aparallel plate condenser is completely filledwith two slabs of dielectric constants K1 and

K2 and each slab having area A

2 and thickness

equal to distance of seperation d as shown inthe figure.

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K1 K2

Fig. The equivalent circuit is as shown

K1

2

A

d K1 d

2

A

a) Capacity of the left half C1= K10A

2d

b) Capacity of the right half C2 = K20A

2d

c) C1 and C2 may be supposed to be connected in

parallel then effective capacity

C = C1 + C2 0 1 2A K K

d 2

C = C0

1 2K K

2

where C0 is capacity of

capacitor without dielectric.

d) Effective dielectric constant 1 2K KK

2

W.E-10: A parallel plate capacitor of area A, plate

separation d and capacitance C is filled withthree different dielectric materials havingdielectric constants K1, K2 and K3 as shownin fig. If a single dielectric material is to beused to have the same effective capacitance asthe above combination then its dielectricconstant K is given by :

Sol: Let 0 AC

d

; 0

1 1 1

A2C K K C

d2

0

2 2 2

A2C K K C

d2

;3 o

3 3K A

C , 2K Cd2

A/2

K1 K2

A

A/2

K3

d

The equivalent circuit as shown

2

A

2

d

2

d

2

d

2

A

A

K3

K2K1

3 1 2

1 1 1

C C C C , 3 1 2

1 1 1

KC 2K C K K C

1 2 3

1 1 1

K K K 2K

W.E-11: Four identical metal plates are locatedin air at equal distance d from one another.The area of each plate is A. Find the equivalentcapacitance of the system between X and Y.

1

2

3

4

X

Y

Sol: Let us give numbers to the four plates. Here X andY are connected to the positive and negativeterminals of the battery (say), then the chargedistribution will be as shown

1

2

3

4

X

Y

Here the arrangement can be represented as thegrouping of three identical capacitors each of

capacity 0 A

d

. The arrangement will be as shown

X( ) ( )

3 42 1

2 3

Y

Now the equivalent capacitance between X and Y

is 0XY

2 A(C C)C 2CC

C C C 3 3d

W.E-12: Find equivalent capacity between X and

Y 1

2

3

4

Y X

Sol: Let us give numbers to the four plates. Here X andY are connected to the positive and negativeterminals of the battery (say) ,


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1

2

3

4

Y X

Here the arrangement can be represented as thegrouping of two identical capacitors each of capacity

0 A

d


1 2

4 2

X Y

Now the equivalent capacitance between X and Yis

0

XYA

C (C C) 2C 2d

W.E-13: Find equivalent capacity X and Y1

2

34

X

Y

Sol: Let us give numbers to the four plates. Here X andY are connected to the positive and negativeterminals of the battery (say).

12

34

X

Y

Here the arrangement can be represented as thegrouping of three identical capacitors each of

capacity 0 A

d


2 1 3 4

2 3X Y

Now the equivalent capacitance between X and Y

is

XY

C C C 3C C C C

C C 2 2

= 0

XYA3 3

C C2 2 d

Types of Dielectrics :

A dielectric is an insulating material in whichelectrons are tightly bound to the nuclei of the atoms.Ex: glass, mica, paper etc. There are two types of dielectrics1) Non-polar dielectrics2) Polar dielectrics In non polar dielectrics the centre of positive chargeand centre of negative charge of each moleculecoincide Under ordinary conditions Non-polar molecule will

have zero dipole moment. When a Non-polar dielectric is subjected toelectric field, the positive charge of each moleculeis shifted in the direction of electric filed andnegative charge in the opposite direction.Ex: oxygen, nitrogen In polar dielectrics the centre of positive chargesand centre of negative charges of each moleculedo not coincide. Each molecule has a permanent dipole moment. When polar dielectric is subjected to externalelectric field, the electric field exerts torque on thedipoles, tending to align them in the direction ofthe field.Ex: CO

2, NH

3,HCl, etc. If a dielectric is charged by induction then induced

charge 1q is less than inducing charge q. Inducedcharge,

1 1

1q qK

where K is dielectric constant. Electric field due to induced charges on the

dielectric is 00 0

11ind p

EE or E E E

K k .

Dielectric Strength of Air : A conductingsphere cannot hold very large quantity of charge.It can hold a maximum charge Q such that theelectric intensity on the surface is equal to dielectric

strength of air 6 13 10 Vm

i.e. 6 1

20

13 10

4

QVm

R Energy stored in a condenser : Energy storedin a charged condenser

221 1

2 2 2

qU C V qV

C

If a condenser is connected across a battery andU is the energy stored in the condenser then thework done by the battery in charging thecondenser is 2U ( W = qV = 2U )For a parallel plate capacitor

2

0 0

1( )

2U A d as E

Energy density

22

00

1

2

UE

V

( here V is volume i.e. Ad)

a) When three capacitors are in series, the ratioof energies is

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2 2 2

1 2 31 2 3 1 2 3

Q Q Q 1 1 1U : U : U : : : :

2C 2C 2C C C C

b) When three capacitors are in parallel, theratio of energies is

2 2 21 2 3 1 2 3 1 2 3

1 1 1U : U : U V : V : V C : C : C

2 2 2C C C

c) Energy density ( ) = energy/ volume

2 20

1 1E k E

2 2 (Where K in the dielectric constant of

medium between the plates)

Effect of Dielectric: A parallel plate capacitor is fully charged to a

potential V. Without disconnecting the batteryif the gap between the plates is completely filled bya dielectric medium,capacity increases to k timesthe original capacity. P.D. between the plates remains same. Charge on the plates increases to k times the originalcharge. Energy stored in the capacitor increases to k timesthe original energy. After disconnecting the battery if the gapbetween the plates of the capacitor is filled by adielectric medium,capacity increases to k times theoriginal capacity.

P.D. between the plates decreases to 1

k times the

original potential. Charge on the plates remains same.

Energy stored in the capacitor decreases to 1

ktimes

the original energy. A capacitor is fully charged to a potential 'v'. After

disconnecting the battery, the distance between theplates of capacitors is increased by means ofinsulating handles. Potential difference between the

plates increases. (Q

VC

, Q remains same, and C

decreases) A capacitor with a dielectric is fully charged. Without

disconnecting the battery if the dielectric slab isremoved, then some charge flows back to thebattery.

Mixed Grouping of Capacitors: Number of capacitors in a row

desired potentialn

given potential

Number of such rows desired capacity

m noriginal capacity

Total number of capacitors = m n .

Coalesence of a Charged Oil Drops:There are ‘n’ charged drops of radius ‘r’ and charge‘q’. The drops are merge to form a bigger drop. Ifcapicity of small drop is ‘C’ then

1). capacity of bigger drop is 1

1 3C n C 2) Potential of bigger drop is

11

2/32/3

1/3VQ nq n q

n V.C n .C C

3) Energy of bigger drop is

1

5/3 25/3

1/3

2 2 2

UQ n q n q

n U.2C 2n .C 2C

4) Surface charge density of bigger drop is

1

1/31/3

2 2 /3 2 2Q n q n q

n .4 R 4 n .r 4 r

S.No. Quantity For each charged For the big small drop drop

1. Radius r R=n1/3 r 2. Charge q Q = n × q 3. Capacity C C1 = n1/3 × C 4. Potential V V1 = n2/3 × V 5. Energy U U1 = n5/3 U

6. Surface charge 1/ 3n . INTRODUCTION OF DIELECTRIC IN A

CHARGED CAPACITORA dielectric slab (K) is introduced between theplates of the capacitor

Sno Physical With battery with battery quantity permanently connected disconnected

1 Capacity K times increases K times increases

2 charge K times increases Remains constant

3 P.D Remains constant K times decreases

4 Electric Intensity Remains constant K times decrease

5 Energy stored in K times increases K times decrease

condenser

The distance between the plates of condenser isincreased by n times.

Sno Physical With battery with battery quantity permanently connected disconnected

1 Capacity n times decreases n times decreases


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2 charge n times decreases Remains constant

3 P.D Remains constant n times increases

4 Electric Intensity n times decrease Remainsconstant 5 Energy stored in n times decreases n times increases

condenserREDISTRIBUTION OF CHARGE, COMMON

POTENTIAL AND LOSS OF ENERGYTwo capacitors of capacities C1 and C2 are chargedto potentials V1 and V2 separately and they areconnect so that charge flows. Here charge flowsfrom higher potential to lower potential till bothcapacitors get the same potential

a) Two capacitors are connected in parallel such thatpositive plate of one capacitor is connected topositive plate of other capacitor

Let V be the common potentialThen Q = Q1 + Q2 (charge conservation)

(C1 + C2) V = C1V1 + C2 V2 ; 1 1 2 2

1 2

C V C VV

C C

In this case there will be loss in energy of the system

f iU U U ; where 2 2f 1 2

1 1U C V C V

2 2

2 2i 1 1 2 2

1 1U C V C V

2 2 ;

21 21 2

1 2

1 C CU (V V )

2 C C

b) If positive plate of one capacitor is connected tonegative plate of other capacitor, common potentialis given by

