Electric Charges and Fields - 1 File Download

Level - I

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

SECTION - A

School/Board Exam. Type Questions

Very Short Answer Type Questions :

1. An electrostatic field line cannot be discontinuous. Why?

Sol. An electrostatic line is a continuous curve, because tangent to it at any point represents the direction in which

a test charge kept at that point will experience force. It cannot have sudden breaks, because no abrupt force

acts on a test charge.

2. The distance of the point on the equatorial plane of a small electric dipole is halved. By what factor will the

electric field due to the dipole at the point charge?

Sol.3

1E

r . Since distance is halved so electric field E will become 8 times.

3. In an electric field an electron is kept freely. If the electron is replaced by a proton, what will be the relationship

between the forces experienced by them?

Sol. F = qE, so force on electron –

eF eE� �

and force on proton pF eE� �

–

e pF F� �

4. Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?

Sol. When and�

�

p E are parallel, dipole remains at stable equilibrium.

5. Figure shows three point charges, +2q, –q and +3q. Two charges +2q and –q are enclosed within a surface S.

What is the electric flux due to this configuration through the surface ‘S’?

+2q

–q

+3q

S

Sol. Electric flux = (charge enclosed)/0 =

0

q

Chapter 1

Electric Charges and Fields

Solutions (Set-1)

2 Electric Charges and Fields Solutions of Assignment (Set-1) (Level-I)


6. A metallic sphere is placed in uniform electric field as shown in the figure. Which path is followed by electric field

lines and why?

a

b

c

d

Sol. Line d is correctly drawn, because it is not passing through conductor and it is perpendicular to the surface

from where it starts or meets the conductor.

7. When two electrically charged particles having charges of different magnitude are placed at a distance d from

each other, they experience a force of attraction F. These two particles are put in contact and again placed at the

same distance from each other. What is the nature of new force between them?

Sol. In such cases final force will always be repulsive.

8. Can a charged body attract an uncharged body?

Sol. Yes. By induction.

9. An electric dipole of dipole moment p�

is placed in a uniform electrostatic field E�

. For what angle between p�

and

E�

will the potential energy of the electric dipole be half of its maximum value?

Sol. – cos2

pEpE = 120º

10. What is the line of symmetry of a dipole field?

Sol. About axial line.

11. Find the value of electric field that would exactly balance the weight of electron.

Sol.mg

Ee

= 5.67 × 10–11 N/C (m = 9.1 × 10–31 kg and e = 1.6 × 10–19 C)

12. A small test charge is released at rest at a point in an electrostatic field. Will it travel along the field line

passing through that point?

Sol. If electric line is straight, then it will move on the line. If line is curved then charge will move tangential to it.

13. If Coulomb’s law involved 3

1

r

dependence 2

1instead of

r

, where r is distance between two point changes

would Gauss’s law still be true?

Sol. No

14. A glass rod is rubbed with silk. Will its mass increase or decrease?

Sol. Mass will decrease because it loses electrons.

3Solutions of Assignment (Set-1) (Level-I) Electric Charges and Fields


15. Repulsion is the sure test of electric charge. Explain

Sol. Because if two bodies repel, then they surely have charges and that too of similar nature.

Short Answer Type Questions :

16. A system has two electric charges qA = 2.5 × 10–7C and q

B = –2.5 × 10–7 C located at points A(0, 0, –15 cm)

and B(0, 0, +15 cm) respectively. Calculate the electric dipole moment of the system. What is its direction?

Sol. Magnitude of dipole moment p = ql = (2.5 × 10–7C)(30 × 10–2 m).

= 7.5 × 10–8 Cm

Its direction is from –ve to +ve charge i.e., along positive z-axis.

17. Three point charges of +2 C, –3 C and –3 C are kept at the vertices A, B and C respectively of an

equilateral triangle of side 20 cm as shown in figure. What should be the sign and magnitude of the charge

to be placed at the mid-point (M) of side BC so that the charge at A remains in equilibrium?

2 C

–3 C–3 CB C

A

M

Sol. It should be positive. Resultant force on A by charges at B and C are along AM��

. Force by charge at M should

have a repulsive force on A along MA.

18. The flux of the electrostatic field through the closed spherical surface S is found to be four times that through

the closed spherical surface S. Find magnitude of charge Q. Given : q1 = 1C, q

2 = –2 C and q

3 = 9.854 C

q1

q2

1 m

q3

SS

2 m

Q

Sol.1 2 3

1 2 3

1

4

q q q

Q q q q

Q = 3(q1 + q

2 + q

3)

19. A charge of 17.7 × 10–4C is distributed uniformly over a large sheet of area 200 m2. Calculate the electric

field intensity at a distance 20 cm from it in air.

