KVPY ELECTROSTATS.pdf

Physics for AIEEE Electric Charge and Field

1. ELECTROSTATICS

The branch of physics which deals with charges at rest is called electrostatics.

2. ELECTRIC CHARGE

Charge is scalar physical quantity associated with matter due to which it produces and experiences electrical

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

charged body has excess of electrons while a positively charged body has lost some of its electrons.

+++++++ +++++++++ +++++++++++- - - - - - -+ + + + + +

+ + + + + + - - - - - - -

Electrons = Protons Electrons < Protons Electrons > Protons

Positivelycharged

body

Negativelycharged

body

Properties of Charge

(1) Like charges repel while unlike charges attract each other. The true test of electrification is repulsion

(2) Charge is a scalar.

(3) Charge is always associated with mass :

In charging, the mass of a body changes. If electrons are removed from the body, the mass of the body

will decrease and the body will become positively charged. If electrons are added to a body, the mass ofthe body will increase and the body will acquire a net negative charge. Due to extremely small mass ofelectron (= 9.1 × 10–31 kg) the change in mass of a body due to charging is negligible as compared to the

mass of the body.

M + m > MNeutral

Body = M

++

++

++

++ + + +

++

++

++

M - m < M

ELECTRIC CHARGE AND FIELD

1


(4) Charge is quantised : When a physical quantity has only discrete values the quantity is said to be quantised.

Milikan oil drop experiment established that the smallest charge that can exist in nature is the charge of

an electron. If the charge of an electron (e = 1.6 × 10–19 C) is taken as the elementary unit, i.e., quanta

of charge, the charge on a body will be an integral multiple of e

i.e., q = ± ne with n = 1, 2, .................

(5) Charge is invariant : This means that charge is independent of frame of reference, i.e., charge on a body

does not change with speed. The charge density or mass of a body depends on speed and increases with

increase in speed.

(6) Unit of Charge :

[a] S.I. : coulomb. [1 Coulomb = 1 ampere × 1 second]

[b] C.G.S. : Static coulomb or frankline

1 coulomb = 3 × 109 static coulomb

1 coulomb = 3 × 109 esu of charge =101

emu of charge

[esu = electrostatic unit] [emu = electro magnetic unit]

Practical units of charge are amp × hr (= 3600 coulomb) and faraday (= 96500 coulomb)

3. CHARGING OF A BODY

(a) Friction : In friction when two bodies are rubbed together, electrons are transferred from one body to the other.

This makes one body become positively charged while the other become negatively charged, e.g., when a glass

rod is rubbed with silk, the rod becomes positively charged while the silk is negatively charged. Clouds are also

charged by friction. Charging by friction is in accordance with conservation of charge. The positive and negative

charges appear simultaneously in equal amounts due to transfer of electrons from one body to the other.

(b) Induction : If a charged body is brought near a neutral body, the charged body will attract opposite charge

and repel similar charge present in the neutral body. This makes one side of the neutral body become positively

charged while the other side negative.


Charg

ingbo

dy

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

q'=

0V' =

+ ive

Charg

edbo

dyis

brou

ght n

ear

unch

arge

d body

+++

+++

+

q' = -ive

V' = 0

Uncharged body is

connected to the Earth

+++

+++

+

q' = -iveV' = 0Uncharged body isdisconnected

from the earth

q' = -iveV' = -iveCharging bodyis removed

Charging a body by induction

e

Important Points

(i) Inducing body neither gains nor loses charge.

(ii) The nature of induced charge is always opposite to that of inducing charge.

(iii) Induced charge can be lesser or equal to inducing charge (but never greater) and its maximum value is

K11q'q

where q is the inducing charge and K is the dielectric constant of the material of the uncharged body.

(iv) For metals, K = and so q' = –q i.e., in metals induced charge is equal and opposite to inducing charge.

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

(c) Conduction : When an insulated conductor is brought in contact with a charged body and it gets the same

charge as the charged body then conduction takes place. Conduction is only possible in conductors and not in

insulators.

4. COULOMB'S LAW

The force of attraction or repulsion between two stationary point charges is directly proportional to the product

of charges and inversely proportional to the square of distance between them. This force acts along the line

joining the centre of two charges.

If q1 & q

2 are charges, r is the distance between them and F is the force acting between them

Then, F q1 q

2, F 1/r²

F 221

rqq

or2

21

rqqCF

q1 q2

r


C is const. which depends upon system of units and also on medium between two charges

²C/²Nm1094

1C 9

0

(In SI unit)

C = 1 in electrostatic unit (esu)

0 = 8.85 × 10–12 C²/Nm² = permittivity of free space or vacuum

Effect of medium

The dielectric constant of a medium is the ratio of the electrostatic force between two charges separated

by a given distance in air to electrostatic force between same two charges separated by same distance in that

medium.

Fair

= 221

0 rqq

41

and Fmedium

= 221

r0 rqq

41

rair

medium 1F

F = K

r or K = Dielectric constant or Relative permittivity or specific inductive capacity of medium.

(i) Permittivity : Permittivity is a measure of the ability of the medium surrounding electric charges to allow electric

lines of force to pass through it. It determines the forces between the charges.

(ii) Relative Permittivity : The relative permittivity or the dielectric constant (r or K) of a medium is defined

as the ratio of the permittivity of the medium to the permittivity 0 of free space i.e.

r or

0K

Dimensions of permittivity 2

2

0 lengthFQ

22

22

LMLTAT = M–1 L–3 T4A2

The dielectric constants of different mediums are

Medium Vacuum Air Water Mica Teflon Glass PVC Metal

r

1 1.00059 80 6 2 5-10 4.5

Coulomb's law in vector form

The direction of the force acting between two charges depends on their nature and it is along the line joining

the centre of two charges.