1 1 2 2

1 2

C V ~ C VV

C C

Here charge flow takes place if 1 2V VIn this case, the loss of energy

21 21 2

1 2

1 C CU (V V )

2 C C Charge transfered is 1 1

1 1 2 2q q or q q 1 1 1 2 2 2( )CV CV or C V C V 1 1 2 2C V V or C V V

WE.14: A capacitor of capacitance Co is chargedto a potential Vo and then isolated. A smallcapacitor C is then charged from Co,discharged and charged again, the processbeing repeated n times. Due to this, potentialof the large capacitor is decreased to V. Findthe capacitance of the small capacitor:

Sol: When key is closed, common potential o o1

o

C VV

C C

charge left on large capacitor after first sharing ofcharges 1

o o 1Q C Vcommon potential after second sharing of charges

in 0

2 10

CV V

C C ; =

( + )

2o 0

2 2o

C VV

C C

after nth sharing charges =+

no

n 0o

CV V

C C

But nV V ;

n0

00

CV V

C C

; 1/ n

oo

VC C 1

V

W.E-15: In the circuit shown in figure 1 1C F

and 2 2C F . The capacitor C1 is charged to100V and the capacitor C2 is charged to 20V.After charging then are connected as shown.When the switches 1 2 3,S S and S are closed, thecharge flowing through 1S is

V =1001 V =20V2

S1 S3 S2C1 C2

+ - + -

Sol: When 1 2 3S ,S and S are closed, both the capacitorsare in parallel with unlike charged plates together.So, they attain a common potential.Before closing the switches,Charge on 1C is 1q 100 1 100 C Charge on 2C is 2q 20 2 40 C After closing the switches

Common potential 1 2

1 2

q q 100 40V 20V

C C 3

Now final charges1

1 1q C V 1 20 20 C 1

2 2q C V 2 20 40 C The charge that flows through 1S is

q 100 20 80 C Application :

a) Redistribution of charges when two con-ductors are connected by conduting wireIn charging a conductor, work is required to bedone. This work done is stored up as the potentialenergy of the conductor.Energy of a charged conductor,

221 1 Q

U CV QV2 2 2C

When two charged bodies are connected by aconducting wire then charge flows from a conductorat higher potential to that at lower potential untiltheir potentials are equal.Let the amounts of charge on two conductors Aand B are Q1 and Q2 their capacities are C1 andC2 and their potentials are V1 and V2 respectively,

83 PINEGROVE


then 1 1 1Q C V and 2 2 2Q C VLet the amount of charge after the conductors areconnected, are |

1Q and |2Q respectively, then

1 11 1 2 2Q C V;Q C V

Charges are redistributed in the ratio of their ca-pacities. I I

1 2 1 2Q :Q C :C (since V is same)In case of spherical conductors, 0C 4 r so, I I

1 2 1 2Q : Q r : rVan De Graff GeneratorVan De Graaff generator is used to develop veryhigh voltages and resulting large electric fields andused to accelerate charged particles to high energiesPrinciple :- Whenever a charge is given to ametal body it will spread on the outer surface of it.if we put a charged metal body inside the hollowmetal body and the two are connected by a wire,whole of the charge of innner body will flow to theouter surface of the hallow body. No matter howlarge the charge is on the inner body.

Consider a sperical conductor 1 of radius 1r holding

charge 1q uniformly distributed on it. it is kept inside

a hollow conductor 2 of radius 2r which isuncharged.

Electric potential of inner sphere is

11

o 1

q1V

4 r

Electric potential of outer sphere is

12

o 2

q1V

4 r

potential difference between the twoconductors

11 2

o 1 2

q 1 1V V

4 r r

If ‘ 2q ’ charge is on the outer shell

1 21

o 1 2

q q1V

4 r r

q 2

q 1

r2

r1

1 22

o 2

q q1V

4 r

1

1 2o 1 2

q 1 1V V

4 r r

potential difference 1 2V V will remain the

same for any value of 2q

C.U.Q.1. A condenser stores

1) potential 2) charge3) current 4) energy in magnetic field

2. Out of the following statementsA) The capacity of a conductor is affected dueto the presence of an uncharged isolatedconductorB) A conductor can hold more charge at thesame potential if it is surrounded by dielectricmedium.1) Both A and B are correct2) Both A and B are wrong3) A is correct and B is wrong4) A is wrong and B is correct

3. If an earthed plate is brought near positivelycharged plate, the potential and capacity ofcharged plate1) increases, decreases 2) decreases, increases3) decreases, decreases 4) increases, increases

4. The plates of charged condenser are connectedby a conducting wire. The quantity of heatproduced in the wire is1) Inversely proportional to the capacity of thecondenser.2) Inversely proportional to the square of thepotential of the condenser.3) proportional to the length of wire4) independent of the resistance of the wire

5. A capacitor works in1) A.C. circuits only 2) D.C. circuits only3) both A,C & D.C4) neither A.C. nor in D.C. circuit.


84 PINEGROVE

6. In order to increase the capacity of a parallelplate condenser one should introduce betweenthe plates a sheet of (assume that the space iscompletely filled)1) Mica 2) Tin3) Copper 4) Stainless steel

7. In a parallel plate capacitor, the capacitance1) increases with increase in the distance betweenthe plates2) decreases if a dielectric material is put betweenthe plates3) increases with decrease in the distance betweenthe plates4) increases with decrease in the area of the plates

8. When a dielectric material is introducedbetween the plates of a charged condenser,after disconnecting the battery the electric fieldbetween the plates1) decreases 2) increases3) does not change 4) may increase or decrease

9. A parallel plate capacitor is charged and thecharging battery is then disconnected. If theplates of the capacitor are moved further apartby means of insulating handles1) the charge in the capacitor becomes zero2) the capacitance becomes infinite3) the charge in the capacitor increases4) the voltage across the plates increases

10. The ratio of charge to potential of a body isknown as1) conductance 2) capacitance3) inductance 4) reactance

11. A parallel plate capacitor filled with a materailof dielectric constant K is charged to a certainvoltage and is isolated. The dielectric materialis removed. Thena)The capacitance decreases by a factor Kb) The electric field reduces by a factor Kc) The voltage across the capacitor increasesby a factor Kd) The charge strored in the capacitor increasesby a factor K1) a and b are true 2) a and c are true3) b and c are true 4) b and d are true

12. Force acting upon a charged particle keptbetween the plates of a charged condenser isF. If one of the plates of the condenser isremoved, force acting on the same particle willbecome1) zero 2) F/2 3) F 4) 2F

13. A condenser is charged and then battery isremoved. A dielectric plate is put between the

plates of condenser, then correct statement is1) Q constant V and U decrease2) Q constant V increases U decreases3) Q increases V decreases U increases4) Q, V and U increase

14. If an uncharged capacitor is charged byconnecting it to a battery, then the amount ofenergy lost as heat is1) 1/2QV 2) QV 3) 1/2QV2 4) 2QV

15. When air is replaced by a dielectric mediumof constant K, the capacity of the condenser1) increases K times 2) increases 2K times3) remains unchanged 4) decreases K times

16. If we increase the distance between two platesof the capacitor, the capacitance will1) decrease 2) remain same3) increase 4) first decrease then increase

17. In a charged capacitor the energy is stored in(r) is less than at B1) both in positive and negative charges2) positive charges3) the edges of the capacitor plates4) the electric field between the plates

18. A metal plate of thickness half the separationbetween the capacitor plates of capacitance Cis inserted. The new capacitance is1) C 2) C/2 3) zero 4) 2C

19. One plate of parallel plate capacitor is smallerthan the other, the charge on the smaller platewill be1) less than other 2) more than other3) equal to other4) will depend upon the medium between them

20. Two condensers of unequal capacities areconnected in series across a constant voltaged.c. source. The ratio of the potentialdifferences across the condensers will be1) direct proportion to their capacities2) inverse proportion to their capacities3) direct proportion to the square of their capacities4) inverse proportion to the square root of theircapacities

21. A parallel plate capacitor is first charged andthen isolated , and a dielectric slab isintroduced between the plates. The quantitythat remains unchanged is1) Charge Q 2) Potential V3) Capacity C 4) Energy U

ENERGY STORED IN A CONDENSERAND TYPES OF CAPACITORS

22. The condenser used in the tuning circuit of radioreceiver is1) paper condenser 2) electrolytic condenser3) leyden jar 4) gang condenser

85 PINEGROVE


23. Space between the plates of a parallel platecapacitor is filled with a dielectric slab. Thecapacitor is charged and then the supply isdisconnected to it. If the slab is now taken outthen1) work is not done to take out the slab2) energy stored in the capacitor reduces3) potential difference across the capacitoris decreased4) potential difference across the capacitoris increased

24. A parallel plate condenser is charged byconnecting it to a battery. The battery isdisconnected and a glas slab is introducedbetween the plates. Then1) potential increases2) electric intensity increases3) energy decreases4) capacity decreases

25. A parallel plate condenser is charged byconnecting it to a battery. Withoutdisconnecting the battery, the space betweenthe plates is completely filled with a mediumof dielectric constant k. Then1) potential becomes 1/k times2) charge becomes k times3) energy becomes 1/k times4) electric intensity becomes k times.