Sol.0 0

2 2

QE

A

=

–4

–12

17.7 10

2 200 8.85 10

55 10 N/C



20. Two similar and equally charged identical metal spheres A and B repel each other with a force of 2 × 10–5 N.

A third identical uncharged sphere C is touched with A and then placed at the mid-point between A and B.

Calculate the net electric force on C.

Sol.1 2

� �

F F 2

2

04

QF

d

BA

Q QdF

1F

2

2

2

3 4 2 2 2

0 0

·1 –2 2

–4 4

2 2

Q QQ

QF F F

dd d

� � �

BA d/2 CF

1F

2

d/2

Q/2F4 F

3

QQ/2

= –5| | 2 10 NF F

�

21. Four point charges of qA = 2 C, q

B = –5 C, q

C = 2 C and q

D = –5C are located at the corners of a square

ABCD of side 10 cm. Find the force on a charge of 1 C placed at the centre of the square.

qB

qA

qD q

C

q = 1 C

Sol. By symmetry, net force on central charge will become zero.

22. A charged wire AB of length l is placed inside a sphere, as shown in figure. Linear charge density of wire is

= kx, where x is the distance measured along the wire from end A and K is constant. Calculate the flux

through the sphere.

BAx

l

Sol.0

Q

where 2

0 02

l lkl

Q dx kx dx

2

0 02

Q kl

23. Two plane sheets of charge densities + and – are kept in air, as shown in figure. What are the electric

field intensities at points A and B?A

B+–

Sol. Consider vertically upward direction as positive y-axis

E

�

at A = 0 0

ˆ ˆ–2 2

j j

0+

–B

A

E

�

at B = 0 0

ˆ ˆ– (– )2 2

j j

0

ˆ–

j



24. A charge particle of charge +q, mass m and moving with a velocity of ûi enters a uniform electric field of

strength Ê Ej�

. Find magnitude of velocity and magnitude of displacement of the particle after time t.

Sol. Initial velocity of the charge û ui�

acceleration of the charge ˆ

F qE qEa j

m m m

� �

�

velocity at time t

v u at ��

= ˆ ˆ

qEtui j

m

∴ Magnitude of velocity =

2

2

qEtV u

m

Displacement at time t,

21

2S ut at �

��

= 21ˆ ˆ( )

2

qEut i t j

m

Magnitude of displacement = 2

22 1

2

qEtS ut

m

25. A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of 100 C/m2.

Calculate the

(i) Charge on the sphere.

(ii) Total electric flux coming out a Gaussian surface just enclosing the outer surface of the sphere.

Sol. (i) Surface charge density 2

4

Q

R

Q = ·4R2

= 100 × 10–6 2

2

C 224 (2.5)

7m

= 7.9 × 10–3C = 7.9 mC

(ii) Net flux coming out of the sphere = 0

charge enclosed

⇒

0

Q

26. A positive point charge (+q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines

originating from the point on to the surface of the plate.

Sol. +q

Conductingplate



27. A spherical conducing shell of inner radius r1 and outer radius r

2 has a charge Q. Another charge q is placed

at the centre of the shell.

(a) What is the surface charge density on the

(i) Inner surface?

(ii) Outer surface of the shell?

(b) Derive the expression for the electric field at a point x > r2 from the centre of the shell.

Sol. (a) (i) Let the charge on the inner surface of spherical shell is q0. Applying Gauss law to Gaussian surface

S1, intermediate to outer and inner surface.

1

0

0

·

s

q qE ds� ��

� S2

S1

q1

q

q0

x

r2

r1

⇒ 0

0

0q q

[ 0E �

because S1 is inside metal]

q0 = –q

Surface charge density on inner surface 1 =

2

1

–

4

q

r

(ii) Let the charge on the outer surface is q1

Applying conservation of charge q1 + q

0 = Q

q1 = Q – q

0 = Q + q

Surface charge density on outer surface 2 =

2

24

Q q

r

(b) Applying Gauss law to the Gaussian surface S2 at distance x from centre

2

0

·

S

Q qE ds

� ��

�

⇒ E·4x2 = 0

Q q

2

04

Q qE

x

28. Show that the electric field at the surface of a charged conducting sphere is given by 0

ˆ

E n�

, where is

the surface charge density and ˆn is a unit vector normal to the surface in the outward direction.

Sol. If is the surface charge density on the sphere, then charge on the sphere,

Q = × 4R2 = 4R2

Consider a Gaussian surface S coincident with the outer

surface of metal sphere.