21F = force on q2 due to q

1 12212

21

r021 r

rqq

41F

12F = Force on q1 due to q

2 21221

21

r012 r

rqq

41F

r12

F21F12

q1 q2

2112 FF (as 2112 rr )

or 0FF 2112

5. ELECTRIC FIELD

To explain 'action at a distance', i.e., 'force without contact' between charges we assume that a charge or charge

distribution produces a field in space surrounding it. The region surrounding a charge or charge distribution in

which its electrical effects are perceptable is called the electric field of the given charge. Electric field at a point

is characterised either by a vector function of positionE called electric intensity or by a scalar function of

position V called electric potential. The electric field in a certain space is also visualised graphically in terms

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

field

Electric field IntensityE

The electric field intensity at a point in an electric field is defined as the force experienced

by a unit positive point charge called test charge supposed to be placed at that point. The

test charge does not affect the source charge or charge distribution producing the field. If

a test charge q0 at a point P in an electric field experiences a force

F , then electric field

)q/F(E 0

If the field is produced by a point charge q, then from Coulomb's law

r

rqq

41F 3

0

0

field due to point-charge

q at positionr in free space

20

300 r

q4

1Eorrrq

41

qFE

If field is produced by a charge distribution, then by 'principle of superposition' field

is given as

Source ChargeqO

r

Pq0

q1

r1

P

q2q4

q2


n

1ii21 E.......EEE with

i3

i

i

0i r

rq

41E

while for continuous charge distribution (treating small charge element as a point charge),

,rrdq

41Ed 3

0

i.e.,

r

rdq

41E 3

0

Important points

(1) It is a vector quantity having dimensions13

2

AMLTAT

MLTqFE

The SI unit is N/C or V/m as

mV

mCJ

mCmN

CN VC/JJmN

(2) By definition ,q/FE 0

or

EqF 0

A charged particle in an electric field experiences a force whether it is at rest or in motion. The direction of

force is along the field if it is positive and opposite to the field if it is negative.

+F = +qE

E

F = -qE

E

(3) In free space Electric field is 20

0 rq

41E

In a medium of permittivity field is 2rq

41E

So, K1

EE 0

0

[as =

0 K]

or, E = E0/K

In presence of a dielectric, electric field decreases and becomes 1/K times of its value in free space.

6. ELECTRIC LINES OF FORCE

The idea of lines of force was introduced by Michel Faraday. A line of force is an imaginary

curve the tangent to which at a point gives the direction of intensity at that point and the

number of lines of force per unit area normal to the surface surrounding that point gives

the magnitude of intensity at that point.

A

EB

BE A


Important points

(1) Electric lines of force usually start or diverge out from positive charge and end or converge on negative charge.

+ –

(2) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In SI

units 1/0 shows electric lines associated with unit (i.e., 1 coulomb) charge. So if a body encloses a charge q,

total lines of force or flux associated with it is q/0. If the body is cubical and charge is situated at its centre

the lines of force through each face will be q/60

.

+q

Total lines of force (q/ )while through each face (q/6 )

0

0

Charge A is positive whileB is negative and q > qA B

A B

(3) Lines of force never cross each other because if they cross then intensity at that point will have two directions

which is not possible.

(4) In electrostatics the electric lines of force can never be closed loops, as a line can never start and end on

the same charge. If a line of force is a closed curve, work done round a closed path will not be zero and

electric field will not remain conservative.

(5) Lines of force have tendency to contract longitudinally

like a stretched elastic string producing attraction

between opposite charges and repel each other laterally

resulting in, repulsion between similar charges and

'edge-effect' (curving of lines of force near the edges

of a charged conductor)Electric lines of force for a system

of two positive charges

–

AttractionElectric lines of force for a dipole

+

Repulsion

+ +


(6) If the lines of force are equidistant straight lines the field is uniform and if lines of force are not equidistant

or straight lines or both, the field will be non-uniform. The first three represent non-uniform field while last

shows uniform field.

Magnitude isnot constant

Direction isnot constant

Both magnitude anddirection not constant

Both magnitude anddirection constant

(7) Electric lines of force end or start normally on the surface of a conductor. If a line of force is not

normal to the surface of a conductor, electric intensity will have a component along the surface of

the conductor and hence conductor will not remain equipotential which is not possible as in electrostatics

conductor is an equipotential surface.

Fixed point charge nearinfinite metal plate

(A)

+

d+q

Parallel metal plateshaving dissimilar charges

(B)

+ + –+++++

––––––

E = 0 E = 0

Edge Effect

E = 0

Uniformfield

Parallel metal plateshaving similar charges

+++++++

E = 0

++++++

+

(C)

E = 0

E = 0

(8) If in a region of space, there is no electric field there will be no lines of force. This is why inside a

conductor or at a neutral point where resultant intensity is zero there is no line of force.

(9) The number of lines of force per unit normal area at a point represents magnitude of electric field intensity.

The crowded lines represent strong field while distant lines shows a weak field.

(10) The tangent to the line of force at a point in an electric field gives the direction of intensity. It gives

direction of force and hence acceleration which a positive charge will experience there (and not the direction

of motion). A positive point charge free to move may or may not follow the line of force. It will follow

the line of force if it is a straight line (as direction of velocity and acceleration will be same) and will

not follow the line if it is curved as the direction of motion will be different from that of acceleration.

The particle will not move in the direction of motion or acceleration (line of force) but other than these

which will vary with time as atuv .


7. ELECTRIC-FLUX

Electric flux through an elementary area ds is defined as the scalar product of area and field, i.e.,

cosEdsds•Ed E i.e.,

ds•EE

It represents the total lines of force passing through the given area. Here area is treated as a vector. The direction

of area vector is given by direction of normal to the surface.

Important points

(1) It is a real scalar physical quantity with units volt × m and dimensions

13322

E ATMLLAT

MLTdsqFEds

(2) It will be maximum when cos is max = 1, i.e., = 0°, i.e., electric field is normal to the area with (dE)

max

= E ds

(3) It will be minimum when cos is min = 0, i.e., = 90°, i.e. field is parallel to the area with (dE)

min = 0

(4) For a closed body outward flux is taken as positive while inward flux is taken as negative.

nE

Positive - flux

Body

Negative - flux

En

Enn

Cylinder in a uniform field

ER2E 0E ER2

E

Body n^E


8. GAUSS'S LAW

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

According to it, the total flux linked with a closed surface is 1/0 times the charge enclosed by the closed surface,

Mathematically0s

qds•E

Applications of Gauss law

(1) Electric field due to a line charge : Gauss law is useful in calculating electric field intensity due to symmetrical

charge distributions.

We consider a gaussian surface which is a cylinder of radius r which encloses a line charge of length h with

line charge density .