26. A parallel plate capacitor of capacity Co ischarged to a potential Vo.A) The energy stored in the capacitor whenthe battery is disconnected and the plateseparation is doubled is E1B) The energy stored in the capacitor whenthe charging battery is kept connected and theseparation between the capacitor plates is

doubled is 2E . Then 1

2

E

E value is

1) 4 2) 3/2 3) 2 4) 1/227. Select correct Statements

a) Charge cannot be isolatedb) Repulsion is the sure test to know thepresence of chargec) Waxed paper is dielectric in paper capacitord) Variable capacitor is used in tuning circuitsin radio1) a, b only 2) a, c only3) a, b, c only 4) b,c,d only

28. A variable parallel plate capacitor and anelectroscope are connected in parallel to abattery. The reading of the electroscope wouldbe decreased by1) increasing the area of overlap of the plates

2) placing a block of paraffin wax between theplates3) decreasing the distance between the plates4) decreasing the battery potential

29. Three identical capacitors are connectedtogether differently. For the same voltage toevery combina-tion, the one that storesmaximum energy is1) the three in series 2) the three in parallel3) two in series and the third in parallel with it4) two in parallel and the third in series with it

30. The magnitude of electric field E in the annularregion of a charged cylindrical capacitor1) is same throughout2) is higher near the outer cylinder than near theinner cylinder3) varies as 1/r where r is the distance from the axis4) varies as r where r is the distance from the axis

31. Two identical capacitors are joined in parallel,charged to a potential V, separated and thenconnected in series i.e., the positive plate ofone is connected to the negative plate of other.1) the charges on the free plates are enhanced2) the charges on the free plates are decreased3) the energy stored in the system increases4) the potential difference between the free platesis 2V

32. Two parallel plate air capacitors areconstructed, one by a pair of iron plates andthe second by a pair of copper plates of samearea and same spacings. Then1) the copper plate capacitor has a greatercapacitance than the iron one2) both capacitors will have equal non zerocapacitances, in the uncharged state3) both capacitors will have equal capacitances onlyif they are charged equally4) the capacitances of the two capacitors areunequal even they are unequally charged

33. Select correct statement for a capacitor havingcapacitance C, is connected to a source ofconstant emf E1) Almost whole of the energy supplied by thebattery will be stored in the capacity, if resistanceof connecting wire is negligibly small2) Energy received by the capacitor will be half ofenergy supplied by the battery only when thecapacitor was initially uncharged3) Strain energy in the capacitor must increases evenif the capacitor had an initial charge4) Energy stored depends on type of the source ofemf


86 PINEGROVE

34. Van de Graff genetor is used to :1) supply electricity for industrial use2) produce intense magnetic fields3) generate high voltage4) obtain highly penetrating X-rays

35. A number of spherical conductors of differentradii have same potential. Then the surfacecharge density on them1) is proportional to their radii2) is inversely proportional to their radii3) are equal4) is proportional to square of their radii

36. Three charged particles are initially in position1. They are free to move and they come inposition 2 after some time. Let U1 and U2 bethe electrostatic potential energies in position1 and 2. Then1) U

1 > U

22) U

2 > U

13) U

1 = U

24) 2 1U U

37. An insulator plate is passed between the platesof a capacitor. Then current

A B

1) always flows from A to B2) always flows from B to A3) first flows from A to B and then from B to A4) first flows from B to A and then from A to B

38. Read the following statementsa) Non polar molecules have uniform chargedistributionb) Polar molecules have non - uniform chargedistributionc) Polar molecules are already polarizedd) Molecules are not already polarized withoutelectric field in Non - polar molecules1)only a & b are correct 2)only c & d are correct3) only c is wrong 4) all are correct

39. The capacitance of a capacitor depends on1) the geometry of the plates2) separation between plates3) the dielectric between the plates4) all the above

40. The electric field E

between two parallelplates of a capacitor will be uniform if1) the plate separation (d) is equal to area of theplate (A)2) the plate separation (d) greater when comparedto area of the plate (A)3) the plate separation (d) is less when comparedto area of the plate (A)4) 2 (or) 3

41. For metals the value of dielectric constant (K)is1) One 2) Infinity 3) Zero 4) Two

42. A capacitor C is connnected to a battery circuithaving two switches 1 2S and S and resistors

1 2R and R . The capacitor will be fully chargedwhen

C

R2

R1 S1 S2

1) both 1 2S and S are closed

2) 1S is closed and 2S is open

3) 1S is open and 2S is closed4) any one of the above

43. Figure shows two capacitors connected inseries and joined to a cell. The graph showsthe variation in potential as one moves fromleft to right on the branch containingcapacitors.

1) 1 2C C 2) 1 2C C 3) 1 2C C4) data insufficient to conclude the answer

44. Two condensers of unequal capacities areconnected in parallel across a constant voltaged.c. source. The ratio of the charges stored inthe condensers will be1) direct proportion to their capacities2) inverse proportion to their capacities3) direct proportion to the square root of theircapacities4) inverse proportion to the square of their capacities

45. A parallel plate capacitor is charged and thenisolated. Regarding the effect of increasing theplate separation, select the appropriatealternative. Charge Potential Energy1) decreases constant decreases2) increases increases increases3) constant decreases decreases4) constant increases increases

46. A parallel plate capacitor is charged byconnecting its plates to the terminals of abattery. The battery remains connected to thecondenser plates and a glass plate isinterposed between the plates of the capacitor,then

87 PINEGROVE


1) the charge increases while the potential differenceremains constant2) the charge decreases while the potential differenceremains constant3) the charge decreases while the potential differenceincreases4) the charge increases while the potential differencedecreases

47. A parallel plate capacitor is charged to a fixedpotential and the charging battery is thendisconnected. If now, the plates of the capacitorare moved further apart, then1) the charge on the capacitor increases2) the voltage across the capacitor increases3) the energy stored in the capacitor decreases4) the capacitance increases

48. A parallel plate air condenser is charged andthen disconnected from the charging battery.Now the space between the plates is filled witha dielectric then, the electric field strengthbetween the plates1) increases while its capacity increases2) increases while its capacity decreases3) decreases while its capacity increases 4 )decreases while its capacity decreases

49. When two identical condensers are connectedin series choose the correct statementregarding the working voltage (the maximump.d. that can be applied to a condenser) andthe capacity1) working voltage increases, capacity increases2) working voltage increases, capacity decreases3) working voltage decreases, capacity increases4) working voltage decreases, capacity decreases

50. Two unequal capacitors, initially uncharged, areconnected in series across a battery. Which ofthe following is true1) The potential across each is the same2) The charge on each is the same3) The energy stored in each is the same4) The equivalent capacitance is the sum of the twocapacitances

51. Which of the following will not increase thecapacitance of an air capacitor?1) adding a dielectric in the space between theplates2) increasing the area of the plates3) moving the plates closer together4) increasing the voltage

52. In a parallel-plate capacitor, the regionbetween the plates is filled by a dielectric slab.The capacitor is connected to a cell and theslab is taken out. Then1) some charge is drawn from the cell2) some charge is returned to the cell

3) the potential difference across the capacitor isreduced4) no work is done by an external agent in takingthe slab out

53. Which of the following statements are correct?a) When capacitors are connected in parallel theeffective capacitance is less than the individualcapacitancesb) The capacitances of a parallel plate capacitorcan be increased by decreasing the separation ofplatesc) When capacitors are connected in series theeffective capacitance is less than the least of theindividual capacitiesd) In a parallel plate capacitor the electrostaticenergy is stored on the plates1) (a) and (b) 2) (a) and (c)3) (c) and (d) 4) (b) and (c)

54. Three identical condensers are connectedtogether in four different ways. First all of themare connected in series and the equivalentcapacity is C1. Next all of them are connectedin parallel and the equivalent capacity is C2.Next two of them are connected in series andthe third one connected in parallel to thecombination and the equivalent capacity is C3.Next two of them are connected in parallel andthe third one connected in series with thecombination and the equivalent capacity is C4.Which of the following is correct ascendingorder of the equivalent capacities?1) 1 3 4 2C C C C 2) 1 4 3 2C C C C 3) 2 3 4 1C C C C 4) 2 4 3 1C C C C

55. On a capacitor of capacitance 0C followingsteps are performed in the order as given incolumn I.(A) Capacitor is charged by connecting it acrossa battery of emf 0E(B) Dielectric of dielectric constant K andthickness d is inserted(C) Capacitor is disconnected from battery(D) Separation between plates is doubledColumn-I Column-II(Steps performed) (Final value of

Quantity (Symbols have usual meaning)

(a) (A)(D)(C)(B) (p) 0 0

2

C EQ

(b) (D)(A)(C)(B) (q) 0 0

1

KC EQ

K


88 PINEGROVE

(c) (B)(A)(C)(D) (r) 0

1

KCC

K

(d) (A)(B)(D)(C) (s) 0( 1)

2

E KV

K

1) a-p,r,s, b-p,r,s, c-r, d-q,r2) a-p, b-p,r c-r, d-q,3) a-p,s, b-r,s, c-r, d-q,4) a-r,s, b-s, c-r, d-q,r

56. In the circuit, both capacitors are indentical.Column I indicates action done on capacitors1 and Column II indicates effect on capacitor2

(1) (2)

Column-I Column-II(a) Plates are moved (p)Amount of charge

further apart on left plateincreases

(b) Area increased (q) Potential differenceincreases

(c) Left plate is earthed (r) Amount of chargeon right platedecreases

(d) It's plates are short (s) None of the above circuited effects1.a-r, b-p,q, c-s, d-p,q 2.a-r, b-p, c-s, d-q3.a-r, b-p, c-r, d-q 4.a-s, b-,q, c-s, d-q

57. The potential across a 3 F capacitor is 12 Vwhen it is not connected to anything. It is thenconnected in parallel with an uncharged 6 Fcapacitor. At equilibrium, the charge andpotential difference across the capacitor 3 Fand 6 F are listed in column I. Match it withcolumn III.