Applying Gauss law to the surface

++++++

+++++++++++

++++

++++++

+++++++++++

++++++

–

–

–

–

–

–

–

– –

–

–

–

–

–

–

–

–

– –

–

–

–

–

–

–

–

– –

–

–

–

–

–

–

–

–

– –

R

S

dsE

in

0

·

S

QE ds

� ��

�

0

cos0

S

QE ds

�



0

S

QE ds

�

2

0

4Q

E R

2

004

QE

R

Electric field is directed radially outward so

0

ˆn

E�

29. Two small identical electric dipoles AB and AC, each of dipole moment p are kept at an angle of 120º as

shown in figure. What is the resultant dipole moment of this combination? If this system is subjected to electric

field E�

directed along +x direction, what will be the magnitude and direction of torque acting on this?

C

A

B

y

x120º

Sol. Resultant dipole moment

x

D

C

E

B

A

p2

p1

30º

30º

p

y

1 2p p p � � �

Using parallelogram law of vector addition, resultant dipole moment

p�

is directed along 30º with positive x-axis (∵ BAD = DAC = 60º)

Magnitude of 2 2

1 2 1 22 cos120p p p p p

�

= 2 2 2 1

2 –2

p p p

= p

Electric field is acting along positive x-axis

So angle between E�

and p�

= 30º

Torque on the dipole, p E �

��

= pEsin (Normally into the paper (i.e., along negative z-axis))

= pEsin30

= 2

pE normally into the paper or along negative z-axis.



30. Charges of magnitudes 2Q and –Q are located at points (a, 0,0) and (4a, 0, 0). Find the ratio of the electric

flux due to these charges through concentric spheres of radii 2a and 5a centered at the origin.

Sol. Electric flux through sphere S1 of radus 2a

1

1·

S

E ds � ��

�

in

0 0

2Q Q

Electric flux through sphere S2 of radius 5a

2Q

S1

S2

x a = 5x a = 4

–Qx a= –2

x a= –5

x a = 2

x a = x = 0

1

2·

S

E ds � ��

�

in

0 0 0

2 –Q Q Q Q

Long Answer Type Questions :

31. (i) State Coulomb’s law and write the formula of electrostatic force between two charges, separated by certain

distance. Under which condition this formula is applicable?

(ii) Two positive point charges which are 0.1m apart repel each other with a force of 18 N. If the sum of the

charges be 9 C, calculate their separate values of charges.

Sol. (i) Scalar form:

Coulomb’s law is a quantitative statement about the force between two point charges. Coulomb measured

the force between two point charges and found that it varied inversely as the square of the distance between

the charges and was directly proportional to the product of magnitude of the two charges and acted along the

line joining the two charges.

If two point charges Q1 and Q

2 at rest are separated by a distance r in vacuum, the magnitude of force

between them is given by 1 2

2

k QQF

r . The constant k is usually put as

0

1

4k

, where

0 is called the

permittivity of free space and has the value 0 = 8.854 × 10–12 C2/Nm2. For all practical purposes we will take

9 2 2

0

19 10 Nm / C

4

. The choice of k determines the size of the unit of charge. SI unit of charge is

defined to be 1C. So 1C is the charge that when placed at a distance of 1 m from another charge of the same

magnitude in vacuum experience an electrical force of repulsion of magnitude 9 × 109 N.

Coulomb’s law is strictly applicable for point charges.

(ii) Using Coulomb’s law, 1 2

2

kQQF

r

or, 9 –12

1 2

2

9 10 1018

(0.1)

QQ

Q1Q

2 = 20, again Q

1 + Q

2 = 9

Q1 = 4 C and Q

2 = 5 C



32. (i) Express Coulomb’s law in vector form.

(ii) Two point charges q1 = 5 × 10–6 C and q

2 = 3 × 10–6 C are located at (3, 5, 1) m and (1, 3, 2) m. Find

12F

��

and 21

F

��

using vector form of Coulomb’s law.

Sol. (i) Coulomb’s law in vector form

Since force is a vector, Coulomb’s law in the vector notation will be written as follows. Let the position vector

of charges q1 and q

2 be

1r

��

and 2r

��

respectively (figure). We denote force on q1 due to q

2 by

12F

��

and force or

q2 due to q

1 by

21F

��

.

q2

q1

r 1 r 2

F12

F21

r

r

r

2

2

1

= –

O

y

x

The two point charges q1 and q

2 have been numbered 1 and 2 and the vector leading from 1 to 2 is denoted by

21 2 1, . ., –r i e r r

��

. In the same way, the vector leading from 2 to 1 is denoted by 12r

��

. So, 112 2 21

– –r r r r ��

. The

magnitude of the vectors 21r

��

and 12r

��

is denoted by r21

and r12

, respectively and they are equal, i.e., r12

= r21

.