According to Gauss law ds.E =

0

inq

surfacelCylindrica

ds.E

+surface

circulards.E

I

+surface

circulards.E

II

=0

h

sufacelCylindrica

º0cosEds +

surfacecircular

cosEdsI 2

+

surfacecircular

cosEdsII 2

=

0

h

E(2 r h) =0

h

So E = r2 0

)r1E(

(2) Electric field due to an infinite plane thin sheet of charge :

To find electric field due to the plane sheet of charge at any point P distant

r from it, choose a cylinder of area of cross-section A through the point P

as the Gaussian surface. The flux due to the electric field of the plane sheet

of charge passes only through the two circular caps of the cylinder. Let surface

charge density =

According to gauss law

0in /qdS.E

0surface

lcylindricasurface

circularIIsurfacecircularI

AcosdsEcosdsEcosdsE or EA + EA + 0 =

02A

or E =02

r

E

+ +++++

+ ++++++++

EP

E Q

rPlane sheetof charge

GaussianSurface


(3) Electric field intensity due to uniformly charged spherical shell :

We consider a thin shell of radius R carrying a charge Q on its surface

(i) at a point P0 outside the shell (r > R)


1S0 ds.E =

0

Q or E0 (4r2) =

0

Q

E0 = 20r4

Q =

0

2

2

rR

where the surface charge density = areasurfaceeargchtotal

= 2R4Q

The electric field at any point outside the shell is same as if the entire charge is concentrated at centre of

shell.

(ii) at a point Ps on surface of shell (r = R)

ES = 20R4

Q =

0

(iii) at a point Pin inside the shell (r < R)


2Sds.E =

0

inq

As enclosed charge qin = 0

So Ein = 0

The electric field inside the spherical shell is always zero.

(4) Electric field intensity due to a spherical uniformly charge distribution :

We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributedthroughout the volume.

The charge density = volumetotaleargchtotal

=3R

34

Q

= 3R4

Q3

(i) at a point P0 outside the sphere (r > R)

according to gauss law ds.E0 =

0

Q or E0 (4r

2) =0

Q

or E0 = 20r4

Q =

03

2

3

rR

E = 0E = 0

distance from centre (r)

E = Q/4 Rmax 02

E

O r < R r = R r > R

2r

1E

2r

1E

ds

+

R

O

Pin PS P0

++

++ +

++ +

+

+ +

++++

+ + ++++

r


(ii) at a point Ps on surface of sphere (r = R)

Es = 20R4

Q =

03

R

(iii)at a point Pin inside the sphere (r < R)


ds.E in =

0

inq =

0

1 .

34r3 = 3

0

3

RQr

Ein(4r2) = 30

3

RQr

or Ein = 30R4

Qr =

03

r (Ein r)

SOLVED EXAMPLES

Ex. How many electrons are present in 1 coulomb charge.

Sol. q = ne q = 1C e = 1.6 × 10–19 C So n = q/e = 6.25 × 1018 electrons.

Ex. A copper sphere of mass 2.0 g contains about 2 × 1022 atoms. The charge on the nucleus of each atom is 29e.

(i) How many electrons must be removed from the sphere to give it a charge of +2µC?

(ii) Determine the fraction of electrons removed.

(iii) Is there any change in mass of sphere when it is given positive charge?

Sol. (i) Number of electrons to be removed 19

6

106.1102

eQn

= 1.25 × 1013

(ii) Total number of electrons in the sphere = 29 × 2 × 1022 = 5.8 × 1023

Fraction of electrons removed11

23

13

1016.2108.51025.1

Thus 2.16 × 10–9 % of electrons are to be removed to give the sphere a charge of 2µC.

(iii) Yes mass decreases, when body is given a positive charge.

Decrease of mass m = 9 × 10–31 × 1.25 × 1013 = 1.125 × 10–17 kg

Ex. Consider four equal charges placed on the corners of a square with side a. Determine the magnitude and direction

of the resultant force on the charge on lower right corner.

O r < R r = R r > R

E

2r

1E 2r

1E

Er

Er


Sol. The forces on the charge on lower right corner due to charges 1, 2, 3 are F1 = kq²/a², F

2 = kq²/a², F

3 = kq²/2a².

The resultant of F1 and F

2 is

90cosFF2FFF 2122

2112 = 22 a/kq2 .

This is in the direction parallel to F3. Therefore the total force on the said charge is F = F

12 + F

3

221akq

21F 2

2

The direction of F is, 45° below the horizontal line.

Ex. Three identical spheres each having a charge q and radius R, are kept in such a way that each touches theother two. Find the magnitude of the electric force on any sphere due to other two.

Sol. For external points a charged sphere behaves as if the whole of its charge was concentrated at its centre.

Force on A due to B is

BAalongR4q

41

R2qq

41F 2

2

02

0AB

Force on A due to C.

CAalongR4q

41

R2qq

41F 2

2

02

0AC

Now, angle between BA and CA is 60° and FAB

= FAC

= F.

F360cosFF2FFF 22A

2

0 Rq

43

41

Ex. Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 30° witheach other. When suspended in a liquid of density 0.8 gm/cc, the angle remains the same. What is the dielectric

constant of the liquid? ( = 1.6 gm/cc is the density of the sphere)

Sol. The forces acting on each ball are tension T, weight mg and electric force F, for its equilibrium along vertical,

T cos = mg

and along horizontal T sin = F

Dividing we havemgFtan ....(1)

When the balls are suspended in a liquid of density and dielectric constant K, the electric force will become(1/K) times, i.e., F' = (F/K) while weight mg' = mg – Th = mg – Vg [as Th = Vg]

i.e.

1mg'mg

mVas

So for equilibrium of ball, /1Kmg

F'mg

'F'tan ....(2)

According to given problem ' = ; so from eqn. (1) & (2), we have 28.06.1

6.1K


Ex. An infinite plane of positive charge has a surface charge density . A metal ball B of mass m and charge qis attached to a thread and tied to a point A on the sheet PQ. Find the angle which AB makes with the planePQ.

Sol. Due to positive charge the ball will experience electrical force Fe = qE horizontally away from the sheet while

the weight of the ball will act vertically downwards and hence if T is the tension in the string, for equilibrium

of ball:

Along horizontal, T sin = qE

And along vertical, T cos = mg

So tan =mgqE

and T = [(mg)2 + (qE)2]1/2

The field E produced by the sheet of charge PQ having charge density is

02E

So,mg2

qtan0

i.e.,

mg2qtan0

1

Ex. A point charge q is placed at one corner of a cube of edge a. What is

the flux through each face of the cube?

Sol. At a corner, 8 cubes can be placed symmetrically, flux linked with each

cube due to a charge q at the corner will be q/80

For the faces passing through the edge A, electric field E will be parallel

to area of face and so flux through these three faces will be zero.