Column-I Column-II(a) charge on 3 F capacitor (p) 12 C(b) charge on 6 F capacitor (q)24 F(c) potential difference across 3 F (r) 8 V capacitor(d) potential difference across 6 F (s) 4 V capacitor

1) a-r, b-p, c-s, d-q 2) a-p, b-q, c-s, d-s3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q

58. Some events related to a capacitor are listedin column-I. Match these with their effect(s)in column - II

Column-I Column-II(a) Insertion of dielectric (p) Eelctric field while battery between

plates changesremain attached

(b) Removal of dielectric (q) Charge present on while battery plates changes

is not present(c) Slow decrease in (r) Energy stored in separation between capacitor increases

plates while battery isattached

(d) Slow increase of (s) Work done by separation between capacitor agent is plates while battery positive is not present

1) a-r, b-p,, c-p,q,s, d-q2) a-p, b-p,, c-r,s, d-s3) a-q,r, b-p,r,s c-p,q,r, d-r,s4) a-r, p,b-q,, c-s, d-q

59. The effective capacity of the followingcapacitors is ________

e) 2

3

c

f) 2C

g) 3C

h) 5

2

C

i) 3

2

C

1) , , ,a g b f c e d i 2) , , ,a g b h c e d i 3) , , ,a i b h c e d g 4) , , ,a g b e c h d i

60. The circuit involves two ideal cells connectedto a 1 F capacitor via key K. Initially the keyK is in position 1 and the capacitor is charged

89 PINEGROVE


fully by 2V cell. The key is pushed to position2. Column I gives physical quantities involvingthe circuit after the key is pushed from position1. Column.II gives corresponding results.Match the column-I with Column-II

C = 1mF

4V2V

K

1 2

Column-I Column-II(a) The net charge crossing the4 volt cell in C is (p) 2(b) The magnitude of work doneby 4 volt cell in J is (q) 6(c) The gain in potential energy ofcapacitor in J is (r) 8(d) The net heat produced incircuit in J is (s) 161) a-r, b-p, c-s, d-q 2) a-p, b-r, c-q, d-p3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q

61. Two identical capacitors A and B are connectedto a battery of emf E as shown in figure. Nowa dielectric slab is inserted between the platesof capacitor B while battery remainsconnected. Due to this inserting some physicalquantities may change which are mentioned inColumn-I and the effect is mentioned inColumn-II.Match the Column I with Column-II

A

e

BK

Column-I Column-II(a) Charge on A (p) Increases(b) Charge on B (q) Decreases(c) Potential differenceacross A (r) Remains constant(d) Potential differenceacross B (s) Will change1) a-r, b-p, c-s, d-q 2) a-p,s b-q,s, c-q,s d-q,s3) a-r, b-p, c-q, d-q 4) a-r, b-q, c-s, d-q

62. Match the followingSet -I Set-IIa) Electrolyte e) Radio circuitsCapacitors & cheap in costb) Paper Capacitor f) Proper Polarity

high capacitanceof order 310 F

c) Multiple g)High frequencyCapacitor oscillating circuitsd) Variable h) Tuning circuitsCapacitor in radio & T.V

receivers1) , , ,a f b g c h d e 2) , , ,a g b f c e d h 3) , , ,a f b e c g d h 4) , , ,a h b e c f d g

63. Column - I Column - IIA) electrical potential p) vector

B) energy stored in a q)21

2CV

condenserC) force between two r) scalar

D) electric capacity s) 2

0

12

E Acapacitor plates

A B C D1) r q,r p,s r2) r q,r p,q s3) q,r p,q r,s s4) p,q r q,r sASSERTION & REASONING

1) Both A and R false2) Both A and R true and R is not correctreason for A3) A is true and R is false4) Both A and R are true and R is correctreason of A.

64. Assertion (A) : The strength of electric filedin the charged and isolated capacitor isdecreased when the dielectric slab is inserted.Reason(R): When the dielctric slab is insertedbetween the plates of a charged capacitor,electricfield produced due to induced charge,opposite to the external field.

65. Assertion: If temperature is increased, thedielectric constant of a polar dielectricdecreases whereas that of a non-polardielectric does not change significantlyReason: The magnitude of dipole moment ofindividual polar molecule decreasessignificantly with increase in temperature.

66. Assertion: The heat produced by a resistor inany time t during the charging of a capacitorin a series circuit is half the energy stored inthe capacitor by that time.Reason: Current in the circuit is equal to the


90 PINEGROVE

rate of increase in charge on the capacitor.67. Assertion: A dielectric is inserted between the

plates of an isolated fully-charged capacitor.The dielectric completely fills the spacebetween the plates. The magnitude ofelectrostatic force on either metal platedecreases, as it was before the insertion ofdielectric medium.Reason: Due to insertion of dielectric slabin an isolated parallel plate capacitor (thedielectric completely fills the space betweenthe plates), the electrostatic potential energyof the capacitor decreases.

68. Assertion: If the potential difference across aplane parallel plate capacitor is doubled thenthe potential energy of the capacitor is doubledthen the potential energy of the capacitorbecomes four times under all conditionsReason: The potential energy U stored in the

capacitor is 21

2U CV , where C and V have

usual meaning.69. Assertion: A parallel plate capacitor is charged

to a potential difference of 100V, anddisconnected from the voltage source. A slabof dielectric is then slowly inserted betweenthe plates. Compared to the energy before theslab was inserted, the energy stored in thecapacitor with the dielectric is decreased.Reason: When we insert a dielectric betweenthe plates of a capacitor, the induced chargestend to draw in the dielectric into the field (justas neutral objects are attracted by chargedobjects due to induction). We resist this forcewhile slowly inserting the dielectric, and thusdo negative work on the system, removingelectrostatic energy from the system.

70. Statement ' 'A : The energy stored getsreduced by a factor ' 'K when the battery isdisconnected after charging the capacitor andthen the dielectric is introducedStatement ' 'B : The energy stored in thecapacitor increases by a factor ' 'k when adielectric is introduced between the plates withthe battery present in the circuit

71. Assertion (A): A metallic sheild in form of ahollow shell may be built to block an electricfield.Reason (R): In a hollow spherical sheild, theelectric field inside it is zero at every point.

72. Assertion (A): When two spheres carryingsame charge but a different radii areconnected by a conducting wire, the charge

flows from smaller sphere to large sphere.Reason (R): Smaller sphere is at high potentialwhen equal charges are imparted to both thespheres

73. Assertion (A): Two capacitors are connectedin par allel to a battery. If a dielectric mediumis inserted between the plates of one of thecapacitors then the energy stored in the systemwill increaseReason (R): On inserting dielectric mediumbetween the plates of a capacitors, its capacityincreases

74. Assertion (A): When a charged capacitor isdischarged through a resistor, heat is producedin the resistorReason (R): In charging a capacitor, energyis stored in the capacitor.

75. Assertion (A): A capacitor of capacitance C isconnected across a battery of potentialdifference V. The energy stored in the

capacitor is 21

2CV

Reason (R): The energy supplied by the

battery is 21

2CV

76. Assertion (A): Two metal plates each of areaA form a parallel plate capacitor. Now oneplate is displaced up, then the capacitance ofcapacitor decreases.Reason (R): Due to displacing one plate, theoverlapping area decreases, capacitance

0AC

d

decreases.

77. Assertion (A): Two plates of a parallel platecapacitor are drawn apart, keeping themconnected to a battery. Next the same platesare drawn apart from the same initial condition,keeping the battery disconnected, then thework done in both cases are same.Reason (R): Capacitor plates have samecharge in both cases and displacements ofplates in both cases are also same.

78. Assertion (A) : Two metallic plates placed sideby side form three capacitors.Reason (R) : The infinity and first face of firstplate is one capacitor, the second face of firstplate and first face of second plate formssecond capacitor and the second face of secondplate and infinity forms the third capacitor,but the capacitance of first and third

91 PINEGROVE


capacitance are extremely small79. Statement ' 'A : The energy stored gets

reduced by a factor ' 'K when the battery isdisconnected after charging the capacitor andthen the dielectric is introducedStatement ' 'B : The energy stored in thecapacitor increases by a factor ' 'k when adielectric is introduced between the plates withthe battery present in the circuit

C. U.Q - KEY1) 2 2) 1 3) 2 4) 4 5) 3 6) 17) 3 8) 1 9) 4 10) 2 11) 2 12) 213) 1 14) 1 15) 1 16) 1 17) 4 18) 419) 3 20) 2 21) 1 22) 4 23) 4 24) 325) 2 26) 1 27) 4 28) 4 29) 2 30) 331) 4 32) 2 33) 3 34) 3 35) 2 36) 137) 4 38) 4 39) 4 40) 3 41) 2 42) 243) 3 44) 1 45) 4 46) 1 47) 2 48) 349) 2 50) 2 51) 4 52) 2 53) 4 54) 255) 1 56) 1 57) 2 58) 3 59) 1 60) 261) 2 62) 3 63) 1 64) 4 65) 3 66) 467) 4 68) 4 69) 1 70) 4 71) 1 72) 173) 1 74) 2 75) 3 76) 1 77) 4 78) 179) 4