To denote the direction from 1 to 2 (or from 2 to 1), we define the unit vectors:

21

21

21

ˆ

r

r

r

��

and 12

12

12

ˆ

r

r

r

��

where 21 12ˆ ˆ–r r

So Coulomb’s force law between two point charges q1 and q

2 located at

1r

��

and 2r

��

is then expressed as

1 2

21 212

0 21

1ˆ·

4

q qF r

r

��

.

The equation above is valid for any sign of q1 and q

2 whether positive or negative. If q

1 and q

2 are of the same

sign (either both positive or both negative), 21

F

��

is along 21ˆr , which denotes repulsion, as it should be for like

charges. If q1 and q

2 are of opposite signs,

21F

��

is along 21 12ˆ ˆ–r r , which denotes attraction, as expected

for unlike charges. Thus, we do not have to write separate equations for the cases of like and unlike charges.

Above equation takes care of both cases correctly

F21

F12

r21

(Like charges , > 0)i.e. q q1 2

F21

F12

r21

(Unlike charges , < 0)i.e. q q1 2

q2

q1

Also note that the force 12F

��

on charge q1 due to charge q

2 is obtained from above equation by simply

interchanging 1 or 2, i.e., 1 2

12 12 212

0 12

1ˆ –

4

q qF r F

r

��

Thus, Coulomb’s law agrees with the Newton’s third law.



(ii)12 1 2

–r r r� � �

= ˆ ˆ ˆ ˆ ˆ ˆ3 5 – 3 2i j k i j k

= ˆ ˆ ˆ2 2 –i j k

1 2

12 122

12

ˆ

kq qF r

r

��

= 1 2

123

12

kq qr

r

�

�

=

9 –6 –6

3

2 2 2 2

ˆ ˆ ˆ9 10 5 10 3 10 (2 2 – )N

(2 2 1 )

i j k

= –3

27 5 10 ˆ ˆ ˆ2 2 – N27

i j k

= ˆ ˆ ˆ10 10 – 5 Ni j k

Again 12 21

–F F� �

21ˆ ˆ ˆ– 10 10 – 5 NF i j k

�

33. (i) Show that path of a charge particle projected normal to an uniform electrostatic field is a parabola.

(ii) A particle of mass m and charge +q is thrown at a speed u against a uniform electric field E. How much

distance will it travel before coming to rest?

Sol. (i) Consider a charge +q of mass m is projected with speed u normal to uniform electric field as shown in

figure.

Electric field, ˆ–E Ej

�

Initial velocity, û ui�

acceleration, –

ˆqE

a jm

�

x90º

um+q

y

EInstantaneous position of charge, is

21

2r ut at � ��

= 21 –ˆ ˆ( )2

qEut i j t

m

=

2

ˆ ˆ( ) –2

qEtut i j

m

= ˆ ˆ–xi yj



x-co-ordinate of position x = ut x

tu

y-co-ordinate of position y = –2

2

qEt

m

= –

2

2 4

qE x

m

2

22

qEy x

mu = – cx2

Above equation represent equation of parabola, so path of the charge is a parabola.

(ii)+q

muv = 0E

S

Initial velocity of charge ˆ–u ui

�

acceleration of charge ˆ ˆqE qEi qEi

am m m

�

�

, displacement ˆ–S Si

�

Instantaneous velocity v at displacement S�

2 22 ·v u a S

�

�

2 ˆ ˆ0 2 ·(– )qE

u i Sim

0 = 2

2 –qES

um

2

2

muS

qE

34. (i) Derive an expression for electric field due to a point charge at a distance r from the charge and express

it in vector form.

(ii) Four identical point charges of 4 C each are placed at the corners of a square of each side 0.1m.

Calculate the electric field at the centre of the square.

(iii) Calculate the electric field intensity at the centre, when one of the corner charges is removed.

Sol. (i) Consider a charge Q at orgin O, as shown in figure

PQ ErF

q0O

Electric field at point P, at position vector r�

due to charge Q at origin.



0

FE

q

�

�

(q0 = test charge at P)

=

0

2

0

20 0

ˆ

ˆ4

4

Qqr

r Qr

q r

ˆr indicates the electric field is radially outward from charge Q.

(ii)A

E

�

, B

E

�

, C

E

�

and D

E

�

are electric fields by charges at corners A, B, C and D respectively. O is equidistant

from all charges

So EA = E

B = E

C = E

D4 C

4 C4 C

4 C B

CD

A 0.1 m

0.1 m 0.1 m

0.1 m

EB E

A

ED

EC

0

and –A C

E E� �

and –B D

E E� �

Net electric field at O,

0A B C D

E E E E E � � � � � �

(iii) If all charges are present

0A B C D

E E E E � � � � �

–

A B C DE E E E � � � �

If charge at D is absent, then electric field will be

–

A B C DE E E E � � � �

BE

�

= 2

04

Q

r along OD��

=

9 –69 10 4 10

0.1

2

= –436 2 10 N/C

35. (i) Two charges of a dipole –4 C and +4 C are placed at the points A(1, 0, 4)m and B(2, –1, 5)m located

in an electric field ˆ0.20 V/cmE i�

. Calculate the torque acting on the dipole.