As the cube has six faces and flux linked with three faces (through A) is

zero, so flux linked with remaining three face will be (q/80). The remaining

three faces are symmetrical so flux linked with each of the three faces passing through B will be,

00

q241q

81

31

.

Ex. Flux entering a closed surface is 2000 V-m. Flux leaving that surface is 8000 V-m. Find the charge inside surface.

Sol. Net flux = out

– in = (8000 – 2000) = 6000 V-m

0

q so q = (6000) (8.85 × 10–12) = 0.53 µC

T

A+ P

Q

mg

qE qB

+

+

+

+

+

Z

X

YL

A q

B


1. When two bodies A and B are rubbed with eachother, A gets a positive charge. In this process –(1) Protons are transferred from A to B

(2) Protons are transferred from B to A

(3) Electrons are transferred from A to B

(4) Electrons are transferred from B to A

2. Choose the correct option from following –(1) If a particle possesses mass, then it must posses

charge

(2) If a particle possesses charge, then it mustposses mass

(3) If a particle possesses mass, then it must beuncharged

(4) Both charge and mass for a particle cannot bezero

3. Five balls, numbered 1 to 5, are suspended usingseparated threads. Pairs (1, 2), (2, 4), (4, 1) showelectrostatic attraction, while pairs (2, 3) and (4, 5)show repulsion, therefore ball 1–

(1) must be positively charged

(2) must be negatively charged

(3) may be neutral

(4) must be made of metal

4. An isolated solid metallic sphere is given +Qcharge. The charge will be distributed on thesphere–

(1) uniformly but only on surface

(2) only on surface but non uniformly

(3) uniformly inside the volume

(4) non uniformly inside the volume

5. A soap bubble is given a negative charge then itsradius–

(1) decreases

(2) increases

(3) remains unchanged

(4) nothing can be predicted as information is insufficient

EXERCISE – 1(A)6. A sure test of electrification is–

(1) attraction (2) repulsion

(3) friction (4) induction

7. Mark correct option or options–

(1) like charged bodies always repel each other

(2) like charged bodies always attract each other

(3) like charged bodies may attract each other

(4) none of the above

8. Dimensions of mgke2

is same as that of (here

k = Coulomb’s constant), e is charge of electron, mis mass of proton and g is acceleration due togravity –(1) Area (2) Pressure(3) Energy (4) Volume

9. Two small balls with like charges are suspended bylight strings of equal length L from the same point.When taken to a place where they are in a state ofweightlessness the separation between the balls willbe-

(1) 2L (2)L2

(3)L L( )12

(4) L L( )1

10. A, B, C are three identical small metal balls havingcharges q, –3q and q respectively. When A and Care placed at a certain distance apart electrostaticforce between them is F. If B is touched with A andthen removed, then magnitude of electrostatic forcebetween A and C will be

(1) 4F

(2) F

(3) 2F

(4) 2F

11. Three charge +4q, Q and q are placed in a straightline of length l at points distance 0, l/2 and lrespectively. What should be the value of Q in orderto make the net force on q to be zero ?

(1) –q (2) –2q

(3) –q/2 (4) 4q


12. Two similar very small conducting spheres havingcharges 40 C and –20C are some distance apart.Now they are touched and kept at same distance.

The ratio of the initial to the final force betweenthem is–(1) 8 : 1 (2) 4 : 1

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

13. Charges +Q, +Q and –Q are placed on the verticesA, B, C of a triangle ABC respectively. The side ofequilateral triangle is a. Magnitude of force acting oncharge at A is –

(1) 2

2

akQ

(2) 32

2

akQ

(3) Zero (4) 2

22akQ

14. Two small spheres, each of mass 0.1 gm andcarrying same charge 10

–9 C are suspended by

threads of equal length from the same point. If the

distance between the centres of the sphere is 3 cm.Then find out the angle made by the thread with thevertical.

(1)

50

1tan 1

(2)

100

1tan 1

(3)

150

1tan 1

(4)

200

1tan 1

15. Two identical conducting spheres (of negligibleradius), having charges of opposite sign, attract eachother with a force of 0.108 N when separated by

0.5 meter. The spheres are connected by aconducting wire, which is then removed (whencharge stops flowing), and thereafter repel each

other with a fore of 0.036 N keeping the distancesame. What were the initial charges on thespheres?

(1) ± 1 × 10–6

C, ± 1 × 10–6

C

(2) ± 1 × 10–6

C, ± 3 × 10–6

C

(3) ± 3 × 10–6

C, ± 1 × 10–6

C

(4) ± 3 × 10–6

C, ± 3 × 10–6

C

16. Ten positively charged particles are kept fixed onthe x axis at points x = 10 cm, 20 cm, 30 cm, …,100 cm. The first particle has a charge 1.0 ×1.0

–8C,

the second 8 × 10–8

C, the third 27 × 10–8

C and soon. Find the magnitude of the electric force actingon a 1 C charge placed at the origin.

(1) 2 × 105 N (approx) (2) 3 × 10

5 N (approx)

(3) 4 × 105 N (approx) (4) 5 × 10

5 N (approx)

17. One brass plate is inserted between two charges.The force between two charges will–(1) remain the same (2) increases

(3) decrease (4) fluctuate

18. Electric lines of force can never start from –(1) Proton (2) Conductor

(3) Insulator (4) Electron

19. For the following electric lines of forces, choose thecorrect option –

A

D

B C

(1) EB is maximum (2) EC is maximum

(3) EA > ED (4) EB = EC

20. Two point charges exert on each other a force F whenthey are placed r distance apart in air. When they areplaced R distance apart in a medium of dielectricconstant K, they exert the same force. The distanceR equals-

(1)rK (2)

rK

(3) rK (4) r K21. Two charges q1 and q2 are placed in vacuum at a

certain distance apart and the force acting betweenthem is F. If a medium of dielectric constant 3 isintroduced between them, then the force acting onq1 becomes / remains–

(1) 3F

(2) 32F

(3) F (4) 3F


22. What is dimensional formula of dielectric constant –

(1) M

–1 L3 T–2 A–2 (2) M–1 L–3 T4 A2

(3) M0 L0 T0 (4) M L T–2

23. Dielectric constant of mica is –

(1) One (2) Less than one

(3) More than one (4) Infinite

24. Two point charges placed at a distance r in air exerta force F on each other. The value of distance Rat which they experience force 4F when placed ina medium of dielectric constant K = 16 is–