EXERCISE - I (C.W.)CAPACITANCE

1. The capacity of a parallel plate condenserconsisting of two plates each 10 cm square andare seperated by a distance of 2 mm is (Takeair as the medium between the plates)1) 138.85 10 F 2) 124.42 10 F3) 1244.25 10 F 4) 1388.5 10 F

2. Sixty four spherical drops each of radius 2 cmand carrying 5C charge combine to form abigger drop. Its capacity is

1) 11810

9F 2) 1190 10 F 3) 111.1 10 F 4) 119 10 F

3. A highly conducting sheet of aluminium foil ofnegligible thickness is placed between theplates of a parallel plate capacitor. The foil isparallel to the plates. If the capacitance beforethe insertion of foil was 10 F, its value afterthe insertion of foil will be1) 20 F 2) 10 F 3) 5 F 4) Zero

4. Two metal plates are separated by a distanced in a parallel plate condenser. A metal plateof thickness t and of the same area is insertedbetween the condenser plates. The value ofcapacitance increases by ...... times

1) d t

d

2) 1

t

d 3)

tt

d 4)

1

1t

d

5. A radio capacitor of variable capacitance ismade of n parallel plates each of area A andseparated from each other by a distance d. Thealternate plates are connected together. Thecapacitance of the combination is

1) on A

d

2) 1 on A

d

3) 2 1 on A

d

4) 2 on A

d

6. The radius of the circular plates of a parallelplate condenser is ‘r’. Air is there as thedielectric. The distance between the plates ifits capacitance is equal to that of an isolatedsphere of radius r' is

1) 2

4 '

r

r2)

2

'

r

r3)

'

r

r4)

2

4

r

CAPACITORS IN SERIES AND INPARALLEL

7. When two capacitors are joined in series theresultance capacity is 2.4 F and when the sametwo are joined in parallel the resultantcapacity is 10 F . Their individual capacities are1)7 , 3F F 2)1 , 9F F 3)6 , 4F F 4 8 , 2F F

8. Three condensers 1F , 2F and 3F areconnected in series to a p.d. of 330 volt. Thep.d across the plates of 3 F is1) 180 V 2) 300 V 3) 60 V 4) 270 V

9. The effective capacitance between the point Pand Q in the given figure is

4 F

4 F 10 F 4 F4 F

P Q1) 4 F 2) 16 F3) 26 F 4) 10 F

10. The equivalent capacitance between P and Qis

10 F 10 F 10 F 10 F5 F5 F5 F10 F

Q

P

1) 10 F 2) 20 F 3) 5 F 4) 15 F11. The equivalent capacity between the points X

and Y in the circuit with 1C F (2007M)C

X YC

C

1) 2 F 2) 3 F3) 1 F 4) 0.5 F

12. The equivalent capacitance of the networkgiven below is 1 F. The value of ‘C’ is


92 PINEGROVE

1.5 F3 F 3 F

Q

C

P

1) 3 F 2) 1.5 F 3) 2.5 F 4) 1 F13. Three capacitors of 3 ,2F F and 6 F are

connected in series. When a battery of 10V isconnected to this combination then charge on3 F capacitor will be1) 5 C 2) 10 C 3) 15 C 4) 20 C

ENERGY STORED INACONDENSER,

TYPES OF CAPACITORS14. Two spheres of radii 12 cm and 16 cm have

equal charge. The ratio of their energies is1) 3 : 4 2) 4 : 3 3) 1 : 2 4) 2 : 1

15. A condenser of capacity 10 F is charged to apotential of 500 V. Its terminals are thenconnected to those of an uncharged condenserof capacity 40 F. The loss of energy inconnecting them together is1) 1J 2) 2.5J 3) 10J 4) 12 J

16. A 2 F condenser is charged to 500V and thenthe plates are joined through a resistance. Theheat produced in the resistance in joule is1) 250 10 Joule 2) 225 10 Joule3) 20.25 10 Joule 4) 20.5 10 JouleEXERCISE - I (C.W.) - KEY

1) 2 2) 1 3) 2 4) 4 5) 2 6) 17) 3 8) 3 9) 1 10) 3 11) 1 12) 213) 2 14) 1 15) 1 16) 2

EXERCISE - I (C.W.) - HINTS

1. 0AC

d

2. 1

1 3C n C

4. 0AC

td t

k

;k =

5. Due to n plates n-1 capacitors are formed

6. 2

14o

o

rr

d

2

14

rd

r

7.1 2

1 2

S

C CC

C C ; 1 2 PC C C

8. effQ C V 1Q CV9.

1 1 2

1 2

C CC

C C 11 3 4

3 4

C CC

C C

1 11effC C C

10. From Left 10 10

510 10

C F 1 5 5 10C F 11 10 10

510 10

C F and so on

finally 10 10

510 10effC F

11. 1 2effC C C 12. 1.5 ,c c are in parallel ;

its effective capacitance 1.5 + c1.5+c, 3 , 3F F are in series

13. In series charge constant effQ C V14. U =

2

2

q

C,

1U

r

15. 21 21 2

1 2

1

2

C CE V V

C C

16. Energy Stored = 21

2cv

EXERCISE - I (H.W.)CAPACITANCE

1. The charge stored in a capacitor is 20 C andthe potential difference across the plates is 500V. Its capacity is1) 0.04 F 2) 210 F 3) 62 10 F 4) 250 F

2. An oil condenser has a capacity of 100 F . Theoil has dielectric constant 2. When the oil leaksout its new capacity is1) 200 F 2) 0.02 F 3) 50 F 4) 0.5 F

3. A dielectric of thickness 5cm and dielectricconstant 10 is introduced between the platesof a parallel plate capacitor having plate area500 sq. cm and separation between the plates10cm. The capacitance of the capacitor withdielectric slab is 12 2 2

0 8.8 10 /C N m 1) 4.4pF 2) 6.2pF 3) 8pF 4) 10pF

4. The capacitance of a sphere of radius 10cmsituated in air is approximately1) 611 10 F 2) F1011 9 3) F1011 12 4) ZeroCAPACITORS IN SERIES AND IN

PARALLEL5. The ratio of the resultant capacities when three

capacitors of 2 F , 4 F and 6 F areconnected first in series and then in parallel is1) 1 : 11 2) 11 : 1 3) 12 : 1 4) 1 : 12

93 PINEGROVE


6. A condenser A of capacity 4 F has a charge20 C and another condenser B of capacity10 F has a charge 40 C . If they areconnected parallel, then1) charge flows from B to A till the charges on themare equal.2) charge flows from B to A till common poten tialis reached3) charge flows from A to B till common potentialis reached4) charge flows from A to B till charges on themare equal.

7. A capacitor of 30 F charged to 100 V isconncected in parallel to capacitor of 20 Fcharged to 50 volt. The common potential is1) 75 V 2) 150 V 3) 50 V 4) 80 V

8. The equivalent capacity between the points ‘A’and ‘B’ in the following figure will be

A

1) 3C 2) C/3 3)3/C 4) 1/3C9. Two capacitors with capacitances 1C and 2C

are charged to potentials 1V and 2Vrespectively. When they are connected inparallel the ratio of their respective chargesis

1) 1

2

C

C 2) 1

2

V

V 3) 2

12

2

V

V 4) 2122

C

C

10. The equivalent capacitance between P and Qof the given figure is (the capacitance of eachcapacitor is 1 F )

C

C

C

C

C

OP

OC

1) 2 F 2) 0.5 F 3) 5 F 4) 0.2 F11. The resultant capacity between the points P

and Q of the given figure is4 F 2 FP

Q4 F1) 4 F 2)

16

3F

3) 1.6 F 4) 1 F12. Charge ‘Q’ taken from the batteryof 12V in

the circuit is12V

3 F 6 F

4 F

1) 72 C 2) 36 C

3) 156 C 4)20 C

13. If 3 capacitors of values 1, 2 and 3 F areavailable. The maximum and minimum valuesof capacitances one can obtain by differentcombinations of the three capacitors togetherare respectively .... and

1) 6

6 ,11

F F 2) 11

6 ,6

F F 3) 3 ,1F F 4) 4 ,2F F

ENERGY STORED IN ACONDENSER,TYPES OF CAPACITORS

14. A capacitor of 8 micro farad is charged to apotential of 1000V. The energy stored in thecapacitor is1) 8 J 2) 12 J 3) 2 J 4) 4 J

15. A condenser is charged to a p.d. of 120 volt.Its energy is 51 10 joule . If the battery is thereand the space between plates is filled up witha dielectric medium 5r , its new energy is1) 510 J 2) 52 10 J 3) 53 10 J 4) 55 10 J

16. The plates of a parallel plate capacitor havean area of 290cm each and are separated by2 mm. The capacitor is charged by connectingif to a 400 V supply. Then the density of theenergy stored in the capacitor 12

o 8.8 10 F / m 1) 30.113Jm 2) 30.117 Jm

3) 30.152Jm 4) 30.226Jm

EXERCISE - I (H.W. ) - KEY1) 1 2) 3 3) 3 4) 3 5) 1 6) 37) 4 8) 1 9) 1 10) 2 11) 1 12) 113) 1 14) 4 15) 4 16) 2

EXERCISE - I (H.W.) - HINTS

1.q

CV

2. 'C

CK

3.