(ii) A Gaussian surface is shown in figure. Charges q1 and q

2 are inside and charge q

3 is outside the surface.

Indicate the charges which contributes the electric field appearing in the formula in

0

·

qE ds

��

�.

q1

q2

q3



(iii) The electric field in a region is given by 0 0

3 4ˆ ˆ

5 5E E i E j �

with 3 –1

02 10 NC .E Find the flux of this field

through a rectangular surface of area 0.2 m2 parallel to y-z plane.

Sol. (i) Dipole moment

2p q l �

�

= –64 10 C AB

��

= –6 ˆ ˆ ˆ4 10 C (2 –1) (–1– 0) 5 – 4i j k

= –6 ˆ ˆ ˆ4 10 – Cmi j k

–6ˆ ˆ ˆ4 – 10 Cmp i j k

�

Electric field 2ˆ ˆ0.2 V/Cm, 0.2 10 V/mE i i �

p E �

��

= –6 2ˆ ˆ ˆ ˆ4 10 – (0.2 10 ) Nmi j k i

= –5 ˆ ˆ8 10 Nmk j

(ii) E is due to all the charges q1, q

2 and q

3.

(iii)0 0

3 4ˆ ˆ

5 5E E i E j �

2 ˆ0.2A m i�

(Area vector is normal to surface)

·E A � �

= 0

30.2

5E

= 33

0.2 2 10 Vm5

= 240 Vm

36. Derive an expression of electric field due to uniformly charged infinite plane sheet.

Sol. Z

Y

EE

21

Let be the uniform surface charge density on the sheet.



We take the X axis normal to the given plane. By symmetry the electric field will not depend on Y and Z

coordinates and its direction is always parallel to the X-coordinate. We can take the Gaussian surface to be

a rectangular parallelopiped of cross-section area A. The electric flux is non zero for surface 1 and 2 as shown

in figure and for other surfaces the flux is zero.

Flux = 0

in

q

, by Gauss’s law

So 2 EA = 0

A

E = 02

37. Three charges of same magnitude q are placed at the corners of an equilateral triangle of side length a. Find

the net force on any one of charges.

Sol. Net force on charge is = 60cos22

0

2

0

2

0 FFF

60°

q

qq a

a a

F F0

F0

where F0 = 2

2

04

1

a

q

net force F = 03 F = 2

2

04

3

a

q

38. Find electric field inside a uniformly charged thin spherical shell having charge Q. If a charge q is placed at

the centre of this shell, how the electric field will change?

Sol. For thin spherical shell charge is always on the outer surface.

r

Q

+

+

+

+

+

+

+

+

+

+

+

+

O

Consider a Gaussian spherical surface at distance r

0

in·

q

SdE

0

in24

q

rE

but qin = 0

E = 0

but when we put a charge at the centre

qin = q

E = 0

2

in

4 rq



39. Derive an expression for torque acting on a dipole of dipole moment p , when it is placed in uniform external

field E�

.

Sol. Electric field is in upward direction as shown in figure force on positive charge is in upward direction and that

on negative charge is in downward direction.

qE

– q

a

a

+ q

qEE

Magnitude of torque = qE × 2a sin= 2qaE sin

Its direction is normal to the plane of the paper coming out

of it.

The magnitude of Ep is also pEsin and its direction is

normal to the paper, coming out of it

Therefore, in vector form the torque can be given as

Ep

40. (a) Explain the invariant property of charge.

(b) Draw the electric field lines for a dipole.

Sol. (a) The magntidue of charge on a body is independent of its velocity, it is invariant for all frames of reference

in relative motion. This is not always true for every scalar.

(b)–q+q

41. Derive an expression for electric field inside a solid non-conducting sphere of charge Q, when the charge is

distributed uniformly. Also calculate electric field at the centre of sphere and electric field on the surface of

sphere.

Sol. R is radius of sphere and we want to calculate field at r. Consider a spherical Gaussian surface at distance r

from the centre. By symmetry electric field at this closed surface is always radially outwards.