(1) r (2) r/4

(3) r/8 (4) 2r

25. A proton and an electron are placed in a uniformelectric field-

(1) the electric forces acting on them will be equal

(2) the magnitudes of the forces will be equal

(3) their accelerations will be equal

(4) the magnitudes of their accelerations will be equal

26. On putting salt [NaCl] in air a force F acts betweensodium and chlorine ions at a distance of1 cm from each other. The permittivity of air and thedielectric constant of water are 0 and Krespectively, when a piece of salt is placed in waterthen the force between Na

+ and Cl

–1 ions separated

by a distance of 1 cm will be-

(1) F (2)0

FK

(3)0K

F (4) K

F

27. Two charges of opposite nature having magnitude 10µC are 20 cm apart. The electric field at the centreof line joining these charges will be-

(1) 9 x 106 N/C in the direction of positive charge

(2) 18 x 106 N/C in the direction of negative charge

(3) 18 x 106 N/C in the direction of positive charge

(4) 9 x 106 N/C in the direction of negative charge

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

–Q

A

+2Q

B

(1) somewhere between –Q and 2Q

(2) somewhere on the left of –Q

(3) somewhere on the right of 2Q

(4) somewhere on the right bisector of line joining –Q and 2Q

29. Two point charges +1C and – 4C are placed atpoints having x co-ordinates x = a and x = 3arespectively on x-axis. Neutral point is at-

(1) x = 0 (2) x = –a

(3) x = 2a (4) x =53a

30. The electric field midway between two charges 0.1Cand 0.4 C separated by a distance of 60 cm is-

(1) 5 103 N/C (2) 9 104 N/C

(3) 5 104 N/C (4) 3 104 N/C

31. Two point charges Q and –3Q are placed certaindistance apart. If the electric field at the location of Q

beE , then that at the location of –3Q will be-

(1) 3E (2) 3

E

(3)E / 3 (4)

E / 3

32. In the situation shown the string is ideal and insulating.Mass of the particle is 1mg, It’s charge will be(g = 10ms–2)

45o E = 10 N/C2

(1) 1C (2) 10nC

(3) 100nC (4) 1mC


33. 1C charge when placed at a point in electric fieldexperiences 0.01N force. At the same point it 1mCcharge is placed it will experience force-

(1) equal to 0.01N

(2) equal to 10N

(3) slightly less than 1N

(4) slightly less than 10N

34. An isolated sphere of radius 1cm is placed in airmaximum charge can be given to it without dielectricbreakdown of the surrounding air is-

(1)103

C (2)1003nC

(3)1003

C (4)103nC

35. How many electrons must be added to a sphericalconductor of radius 10cm to produce a field of2 10–3 N/C just above surface?

(1) 1.39 104 (2) 1.6 105

(3) 3 104 (4) 9.1 104

36. Which of the following graph best represents thevariation of electric field intensity due to a chargedsolid sphere of copper?

(1)

rO

E

(2)

rO

E

(3)

rO

E

(4)

rO

E

37. Along x axis at position x = a, x = 2a, x = 4a,x = 8a and so on, charges of strength q areplaced. The electric field at origin will be –

(1) 234

aKq

(2) 254

aKq

(3) 243

aKq

(4) 22aKq

38. A simple pendulum has a length l, mass of bob m.The bob is given a charge q coulomb. Thependulum is suspended in a uniform horizontalelectric field of strength E as shown in figure, thencalculate the time period of oscillation when the bobis slightly displace from its mean position is–

E

l

q m1

(1)g

l2 (2)

mqE

g

l2

(3)

mqE

g

l2 (4)

22

mqE

g

l2

39. Three point charges q0 are placed at three cornersof square of side a. Find out electric field intensityat the fourth corner.

(1) 20

a

kq

2

12

(2) 2

0

a

kq

2

12

(3) 20

a

kq

2

12

(4) 2

0

a

kq

2

11

40. Two insulating spheres of radii 2 cm and 4 cm haveequal volume charge density. The ratio of electricfield on the surfaces of the spheres will be –(1) 1 : 2 (2) 4 : 1

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


41. The maximum electric field intensity on the axis ofa uniformly charged ring of charge q and radius Rwill be–

(1) 20 R33

q

4

1

(2) 20 R3

q2

4

1

(3) 20 R33

q2

4

1

(4) 20 R32

q3

4

1

42 A circular ring of radius a carries a total charge Qdistributed uniformly over its length. A small lengthdL of the wire is cut off. Find the electric field atthe centre due to the remaining wire.

(1) 30

2 a2

Qdl

(2) 30

2 a4

Qdl

(3) 30

2 a6

Qdl

(4) 30

2 a8

Qdl

43. A point charge q and a charge –q are placed at x= –a and x = +a respectively. Which of thefollowing represents a part of E-x graph?

(1)

E

–a O +a x (2)

E

–aO

x

(3)

E

+aOx

(4) all of these

44. A particle of mass m and charge q starts moving fromrest along a straight line in an electric field E = E

0 – ax

where a is a positive constant and x is the distancefrom starting point. Find the distance travelled by theparticle till the moment it came to instantaneous rest-

(1)2 0Ea

(2)Ea0 (3)

E qm0 (4)

Eq0

45. A very long charged rod is placed along y = –xstraight line as shown in figure, it carries chargeper unit length. Electric field at point P is–

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

45º

y (m)

x (m)

(0, 6)P

(1)012

(2) 22 0

(3)012

2

(4)06


1. Dimensions of pE where p is electric dipole momentand E is electric field –(1) MLT–2 (2) ML2T–2

(3) MLT–1 (4) ML–1T–2

2. If

1E and

2E are electric field at two points onequatorial line at distance r and 2r from short dipole,then E1/E2 –

(1) 41

(2)1

8

(3) 81

(4) 11

3. Due to an electric dipole shown in figure, the electricfield intensity is parallel to dipole axis :

YQ

–q +q Px

equitarial

(1) at P only (2) at Q only

(3) both at P and at Q (4) neither at P nor at Q

4. Three charges are arranged on the vertices of anequilateral triangle as shown in figure. Find the dipolemoment of the combination.