0AC

td t

k

4. 04C R

5.1 2 3

1 1 1 1

Cs C C C

1 2 3pC C C C 6. In parallel potential constant

1 1 2 2

1 2

C V C VV

C C

then find charges 11 1q C V , 1

2 2q C V7.

1 1 2 2

1 2

C V C VV

C C


94 PINEGROVE

8. Capacitors are in parallel9. Parallel potential constant and Q C10. 1C C C ; 2C C ; 3C C C

1 2 3

1 1 1 1

effC C C C

11. 1

4 4

4 4C

; 2 2C F ; 1 2effC C C 12.

6 34

6 3effC ; effQ C V

13. For maximum capacitors are parallelFor minimum capacirots are in sseries

14. U = 1/2 2CV 15. 1 rE E

16.2

o

1 VU E ,E

2 d

EXERCISE - II (C.W.)CAPACITANCE

1. A parallel plate condenser has initially airmedium between the plates. If a slab of dieletricconstant 5 having thickness half the distanceof seperation between the plates is introduced,the percentage increase in its capacity is1) 33.3% 2) 66.7% 3) 50% 4) 75%

2. When a dielectric slab of thickness 4 cm isintroduced between the plates of parallel platecondenser, it is found the distance between theplates has to be increased by 3cm to restoreto capacity to original value. The dielectricconstant of the slab is1) 1/4 2) 4 3) 3 4) 1

3. The area of the positive plate is 1A and the

area of the negative plate is 2A ( 2A < 1A ) .They are parallel to each other and areseparated by a distance d . The capacity of acondenser with air as dielectric is

1) 0 1A

d

2)

0 2A

d

3)

0 1 2A A

d

4)

0 1

2

A

A d

4. The cross section of a cable is shown in fig.

The inner conductor has a radius of 10 mm andthe dielectric has a thickness of 5 mm. Thecable is 8 km long. Then the capacitance of

the cable is elog 1.5 0.41) 3.8 F 2) 1.1 F 3) 104.8 10 F 4) 3.3 FCAPACITORS IN SERIES AND IN

PARALLEL5. Two condensers of capacity C and 2C are

connected in parallel and these are chargedupto V volt. If the battery is removed and

dielectric medium of constant K is put betweenthe plates of first condenser, then the potentialat each condenser is

1) 2

V

k 2) 23

k

V 3)

2

2

V

k 4) 3

2

V

k 6. Given a number of capacitors labelled as C,V.

Find the minimum number of capacitors neededto get an arrangement equivalent to ,net netC V

1) 2

2net netC V

nC V

2) 2

2net net

C Vn

C V

3) net net

C Vn

C V 4) netnet VC

nC V

7. Two capacitors of capacities 3 F and 6 F

are connected in series and connected to 120V.The potential difference across 3 F is 0V and

the charge here is 0q . We have

A) 0 40q C B) 0 60V VC) 0 80V V D) 0 240q C1) A, C are correct 2) A, B are correct3) B, D are correct 4) C, D are correct

8. n Capacitors of 2 F each are connected inparallel and a p.d of 200v is applied to thecombination. The total charge on them was 1cthen n is equal to1) 3333 2) 3000 3) 2500 4) 25

9. An infinite number of identical capacitors eachof capacitance 1 mF are connected as shownin the figure. Then the equivalent capacitancebetween A and B is

16 Capacitors

8 Capacitors

A B

1) 1 mF 2) 2 mF

3) ½ mF 4) 0.75 mF

ENERGY STORED IN A CONDENSERTYPES OF CAPACITORS

10. Two capacitors of capacites 1F and C Fare connected in series and the combination ischarged to a potential difference of 120 V. Ifthe charge on the combition is 80 ,C theenergy stored in the capacitor C in micro joulesis :1) 1800 2) 1600 3) 14400 4) 7200

11. A parallel capacitor of capacitance C ischarged and disconnected from the battery.

95 PINEGROVE


The energy stored in it is E. If a dielectricslab of dielectric constant 6 is inserted betweenthe plates of the capacitor then energy andcapacitance will become1) 6E, 6C 2) E, C 3) E/6, 6C 4) E, 6C

12. In the circuit diagram given below, the valueof the potential difference across the plates ofthe capacitors are

12kv

7 F13kv

3 F

1) 17.5 KV, 7.5 KV 2) 10 KV, 15 KV3) 5 KV, 20 KV 4) 16.5 KV, 8.5KV

13. The equivalent capacity of the infinite net workshown in the figure (across AB) is (Capacityof each capacitor is 1 F)

1) 2) 1 F 3) 3 1

2F 4)

3 1

2F

14. The extra charge flowing through the cell onclosing the key k is equal to k

C

CC

C

V

1) 4

CV2) 4 CV

3) 4

3CV 4)

3

4CV

EXERCISE - II (C.W) - KEY1) 2 2) 2 3) 2 4) 1 5) 4 6) 17) 4 8) 3 9) 2 10) 2 11) 3 12) 113) 3 14) 1

EXERCISE - II (C.W) - HINTS

1. 00

AC

d

; 0

1 1/

AC

d t k

0

0

% 100%C C

CC

2. 0 0

1 1/ '

A AC

d t k d d

3. Effective area only 2A 4. oK 2 l

Cln b / a

5. Q = constant , CV + 2CV = | |2KCV CV

6.2

0 02 2Q Q

FA

7.

1 2

1 2

C CQ V

C C

8. Q nCV9. R

C CC C ......

2 4 10. 1 1 2 2CV C V

11.2 2

,2 2f

q qU U

C KC

12. By Kirchoff loop theorem

12 13 03 7

q q ; 3 4,3 7

q qV V

13. Between D & E effective capacitance is x1

1 11

xx

14.

3

4netC C ; when key was open 3

4q CV

when key was closed 3C becomes short circuited.Net charge on C is now 'q CV

'

4

CVq q q

EXERCISE - II (H.W)CAPACITANCE

1. The capacity of a condenser A is 10 F and itis charged to a battery of 100 volt. The batteryis disconnected and the condenser A isconnected to a condenser B the commonpotential is 40V. The capacity of B is1) 8 F 2) 15 F 3) 2 F 4) 1 F

2. A parallel plate capacitor has the spacebetween its plates filled by two slabs of

thickness 2

d each and dielectric constant 1K

and 2K . d is the plate separation of thecapacitor. The capacitance of the capacitor is

1) 0 1 2

1 2

2 A K K

d K K

2) 01 2

2 AK K

d

3) 0 1 2

1 2

2 A K K

d K K

4) 0 1 2

1 2

2 d K K

A K K

3. An isolated capacitor of capacitance ‘C’ is

charged to a potential ‘V’. Then a dielectricslab of dielectric constant K is inserted asshown in fig. The net charge on four surfaces1,2,3 and 4 would be respectively.

1) 0, CV,-CV, 0 2) 0, CV CV

,K K

, 0


96 PINEGROVE

3) CV, 0, 0, -CV 4) CV CV

CV, , , CVK K

CAPACITORS IN SERIES AND IN

PARALLEL4. A parallel plate capacitor with plate area ‘A’

and separation ‘d’ is filled with two dielectricsof dielectric constants 1K and 2K . If thepermittivity of free space is 0 , thecapacitance of the capacitor is given by

d K1 K2

1) 01 2

AK K

d

2) 01 2

2AK K

d

3) 01 22

AK K

d

4) 0 1 2

1 2

2.

A K K

d K K

5. ‘A’ and ‘B’ are two condensers of capacities 2 F and 4 F. They are charged to potentialdifferences of 12V and 6V respectively. If theyare now connected (+ve to +ve), the chargethat flows through the connecting wire is1) 24 C from A to B 2) 8 C from A to B3) 8 C from B to AA 4) 24 C from B to AA

6. Force of attraction between the plates of aparallel plate capacitor is

1) 2

02q

A 2) 2

0q

A 3) 02

q

A 4) 2

202q

A

7. Seven capacitors each of capacitance 2 F areto be connected in a configuration to obtain an

effective capacitance of 10

11F . Which of the

combination shown in figure will achieve thedesired result

1) 2)

3) 4)

8. The equivalent capacitance between ‘A’ and‘B’ in the adjoining figure is

3 F

9 F 9 F9 F

BA

1) 51

30F 2) 6 F 3) 30 F 4) 12 F

NERGY STORED IN A CONDENSERTYPES OF CAPACITORS

9. A capacitor4 F charged to 50V is connectedto another capacitor 2 F charged to 100V .The total energy of combination is

–

–

+

+

1) 313.3 10 J 2) 320 10 J3) 35 10 J 4) 310 10 J10. A4 F capacitor is charged by a 200V battery..

It is then disconnected from the supply and isconnected to another uncharged 2 Fcapacitor. During this process, Loss of energy(in J) is1) Zero 2) 5.33 x 10-23)4 x 10-2 4) 2.67 x 10-2

11. A capacitor of capacitance C has charge Qand stored energy W . If the charge isincreased 2Q the stored energy would be

1) 4W 2) 2

W 3) 2W 4) 4W12. The equivalent capacitance between points M

and N isC1

C2

N

M1) Infinity 2) 2

11

CC

C

3) 1 2

1 2

C C

C C 4) 1 2

1 2

C C

C CEXERCISE - II (H.W) - KEY

1) 2 2) 3 3) 2 4) 3 5) 2 6) 17) 1 8) 2 9) 1 10) 4 11) 4 12) 1

EXERCISE - II (H.W) - HINTS

1.1 1 2 2

1 2

c v c vV

c c

2.