Flux through this surface is

0

24

inq

rE ...(1)

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+ Or

R

charge Q is distributed uniformly

charge per unit volume = 3

3

4R

Q



charge in sphere of radius r is

qin =

3

3

4

4 3

3

Qr

R

qin =

3

3

R

rQ

Put value of qin in equation 1

E4r2 = 3

0

3

R

Qr

E = 304

1

R

Qr

At the centre r = 0 E = 0

On the surface, r = R 2

04 R

QE

42. Derive the expression for electric field on the axis of an electric dipole.

Sol. Let the point P be at distance r from the centre of the dipole on the side of the charge q as shown in figure.

Then

2a

Pr

+ qp– q

qE–

�

= par

qˆ

)(4

–2

0

where pˆ is the unit vector along the dipole axis (from –q to +q). Also

qE

�

= par

qˆ

)–(4 2

0

Total field

E

�

= E+q

+ E– q

= par

arqˆ

)–(

4

4 2220

For r >> a

E

�

= 3

0

4ˆ

4

qap

r

3

0

2ˆ

4

pp

r



43. What is meant by the statement that the electric field of a point charge has spherical symmetry whereas that

of a linear charge of large length is cylindrically symmetrical?

Sol. Consider a charge q at the centre of a sphere of radius r. The magnitude of electric field at all points on the

surface of the sphere is given by

E = 204

1

r

q

.

So the electric field due to a point charge is spherically symmetric.

In the case of a line charge of large length the magnitude of electric field at a distance r from the line is given

by

E = r02

, where is linear charge density.

Now, imagine a cylinder of radius r drawn with the line charge as axis. The electric field, due to the line charge,

at all points on the surface of the cylinder will be the same. So the electric field due to the linear charge has

cylindrical symmetry.

44. (a) A copper sphere of mass 2 g contains nearly 2 × 1022 atoms. The charge on the nucleus of each atom is

29 e. What fraction of the electrons must be removed from the sphere to give it a charge of +2.9 C?

(b) What is an electric line of force? What is its importance?

Sol. (a) Total number of electrons = 29 × 2 × 1022

The number of electrons removed = 19–

6–

106.1

109.2

Fraction of electrons removed = 2219–

6–

10229106.1

109.2

= 3.125 × 10–11

(b) An electric line of force is an imaginary straight or curved line drawn in such a way that the tangent on it

gives the direction of electric force experienced by a positive charge at all points.

The tangent at a point on an electric line of force also gives the direction of the electric field at that point.

The relative closeness of electric lines of force in a certain region provides us an estimate of the electric

field strength in that region.

45. (a) Two infinite parallel planes have uniform charge densities +and –. What is the electric field between

the planes?

(b) How can you charge an ‘uncharged insulated conductor’ negatively by electrostatic induction?

Sol. (a) At a point P between the two planes, the electric fields due +ve and –ve charged planes are respectively

0

–

0 2;

2

EE both are in the same direction.

–P

Net field = 000 22

.



(b) First of all, bring a positively charged glass rod near the given conductor AB. The end A of the conductor

will be charged negatively while the end B will be charged positively as shown in figure.

+ +

+ +

+

+

+

– –

–

A B

– –

+

+

Keeping the glass rod near the conductor, the conductor is earthed as shown in figure.

+ +

+ ++ +

– –

–

A

– –

– – –

– –

–

A B

– –

–

– –

–

Finally, the glass rod and earth-connection are removed. The conductor AB acquires a negative charge.

SECTION - B

Model Test Paper

1. Is the force between two point electric charges q1 and q

2 kept at some distance apart in air, attractive or repulsive

when (i) q1q2 > 0, (ii) q

1q2 < 0?

Sol. (i) Repulsive, like charges repel each other.

(ii) Attractive, unlike charges attract each other.

2. Name the physical quantity which has its S.I. unit volt-metre

Sol. Electric flux, = EA

= 2V

m Vmm

3. If F is the magnitude of force experienced by a charge of 2C at a distance of 1 cm from an infinitely large charged

sheet, then what will be the force experienced by the same charge placed at a distance of 2 cm from the same

sheet?

Sol. Same force, 0

ˆ

2F qE q n

� �

. It is independent of distance.

4. A charge Q is placed at the centre of a cube. What is the electric flux coming out from any one face?

Sol. Total flux through cube = 6(flux through each surface) = 0

Q

6Q

0

6

Q



5. What is the angle between the directions of electric fields due to a dipole, at a point on axial line and a point on

equatorial line?

Sol. 180º ; as, 3

0

2

4a

pE

r

�

�

and 3

0

–

4e

pE

r

�

�

6. What is an electric field line? sketch field lines due to two equal positive charges placed at a small distance

in air.

Sol. It is an imaginary line drawn in electric field such that tangent drawn to it at a point represent the direction

of electric field at that point or tangent drawn to it at a point represent the direction of force on a positive test

charge at that point

7. An electric dipole is free to move in a uniform electric field. Explain its motion when it is placed (i) parallel to the

field, and (ii) perpendicular to the field.