–q2qd

–q

(1) qd (2) 3qd

(3) 2qd (4) qd22

5. An electric dipole is placed (not at infinity) in an electricfield generated by a point charge–

(1) the net electric force on the dipole must be zero

(2) the net electric forceon the dipole may be zero

(3) the torque on the dipole due to the field must be zero

(4) the torque on the dipole due to the field may be zero

6. An electric dipole consists of two charges + q at aseparation 2a. It is placed in such a way that it’scentre coincides with origin and dipole momentvector is directed towards + x axis. Electric fieldintensity magnitude at x = 2a is equal to-

(1)kqa2 3 (2)

23

kqa

(3)89 2

kqa

(4)kqa2 2

7. If an electric dipole is kept in a uniform electricfield, then it will experience –(1) Force only

(2) Torque only

(3) Force and torque

(4) No force and no torque8. An electric dipole is placed along the X-axis at the

origin O. A point P is placed at a distance of 20 cm

from this origin such that OP makes an angle3

with

the X-axis. If electric field at P makes an angle with X-axis, the value of is-

(1) 3

(2) 1 3tan

3 2

(3) 23

(4)23tan 1

9. Find the magnitude of the electric field at the pointP in the configuration shown in figure for d >> a.Take 2qa = p.

P

–q +q

a

+q

a

d

(1)22

30

pqd4

1

(2)222

30

pdqd4

1

(3)222

30

pdqd4

1

(4)22

20

pqd4

1

EXERCISE – 1(B)


10. An electric dipole of moment p

is placed at the origin

along the x-axis. The angle made by electric field withx-axis at a point P, whose position vector makes an

angle with x-axis, is (where tan = tan2

1)–

(1) (2)

(3) + (4) + 2

11. A point charge q and a dipole of dipole moment pare placed at some distance as shown in figure. Theforce on dipole is –

q p x

y

r

(1) 32

rkpq

(2) 3rkpq

(3) 34

rkpq

(4) 32

rkpq

12. An electric dipole having dipole moment 2 10–6 Cm isplaced in uniform electric field having magnitude 103

N/C. The dipole can experience maximum torque -

(1) 2 × 10–3 Nm (2) 1 × 10–3Nm

(3) 4 × 10–3 Nm (4) zero

13. A small electric dipole of dipole moment p isperpendicular to electric field. Work done to rotateit through angle 180º in uniform electric field E is –(1) pE (2) 2pE

(3) –2pE (4) Zero

14. An electric dipole consists of two opposite chargeseach of magnitude 1.0 C separated by a distanceof 2.0 cm. The dipole is placed in an external fieldof 1.0 × 10

5 N/C. The maximum torque on the dipole

is–

(1) 0.2 × 10–3

N-m (2) 1.0 × 10–3

N-m

(3) 2.0 × 10–3

N-m (4) 4.0 × 10–3

N-m

15. The force between two short electric dipolesseparated by a distance r is directly proportional to–

(1) r2

(2) r4

(3) r–2

(4) r–4

16. A short electric dipole of dipole moment p is placedat origin with its dipole moment directedalong x-axis. The value of force experienced by a particlehaving charge –q0, placed at (0, a) is

(1) 30

0

24

1apq

along positive x-axis

(2) 30

0

24

1apq

along negative x-axis

(3) 30

041

apq

along negative x-axis

(4) 304

1apq

along positive x-axis


1. Find out the electric flux through an area 10 m2

lying in XY plane due to an electric field

k5j10i2E

.

(1) 25 Nm2/C (2) 50 Nm

2/C

(3) 75 Nm2/C (4) 100 Nm

2/C

2. In a uniform electric field E if we consider an imagi-nary cubical close gaussian surface of side a, thenfind the net flux through the cube?

(1) 0 (2) Ea2

(3) 2 Ea2

(4) 6 Ea2

3. If electric field is uniform, then the electric linesof forces are–(1) divergent (2) covergent(3) circular (4) paraller

4. A surface S = j10 is kept in an electric field E =

k7j4i2 . How much electric flux will comeout through the surface ?(1) 40 unit (2) 50 unit(3) 30 unit (4) 20 unit

5. In a region of uniform electric field E, a hemispheri-cal body is placed in such a way that field is parallelto its base (as shown in figure). The flux linked withthe curved surface is–

–qO

C

E

–q

(1) zero (2) ER– 2

(3) ER 2 (4) E2

R 2

6. A square of side 'a' is lying in xy plane such thattwo of its sides are lying on the axis. If an electric

field kxEE 0

is applied on the square. The flux

passing through the square is–

(1) 30 aE (2)

2

aE 30

(3)3

aE 30 (4)

2

aE 20

7. Figure (a) shows an imaginary cube of edge L/2. Auniformly charged rod of length L moves towardsleft at a small but constant speed v. At t = 0, the leftend just touches the centre of the face of the cubeopposite it. Which of the graphs shown in figure (b)represents the flux of the electric filed throught thecube as the rod goes through it ?

v

LL/2

(a) (b)

flux

ab

c

d

time

(1) a (2) b

(3) c (4) d

8. The electric field in a region is given by E =

,iax where = constant of proper dimensions.

What should be the charge contained inside acube bounded by the surface, x = l, x = 2l,y = 0, y = l, z = 0, z = l ?

(1) 3

0l(2) 3

0l

(3) 30

l

(4) 3

0l2

9. If the electric flux entering and leaving a closedsurface are respectively of magnitude 1 and 2,then the electric charge inside the surface willbe–

(1)0

12

(2) 021 )(

(3) )( 120 (4) )( 120

10. Consider the charge configuration and a spherical

Gaussian surface as shown in the figure. Whencalculating the flux of the electric field over thespherical surface, the electric field will be due

to–

EXERCISE - 1(C)


q2

+ q1

– q1

(1) q2

(2) only the positive charges

(3) all the charges

(4) + q1 and – q1

11. A point charge Q is placed at the centre of ahemishphere. The electric flux passing through flatsurface of hemisphere is–

(1)0

Q

(2) zero

(3)02

Q

(4) none of these

12. A point charge Q is placed at the cnetre of ahemisphere. the ratio of electric flux passing throughcurved surface and plane surface of the hemisphereis–(1) 1 : 1 (2) 1 : 2

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

13. Select the wrong statement–(1) the electric field calculated by Gauss's law

is the field due to the charges inside the Gaussiansurface

(2) the electric field calculated by Gauss's lawis the resultant field due to all the chargesinside and outside the closed surface

(3) the Gauss's law is equivalent to Coulomb'slaw

(4) the Gauss's law can also be applied to calculategravitational field but with some modifications

14. A charge q is placed outside a hemisphere. Fluxthrough curved surface as shown in the figure–

q

(1)02

q(2)