1 2

1 1 1

C C C ;

11 1

2

2o oA Kk A

C kd d

2 2

2 oAC k

d3. Due to polarization, charge on dielectric slab would

97 PINEGROVE


be 1

CV 1K

4.

1 2

1 2

2 2;

o o

A Ak k

C Cd d

C

1 and C

2 connected in parallel..

5 Charge flow 1 1 oQ C V V where oV common potentialt thickness of metal sheet

6. q CV 1 2

1 2

. .C C

VC C

1o

qV

C

7. Verify the cases individually

effC nc ; effQ C V8. Resultant capacitance of 9 , 9F F and 9 F is

in parallel to 3 F .

9.1 1 2 2

1 2

C V C VV

C C

; 21 2

1

2U C C V

10. 21 21 2

1 2

1

2

C CU V V

C C ; 2

1 2

1

2U C C V

11. 2E Q 12. 0eqM N

q qC

V V

EXERCISE - III1. The time in seconds required to produce a P.D

at 20V across a capacitor at 1000 F when itis charged at the steady rate of 200 / secCis1) 50 2) 100 3) 150 4) 200

2. A parallel plate capacitor of capacity 5 F andplate separation 6cm is connected to a 1Vbattery and is charged. A dielectric of dielectricconstant 4 and thickness 4 cm is introducedinto the capacitor. The additional charge thatflows into the capacitor from the battery is1) 2 C 2) 3 C 3) 5 C 4) 10 C

3. The force between the plates of a parallel platecapacitor of capacitance C and distance ofseparation of the plates d with a potentialdifference V between the plates, is

1) 2

2

CV

d2)

2 2

22

C V

d3)

2 2

2

C V

d4)

2V d

C

4. Two identical capacitors are connected as showin the figure. A dielectric slab is introducedbetween the plates of one of the capacitors soas to fill the gap, the battery remainingconnected. The charge on each capacitor willbe (charge on each condenser is q0 ; k =dielectric constant )

I

II

2VB1)

0

1

2

1 k

q

2) 0

11 k

q

3) 02

1

q

k 4) 0

1

q

k5. Two identical capacitors 1 and 2 are connected

in series to a battery as shown in figure.Capacitor 2 contains a dielectric slab ofdielectric constant K as shown. Q1 and Q2 arethe charges stored in the capacitors. Now thedielectric slab is removed and the

corresponding charges are '1Q and '

2Q . Then

B

1) /1

1

1Q K

Q K 2) /2

2

1

2

Q K

Q

3) /2

2

1

2

Q K

Q K 4) /2

2 2Q K

Q

6. A capacitor of capacitance 1 F withstands amaximum voltage of 6 kV, while anothercapacitor of capacitance 2 F withstands amaximum voltage of 4 kV. If they areconnected in series, the combination canwithstand a maximum voltage of1) 3 kV 2) 6 kV 3) 10 kV 4) 9 kV

7. Energy ‘E’ is stored in a parallel plate capacitor‘C 1’. An identical uncharged capacitor ‘C2’ isconnected to it, kept in contact with it for awhile and then disconnected, the energy storedin C2 is 1) E/2 2) E/3 3) E/44 )Zero

8. A parallel plate capacitor has area of each plateA, the separation between the plates is d . It ischarged to a potential V and thendisconnected from the battery. The amount ofwork done in the filling the capacitorCompletely with a dielectric constant k is

1)2

02

1 11

2

AV

d k

2) 2

01

2

V A

kd

3)2

02

1

2

V A

k d

4)

201 1

12

AV

d K


98 PINEGROVE

9. A capacitor of capacitance 10 F is chargedto a potential 50 V with a battery. The batteryis now disconnected and an additional charge200 C is given to the positive plate of thecapacitor. The potential difference across thecapacitor will be1) 50 V 2) 80 V 3) 100V 4) 60 V

10. A capacitor is filled with an insulator and acertain potential difference is applied to itsplates. The energy stored in the capacitor isU. Now the capacitor is disconnected from thesource and the insulator is pulled out of thecapacitor. The work performed against theforces of electric field in pulling out theinsulator is 4U. Then dielectric constant ofthe insulator is1) 4 2) 8 3) 5 4) 3

11. A capacitor of capacitance C is charged to apotential difference V from a cell and thendisconnected from it. A charge +Q is nowgiven to its positive plate. The potentialdifference across the capacitor is now

1) V 2) Q

VC

3) 2

QV

C 4) ,

QV

C ifV CV

12. A parallel plate capacitor with platesseparated by air acquires 1 C of chargewhen connected to a battery of 500V. Theplates still connected to the battery are thenimmersed in benzene [ k = 2.28]. Then acharge that flows from the battery is

1) 1.28 C 2) 2.28 C 3) 1 / 4 C 4) 4.56 C13. An air capacitor with plates of area 1 m2 and

0.01 metre apart is charged with 610 C ofelectricity. When the capacitor is submergedin oil of relative permittivity 2, then the energydecreases by1) 20 % 2) 50 % 3) 60 % 4) 75 %

14. Three uncharged capacitors of capacitiesC1,C2 and C3 are connected as shown in thefigure to one another and the point. A, B andC are at potentials V1,V2 and V3 respectively.Then the potential at O will be

A

C1

C3C2

CB

O1) 1 1 2 2 3 3

1 2 3

V C V C V C

C C C

2) 1 2 3

1 2 3

V V V

C C C

3) 1 2 3

1 2 3

V V V

C C C

4) 1 2 3

1 2 3

VV V

C C C

15. In the given figure the capacitor of plate areaA is charged upto charge q. The ratio ofelongations (neglect force of gravity) in

springs C and D at equilibrium position is

k2 k1

D C1)

1

2

k

k 2) 2

1

k

k

3) 1 2k k 4) 1

2

k

k16. If metal section of shape H is inserted in

between two parallel plates as shown in figureand A is the area of each plate then theequivalent capacitance is

ba1) 0 0A A

a b

2) 0A

a b

3) 0 0A A

a b

4) 0A

a b

17. The equivalent capacitance ABC of the circuitshown in the figure is

A

CC

CCC

CC C

CC

CC

B1)

5

4C 2)

4

5C

3) 2 C 4) C18. A solid conducting sphere of radius 10cm is

enclosed by a thin metallic shell of radius 20cm.A charge q=20 C is given to the inner sphere.The heat generated in the process is1) 12 J 2) 9 J 3) 24 J 4) zero

19. A condenser of capacity 500 F is chargedat the rate of 400C per second. The timerequired to raise its potential by 40V is1) 50 s 2) 100 s 3) 20 s 4) 10 s

20. In the figure shown the effective capacityacross P and Q is (the area of each plate is ‘a’)

d2

d

2

d

P

K2

K3

Q

K1

1) 0 2 31

2 32

a K KK

d K K

2) 0 1 32

1 32 2

a K KK

d K K

3) 0 3 1 2

1 23 2

a K K K

d K K

4) 0 1 1 2

2 32

a K K K

d K K

21. Two capacitors 1 2C F and 2 6C F in

series, are connected in parallel to a thirdcapacitor 3 4C F . This arrangement is thenconnected to a battery of e.m.f.=2 V, as shownin figure. The energy lost by the battery in

99 PINEGROVE


charging the capacitors is C1 C2

C3

2V

1) 622 10 J 2) 611 10 J3)

63210

3J 4)

61610

3J

22. A capacitor is connected with a battery andstores energy U. After removing the battery,it is connected with another similar capacitorin parallel. The new stored energy in eachcapacitor will be

1) 2

U2) U 3)

4

U4)

3

2

U

23. A parallel plate condenser with a dielectric ofdielectric constant K between the plates has acapacity C and is charged to a potential V volts.The dielectric slab is slowly removed frombetween the plates and then reinserted. Thenet work done by the system in this process is

1) 211

2K CV 2) 2 1 /CV K K

3) 21K CV 4) zero24. A fully charged capacitor has a capacitance C.

It is discharged through a small coil ofresistance wire embedded in a thermallyinsulated block of specific heat capacity s andmass m. If the temperature of the block israised by T , the potential difference V acrossthe capacitor is

1)2mC T

s

2)

mC T

s

3)

ms T

C

4)

2ms T

C

25. A parallel plate capacitor of capacity 100 F

is charged by a battery at 50 volts. The batteryremains connected and if the plates of thecapacitor are separated so that the distancebetween them is halved the original distance,the additional energy gives by the battery tothe capacitor in Joules is ......1) 3125 10 2) 312.5 103) 31.25 10 4) 30.125 10

26. The equivalent capacity between the points Aand B in the adjoining circuit will be

C B

A

C

C

C

CC C C

C

1) C 2) 2C

3) 3C 4) 4

27. A parallel plate capacitor with air as mediumbetween the plates has a capacitance of 10 F. The area of the capacitor is divided intotwo equal halves and filled with two mediahaving dielectric constant 1 2K and 2 4K .The capacitance will now be1) 10 F 2) 20 F 3) 30 F 4) 40 F

28. The capacity of a parallel plate condenser withair medium is 60 F having distance ofseperation d . If the space between the plates

is filled with two slabs each of thinckness 2d

and dielectric constants 4 and 8, the effectivecapacity becomes1) 160 F 2) 320 F 3) 640 F 4) 360 F

29. In the adjoining diagram, the condenser C willbe fully charged to potential V if

10S1 S2

C+

–V

5

1) S1 and S

2 both are open

2) S1 and S

2 both are closed

3) S1 is closed and S

2 is open

4) S1 is open and S

2 is closed.