Sol. (i)

+q–q

qE–qE E

When the dipole is placed parallel ( = 0º) to electric field as shown in figures then net force = – 0qE qE � � �

Net torque sinpE

= 0

So the dipole will remain stationary.

(ii)

+q

–q

qE

–qE

E90º

When dipole is kept perpendicular ( = 90º) with electric field, as shown in figure

Net force = (– )0qE qE� �

But the forces qE�

and –qE�

produce a couple as a result it start rotate clockwise about axis normal to

E

�

and P�

as shown in figure.



8. Two point charges of +10 C and +40 C respectively are placed 12 cm apart. Find the position of the point, where

electric field is zero.

Sol.E

1E

2

x 12 – x1 2

12 cm

10 C 40 C

1E

�

electric field due to 10 C and 2

E

�

is due to 40 C. At equilibrium 1 2

0E E � � �

1 2

E E� �

2 2

10 40

(12 – )x x

12

cm3

x

x = 4 cm

9. Four point charges are placed at the four corners of a square in two ways (i) and (ii) as shown in figure. Will the

electric field at the centre of the squares be the same or different in the two configurations and why?

–q –q

–q +q –q

+q–q

+q(i) (ii)

Sol. (i)0 A B C D

E E E E E � � � � �

= A C B DE E E E � � � �

–q –q

+q–q

EA

EB

ED

O

B

CD

A

EC

= 2 · –A B D

E E E� � �

(ii)0 A B C D

E E E E E � � � � �

–q +q

–q+q

EA

EB

ED

O

B

CD

A

EC

= A C B DE E E E � � � �

= 0 0 – and –A C B D

E E E E � � � � � �

= 0�

10. Calculate the total charge enclosed by a closed surface, if the number of electric field lines entering it is 10,000

and leaving is 20,000.

Sol. Flux = Net number of lines leaving = 20,000 – 10,000 = 10,000

Q = 0(10,000) = 104 × 8.854 × 10–12C = 8.854 × 10–8C



11. Three charges are placed on the vertices of an equilateral triangle of side length l. Find the sign and magnitude of

charge q placed at centroid so that system of charges will be in equilibrium.

+Q

q

+Q+Q

Sol. For system of charges to be equilibrium net force on all individual charges q and Q are zero.

From symmetry force on the charge q,

0

FC

FA

FB

q

A B

CQ

QQ

= 0A B C

F F F � � � �

(as FA = F

B = F

C and angle between them are 120º each)

so, q is at equilibrium.

Consider the charge at corner B

Force on charge at corner B = 0

0

0A C

F F F � � � �

0–

A CF F F � � �

q should be negative

Now 0 A CF F F

� �

C

A B 30º30º

FC

F0

q

O

Q

Q

Q FA

F + A

FC

2

| |2 cos30

3

A

kQ qF

l

2

2 2

3 | | 2 3

2

kQ q kQ

l l

| |3

Qq

–

3

Qq

12. Write two basic properties of charge.

Sol. (i) Additivity of charges

If a system contains two point charges q1 and q

2, then the total charge of the system is obtained simply by

adding algebraically q1 and q

2, i.e., charges add up like real numbers or they are scalars like the mass of a

body. Charge has magnitude but no direction, similar to the mass. However there is one difference between

mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative.

Proper signs have to be used while adding the charges in a system.



(ii) Charge is conserved

The total charge of the isolated system is always conserved. It is not possible to create or destroy net charge

carried by any isolated system although the charge carrying particles may be created or destroyed in a

process. For example, a neutron is unstable and after few minute it turns into a proton and an electron. The

proton and electron thus created have equal and opposite charges and the total charge is zero before and

after the creation.

(iii) Quantization of charge

All experiments so far have shown that all free charges are integral multiples of a basic unit of charge

denoted by e. Thus charge Q on a body is always expressed by Q = ne, where n is any integer, positive or

negative.

This basic unit of charge is the charge that an electron or proton carries. By convention, the charge on an

electron is taken to be negative, therefore charge on an electron is written as –e and that on a proton as +e.

The fact that electric charge is always an integral multiple of e is termed as ‘quantization of charge’. The

quantization of charge was first suggested by the experimental laws of electrolysis discovered by Michael

Faraday. It was experimentally demonstrated by Millikan in 1912. In the international system (SI) of units, a

unit of charge is called a ‘coulomb’ and is denoted by the symbol C.

13. What is the relation between electric field intensity and the force? What determines the direction of the force in an

electric field?

Sol. F qE� �

. Direction of force is determined by direction of E�

and nature of charge.