02

q

(3) Zero (4)02

q

15. A point charge q is held, just below the centre ofcurvature of a hemispherical surface as shown infigure. The value of electric flux passing through thesurface is

(1)0

qCentreof Curvature

q

(2)02

q

(3) More than02

q but less than

0q

(4) Less than02

q

16. A charge Q is placed at the centre of an imaginaryhemispherical surface. Using symmetry argumentsand the Gauss's law, Find the flux of the electricfield due to this charge through the surfaces of thehemisphere–

Q

(1)0

Q

(2)02

Q

(3)0

Q2

(4)04

Q

17. A point charge Q is placed on the axis of a cone as

shown in figure. If flux lined to curved surface is ,then what is flux links to base of cone –

Q

(1) 0

q(2)

0q

(3) Zero (4)0

q


18. The electric field intensity at a distance r from aninfinite sheet of charge with surface charge density is-

(1)0

(2)02

(3)0

2

2

(4) r2 0

2

19. If the electric field due to an infinite long line chargeat distance 1m from it is 1 N/C, the charge per unitlength of line charge is –(1) 0 (2) 0(3) 20 (4) 40

20. A point charge Q is placed at the centre of a cir-cular wire of radius R having charge q. The forceof electrostatic interaction betwene poin charge andthe wire is –

q

Q

O

(1) 20R4

qQ

(2) zero

(3) R4

q

0

2

(4) none of these

21. The electric flux passing through the sphere, if anelectric dipole is placed at the centre of a sphere,is–

(1)0

1

(2)0

2

(3) zero (4) none of these22. Two parallel charged plates have a charge density

of + and –. Net force on proton located outsidethe plates at some distance will be–

(1) e0

(2) e0

2

(3) e02

(4) Zero

23. Electric charges are distributed in a small volume.The flux of the electric field through a sphericalsurface of radius 10 cm surrounding the total chargeis 25 V-m. The flux over a concentric sphere ofradius 20 cm will be–(1) 25 V-m (2) 50 V-m(3) 100 V-m (4) 200 V-m

24. Eight point charges (can be assumed as smallspheres uniformly charged and their centres at thecorner of the cube) having values q each are fixedat vertices of a cube. The electric flux throughsquare surface ABCD of the cube is–

A B

CDq

q

q

qq

q

q

q

(1)024

q

(2)012

q

(3)06

q

(4)08

q

25. The charge density of a spherical charge distributionis given by:

r nr r nr n

b g FHGIKJ

RS|T|

0

0

what is the total charge on the distribution?(1) 2 n3

0(2) n3

0

(3) 4/3 n30

(4) 1/2 n30

26. T w o s u r f a c e s S 1 and S2 enclose some charges asshow in figure. What is ratio of flux linked tosurfaces S1 and S2 –

Q2Q

–2QS1

S2

(1) 23

(2) 13

(3) 31

(4) 31


27. Two identical metal plates are having charges Q1

and Q2 (Q1 > Q2). They are placed at distance d,

the electric field in between the plates is–(1) Proportional to Q1

(2) Proportional to Q2

(3) Proportional to Q1 – Q2

(4) Proportional to Q1 + Q2

28. It a negatively charged particle is placed near an

infinitely extended conducting plane, it will-

(1) remain at rest there

(2) start moving parallel to plane

(3) get repelled by the plane

(4) get attracted by the plane

29. Charge Q is uniformly distributed over a triangle

(equilateral) made from thin conducting wire each

of length L. A gaussian sphere is taken with its

centre at one vertex and passing through centroid of

the triangle. Net electric flux linked with the sphere

is-

(1)Q

3 0(2)

23 0

Q

(3)2

3 3 0

Q (4) zero

30. An electron moves along a metal tube with variable

cross-section. The velocity of the electron when it

approaches the neck of tube, is–

v0

(1) greater than v0 (2) equal to v0

(3) less than v0 (4) not defined

31. A square of side ‘a’ parallel to xy-plane is shown in

figure. Co-ordinate of vertex P of square is

(0, 0, a). A point charge qis placed at origin O then

flux crossing the square is –

O

x

y

R

Q

P

S

z

a

(1)06

q(2)

024q

(3)04

q(4)

02q

32. Electric field at point P is given by E

= 0Er

. Thetotal flux through the given cylinder of radius R andheight h is–

P

O

r

(1) hRE 20 (2) hRE2 2

0

(3) hRE3 20 (4) hRE4 2

033. A charge Q is placed at a distance of 4R above

the centre of a disc of radius R. The magnitudeof flux through the disc is . Now a hemisphericalshell of radius R is placed over the disc suchthat it forms a closed surface. The flux throughthe curved surface taking direction of area vectoralong outward normal as positive, is–

4R

Q

R

(1) zero (2)

(3) – (4) 2


One or more correct choice type–

1. A point charge q is placed at origin. Let BA E,E

and CE

be the electric field at three points AA

(1,2,3), B(1, 1, –1) and C(2,2,2) due to chargeq. Then–

(1) BA EE (2) CA EE

(3) CB E4E

(4) CB E8E

2. A particle of charge q and mass m moves rectilinearlyunder the action of an electric field E = –x. Here, and are positive constants andx is the distance from the point where the particlewas initially at rest. Then–(1) the motion of the particle is oscillatory

(2) the amplitude of the particle is

(3) the mean position of the particle is at x=

(4) the maximum acceleration of the particle is

m

q

3. 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 the directionas shown. Block is released with the spring inunstretched position–

E

k

q, m

Smooth

(1) block will execute SHM

(2) time period of oscillation isk

m2

(3) amplitude of oscillation isk

qE

(4) block will oscillate but not simple harmonically

4. A positively charged thin metal ring of radius R isfixed in the x-y plane with its centre at the origin O.A negatively charged particle P is released fromrest at the point (0, 0, z0) where z0 > 0. Then themotion of P is–

(1) periodic for all values of z0 satisfying 0 < z0 <.

(2) simple harmonic for all values of z0 satisfying 0< z0 R

(3) approximately simple harmonic provided z0<<R

(4) such that P crosses O and continues to movealong the negative z-axis towards z = –

5. A nonconducting solid sphere of radius R is uni-formly charged. The magnitude of the electric fielddue to the sphere at a distance r from its centre–

(1) increases as r increases for r < R

(2) decreases as r increases for 0 < r <

(3) decreases as r increases for R < r <

(4) is discontinuous at r = R

6. The electric field intensity at a point in space isequal in magnitude to–

(1) magnitude of the potential gradient there

(2) the electric charge there

(3) the magnitude of the electric force, a unit charge would experience there

(4) the force, an electron would experience there

7. Figure shows a charge q placed at the centre of ahemisphere. A second charge Q is placed at one ofthe positions A, B, C and D. In which position(s) ofthis second charge, the flux of the electric fieldthrough the hemisphere remains unchanged?