30. The capacity between the point A and B in theadjoining circuit wil be

3

1) 1 2 3 1 2

1 2 3

2

2

C C C C C

C C C

2)

1 2 2 3 3 1

1 2 3

C C C C C C

C C C

3) 1 2 3 2 1 3

1 2 33

C C C C C C

C C C

4)

1 2 3

1 2 2 3 3 1

C C C

C C C C C C 31. The capacitance ABC in the given network

10 F5 F

5 F

5 F10 F

A

B

1) 7 F 2) 50

7F

3) 7.5 F 4) 7

50F

32. In the following circuit; find the potentials atpoints A and B is

A B

10V

–+

1) 10V, 0V 2) 6 V, -4V

3) 4V, -6V 4) 5V, -5V33. The potential difference between the points A

and B in the following circuit in steady statewill be


100 PINEGROVE

100V200

10

CA

3 F

3 F1 F

1 FB

1 F

1) ABV 100 volt 2) ABV 75 volt

3) ABV 25 volt 4) ABV 50 volt34. In the following circuit two identical capacitors,

a battery and a switch(s) are connected asshown. the switch(s) is opened and dielectricof constant K 3 are inserted in thecondensers. The ratio of electrostatic energiesof the system before and after filling thedielectric will be

A S

VB

1) 3: 1 2) 5 : 1

3) 3:5 4) 5 : 335. In the given figure a capacitor of plate area A

is charged upto charge q. The mass of eachplate is m2. The lower plate is rigidly fixed.The value of m1 if the system remains inequilibrium is

m11)

2

20

qm

Ag 2) 2m

3) 2

202

qm

Ag 4) 22m

36. One plate of a capacitor is connected to a springas shown in figure. Area of both the plates isA. In steady state; separation between theplates is 0.8d (spring was unstretched and thedistance between the plates was d, when thecapacitor was uncharged). The force constantof the spring is approximately

E 1)

20

3

4 AEd

2) 02

2 AE

d

3)2

03

6 E

Ad

4)

30

32

AE

d

37. A capacitor is made of a flat plate of area A

and second plate having a stair-like structureas shown in figure. The width of each plate is‘a’ and the height is ‘b’. The capacitance ofthe capacitor is

d

3a

1) 02

3

A

d b

2)

2 2

0 3 6 2

3 2

A d bd b

d b d d b

3)

2 20 2

3 2

A d bd b

d d b d b

4)

2 2

02 2

3 2

A d bd b

d d b d b

38. A parallel plate capacitor of capacitance C isconnected to a battery and is charged to apotential difference V. Another capacitor ofcapacitance 2C is similarly charged to apotential difference 2V. The charging batteryis now disconnected and the capacitors areconnected in parallel to each other in such away that the positive terminal of one isconnected to the negative terminal of theother. The final energy of the configurationis

1) zero 2) 23

2CV 3)

235

6CV 4)

29

2CV

39. Two identical capacitors, have the samecapacitance C. One of them is charged topotential V1 and the other to V2.The negativeends are also connected, the decrease inenergy of the combined system is

1) 2 21 21/ 4C V V 2) 2 2

1 21/ 4C V V3) 2

1 21/ 4C V V 4) 2

1 21/ 4C V V40. Consider the situation shown in the figure.

The capacitor A has a charge q on it whereasB is uncharged. The charge appearing on thecapacitor B a long 7 time after the switch isclosed is: q

+++++++

––––––––

S

A B

1) Zero 2) q /2

3) q 4) 2q41. Find the capacitance of a system of two

identical metal balls of radius a if the distancebetween their centres is equal to b, with b>>a.The system is located in a uniform dielectricwith permittivity K.1) 0 Ka 2) 04 Ka 3) 02 Ka 4) 02/3 Ka

EXERCISE - III KEY1) 2 2) 3 3) 1 4) 1 5) 3 6) 4 7) 3 8) 49) 4 10) 3 11) 312) 1 13) 2 14) 115) 2 16) 4 17) 1 18) 2 19) 1 20) 121) 2 22) 3 23) 4 24) 4 25) 1 26) 227) 3 28) 2 29) 3 30) 1 31) 1 32) 233) 3 34) 3 35) 3 36) 1 37) 2 38) 239) 3 40) 1 41) 3

101 PINEGROVE


EXERCISE - III - HINTS

1..dq c dv

dt dt

2. 1 q cv; /

2 /,

t

d tc Kq C Vc d

; 2 1q q q 3. 2 2

QE CV VF

d

4. 0

1eff

C kC

k ; effq C V

5.' '1 2 2

CEQ Q ; Before the slab is removed

1 2C C and C kC ;1net

kC C

k

'2

2

1

2

Q k

Q k

6. 1 max1 2 max 2V V ,V V 7.

/ 2common2 2

1U C V

2

8. Work done = decrease in energy

ie w = 2

20 01 2

1 11

2 2

A AvE E v

d d k

9. 0 500q CV C

0 0 0 0

700 500

2 2 2 2

q q q q

A A A A

600q C ;

qV

C =60V

10.21

2

qU

C ;

2

0

14

2

qU U

C

2

0

15

2

qU

C ;

0

5C

kC

11.0 0

2Q

CVE

A

;'

0

2Q

CVV Ed

A

d

= 22

QCV Q

VC C

12. 0 0 0Q C V ; 0 0 0Q CV KC V 0 0 01Q Q Q K C V 13.

/ UU

K

14. 1 2 3q q q 1 0 1 0 2 2 0 3 3V V C V V C V V C 1 1 2 2 3 3

01 2 3

C V C V C VV

C C C

15. 1 1 2 2eF k x k x 1 2

2 1

x k

x k

16. Net space between metal plates is a-b

17.

C

CCC C

C

C C

CC

C C

C/2

CC

C C C

3

2

3

2

18.2 2

1 22 2i f

q qH U U

C C

1 0 1 2 0 24 , 4C R C R 19.

.dq c dv

dt dt

20.1

2 31 2 3

2; ;

oo o

AK

K A K AC C C

d d d

2 3

12 3

C CC C

C C

21. 212lost effE C V

22.1 1 2 2

1 2

CV C VV

C C ; 2

1 2

12

U C C V 23. On introduction and removal and again on

introduction, the capacity and potential remainsame. So, net work done by the system in thisprocess.

f iW U U ; 2 21 1

02 2

CV CV 24.

21

2E CV

The energy stored in capacitor is

lost in form of heat energy.H ms T 21

2ms T CV

;

1 2 1 2

2 1

2 1

A B

q q q qC

q qV VC C

1 2 3 1 2

1 2 3

2

2

C C C C CC

C C C

; 2ms T

VC

25.

21 1

1

2E C V ;

22 2

1

2E C V ; 2 12C C

26. Wheat stone’s Bridge

27. 01 22

cc k k 28.

1 20

1 2

2K KC C

K K

29. If S1 and S

2 both are closed then charge and

discharge processes with simoultaneously takeplace. Hence to charge the condenser fully the keyS

1must be closed and S

2 must remain open

30. According to the symmetry of the circuit chargeson two condensers of capacity C

1 will be same and

charges on condensers of capacity C2 will be same.


102 PINEGROVE

+q –q11

q –q21

C1 C2

C3

+q –q22

q –q21

+q2 +q1–q2 –q1

C2 C1

P

A B–+

2 2 1 1

2 3 1

0q q q q

C C C

; 1 2 32

1 2 1 3

C C Cq

q C C C

Capacity of whole circuit

31. 1 2 3 1 2

1 2 3

2

2

C C C C CC

C C C

32. P.D across each condenser = 2V

Potential at earth = 0V ; 6 4A BV V V V 33

2

1 2AB

VCV

C C ;

1100 25

4ABV V 34.

2 2 21

1 1

2 2U CV CV CV

22 2

2

1 1 53 3

2 2 9 3

VU CV C CV

1

2

3

5

U

U

35. 1T m g and 2 eT m g F ; 1 2 em g m g F Here,

2

02e

qF

A

36.2

02

qkx

A ;

2

0

0.82

CEk d d

A , 0

0.8

AC

d

2

03

4 AEK

d

37. In Equivalent circuit C

1, C

2, C

3 are parallel

0 0 01 2 3, ,

3 3 3 2

A A AC C C

d d b d b

38. Net charge 2 1Q Q Q potential is lV

01 0

0

CV V

C C

Similarly after nth operation ; 1 21/ 2E C V

39. Q CV ; 21/ 2iU CV21/ 2fU CV work done = i fU U

40 Due to attraction with positive charge, the negativecharge on capacitor A will not flow through theswitch S.

41.

b a

1 2

a

V V 2V 2 E dr ;

1 2

qC

V V

*******

electro statics & capacitors · 2020-04-27 · t q q q 2 1 2 0 2 2 0 1 4 1 4 q f l q f l s s...

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