B

C A

D

q

(1) A (2) B

(3) C (4) D

EXERCISE – 2


8. An electric dipole is placed at the centre of a sphere,mark the correct options.

(1) the flux of the electric field through the sphereis zero

(2) the electric field is zero at every point of thesphere

(3) the electric field is not zero anywhere on thesphere

(4) the electric field is zero on a circle on the sphere

9. An oil drop has a charge –9.6 × 10–19 C and has a

mass 1.6 × 10–15

gm. When allowed to fall, due toair resistance force it attains a constant velocity.Then if a uniform electric field is to be appliedvertically to make the oil drop ascend up with thesame constant speed, which of the followign arecorrect (g = 10 ms

-2) (Assume that the magnitude

of resistance force is same in both the cases)

(1) the electric field is directed upward

(2) the electric field is directed downward

(3) the intensity of electric field is12 C.N10

3

1

(4) the intensity of electric field is15 NC10

6

1

10. An electric dipole is kept in the electric field pro-duced by a point charge.

(1) dipole will experience a force

(2) dipole can experience a torque

(3) dipole can be in stable equilibrium

(4) it is possible to find a path (not closed) in thefield on which work required to move the dipole iszero.

COMPREHENSION BASED QUESTIONS

I. Find out electric field intensity due to unfiormlycharged solid nonconducting sphere of volumec h a r g e d e n s i t y and radius R at following points :

11. At a distance r from the surface of sphere (inside)-

(1) r3

PR

0(2) r

3

)rR(P

0

(3)0

RP

(4)03

RP

12. At a distance r from the surface of sphere (out-side)-

(1) r)Rr(

RP2

0

3

(2) r)Rr(2

RP2

0

3

(3) r)Rr(3

RP2

0

3

(4) r)Rr(4

RP2

0

3

II. Two identical small spheres are suspended from thesame point by threads 1m long. Each sphere is givena charge 120 nC. Consequently, they repel eachother to a distance 40 cm.

13. Find the mass of each sphere–

(1) 400 mg (approx) (2) 500 mg (approx)

(3) 600 mg (approx) (4) 700 mg (approx)

14. Find the tension in each thread–

(1) 2.1 × 10–3

N (2) 3.1 × 10–3

N

(3) 4.1 × 10–3

N (4) 5.1 × 10–3

N

15. Find the angle between the two threads in the con-dition of equilibrium–

(1) sin–1

(1) (2) sin–1

(2)

(3) sin–1

(3) (4) sin–1

(4)

III. A thin rod of length L carries a positive charge thatis uniformly distributed over tis length. Linear chargedensity of the distribution, i.e. charge per unit lengthis . P is a point at a

P

45°

r

Q A BO

Ld

distance ‘r’ from the midpoint of rod and on itsperpendicular bisector. Q is a point along the axis ofthe rod and located at distance ‘d’ from one end.As shown, the line that joins P with any end of therod makes an angle 45° with the perpendicular bi-sector. It appears that we could obtain electric fieldstrength at various points with the help of Gauss’s

Physics for AIEEE Electric Charge and Fieldlaw. It is possible to consider a cylinder of radius ras the Gaussian surface with AB along its axis.Obviously, P will lie on the curved surface of thiscylinder. But field strength at different points on thecurved surface will be different owing to lack ofsymmetry of the distribution for various points. ThusGauss’s law will not be practically useful. Similarly,we may imagine a cylinder with AB along its axissuch that Q lies on one of its circular ends butGauss’s law is again of little practical use. In sucha situation we can obtain field strength by integrat-ing the field strength due to a small element usingsuitable limits. (Take electric potential = 0, at )

16. Let in figure point O be treated as origin and lengthof rod along x axis so that P lies on y axis. xcomponent of electric field strength at P will be–

(1) r2 0

(2) r83.2 0

(3) zero (4) r

L

4

1

0

17. y-component of electric field strength at P is –

(1) r2 0

(2) r83.2 0

(3) zero (4) r

L

4

1

0

18. Electric field strength at Q is (perhaps it could bemore conveninent to shift the origin to Q for thepurpose of field and potential calculation at Q)–

(1) )dL(d

L

4

1

0

(2) )dL(

L

4

1

0

(3) )2/Ld(

L

4

1

0

(4) zero

Assertion and Reasons

(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

(2) Statement-1 is true, Statement-2 is true; Statement-2is not a correct explanation for Statement-1.

(3) Statement-1 is true, Statement-2 is false.

(4) Statement-1 is false, Statement-2 is true.

19. A : A point charge q is placed at centre of sphericalcavity inside a spherical conductor as shown. An-other point charge Q is placed outside the conduc-tor as shown. Now as the point charge Q is pushedaway from conductor, the potential difference(VA – VB) between two points A and B within thecavity of sphere remains constant.

q

BA Q

R : The electric field due to charges on outer sur-face of conductor and outside the conductor is zeroat all points inside the conductor.

20. A : A solid uncharged conducting cylinder moveswith acceleration a (w.r.t. ground). As a result ofacceleration of cylinder, an electric field is producedwithin cylinder.

a

solid conducting cylinder

R : When a solid conductor moves with accelera-tion a, then from frame of conductor a pseudoforce(of magnitude ma; where m is mass of electron)will act on free electrons in the conductor. As aresult some portion of the surface of conductoracquires negative charge and remaining portion ofsurface of conductor acquires positive charge.

21. A : A charge q is placed at the center of a metallicsheel as shown in figure. Electric field at point P onthe shell due to charge q is zero.

P

q

R : Net electric field in a conductor under electrostaticconditions is zero.


22. A : If there exists attraction between two bodies,both of them may not be charged.

R : Due to induction effects a charged body canattract a neutral body.

23. A : When charges are shared between two bodies,there occurs no loss of charge, but there does occura loss of energy.

R : In case of sharing of charges energy of conser-vation fails.

24. A : On going away from a point charge or asmall electric dipole, electric field decrease at thesame rate in both the cases.

R : Electric field is inversely proportional to squareof distance from a point charge.

KVPY ELECTROSTATS.pdf

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