14. Electric field in the given figure is directed along +x direction and given by E = 50x + 10, where E is in NC–1 and x

is in meter. Calculate, (i) the electric flux through the cube (ii) the net charge enclosed within the cube

y

x

z10 cm

Sol. (i)A

1

E

z

x

D

B

A

G

F

O

C

y

E1

E2

A2

Here –1ˆ(50 10) NCE x i �

is tangential to surfaces BCFG, CFED, ADEO and ABGO. So electric flux

through these surfaces is zero 0BCFG CFED ADEO ABGO .

Electric field at surface OEFG, 1

ˆ ˆ(50 0) 10 10 N/CE i i �

and area of OEFG

–2 2 2 –2 2

1ˆ ˆ(10 10 ) – m –10 mA i i

�



2

–1 2 –1

1 1

Nm· –10 –0.1Nm C

COEFG

E A � �

Electric field at ABCD, –1

250 0.1 10 NCE i

�

= –1ˆ15 NCi

Area of ABCD, –2 2 2 –2 2

2ˆ ˆ(10 10 ) m 10 mA i i

�

–2 2 –1 2 –1

2 2· 15 10 Nm C 0.15 Nm C

ABCDE A � �

Net flux through cube = ABCD OEFG BCFG CFED ADEO ABGO (0.15 – 0.1)Nm2C–1

= 0.05 Nm2C–1 = 5 × 10–2 Nm2C–2

(ii) Q = 0 = 8.85 × 10–12 × 5 × 10–2 = 4.425 × 10–13 C

15. Define the term electric dipole moment, and derive an expression for electric field intensity due to a dipole

on its axial line.

Sol.2a

+q–qp

The dipole moment vector p�

of an electric dipole is defined as 2p aq�

; i.e., it a vector whose magnitude is

charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q

to +q. Its unit is coulomb-metre.

Let us calculate electric field at the point P at a distance r from the centre of the dipole on the axial line of the

dipole on the side of the charge q as shown in figure.

E–q

E+q

P q –qp

2a

r

– 2

0

–ˆ

4 ( )q

qE p

r a

�

, where ˆp is the unit vector along the dipole moment vector.

2

0

ˆ4 ( – )

q

qE p

r a

�

The total field at P is –q q

E E E � � �

= 2 2

0

1 1ˆ–

4 ( – ) ( )

qp

r a r a

= 2 2 2

0

4ˆ

4 ( – )

q arp

r a

2 2 2 2 2 2

0 0

ˆ2(2 ) 2

4 ( – ) 4 ( – )

aq rp rpE

r a r a

�

�



16. An electric dipole with dipole moment 4 × 10–9 Cm is aligned at 30º with the direction of a uniform electric field

of magnitude 5 × 104 NC–1. Calculate the magnitude of the torque acting on dipole and also indicate its

direction.

Sol. = pEsin = 4 × 10–9 × 5 × 104 × sin30º

E

P

= 10–4 Nm

Direction is along p E�

�

, i.e., normally into the plane of paper

17. Derive an expression for the magnitude of electric field intensity at any point along the equatorial line of a short

electric dipole. Give the direction of intensity at that point. For a short dipole what is the ratio of magnitude of

electric field intensities at two equidistant points from the centre of dipole, one along the axial line and another on

the equatorial line?

Sol.

P

p–qq

2a

r

E P at

E +q

E –q

The magnitudes of the electric fields due to the two charges +q and –q are given by

2 2

04

q

qE

r a

and – 2 2

04

q

qE

r a

and they are equal.

The directions of E+q

and E–q

are as shown in the figure.

The components normal to the dipole axis cancel away. The components along the dipole axis add up.

The total electric field E�

at P is opposite to dipole moment vector p�

. So, we have

–

– cosq q

E E E �

2 2 1 1

2 2 2 20 2 2

–2· cos

4 ( )( ) ( )

q a a

r ar a r a

∵

= 2 2 3/2 2 2 3/2

0 0

ˆ–2 –

4 ( ) 4 ( )

aq p pE

r a r a

�

�

For short dipole 3

0

2

4a

pE

r

and 3

04

e

pE

r

2

2 :11

a

e

E

E



18. (i) State Gauss’s law in electrostatics and express it mathematically. Using it derive an expression for electric

field at a point near a thin infinite plane sheet of electric charge.

(ii) Two plane sheets of charge having charge densities 1 and

2 are normal to each other. What will be the

magnitude of electric field at point O?

+

+

+ +

1

2

O

Sol. (i) It states that the electric flux through a closed surface S is equal to

0

q

where q is total charge enclosed

by S.

Mathematically : 0

·

S

qE ds

� ��

�

Electric field due to a uniformly charged infinite plane sheet

E

surface

charge density

21

E

zy

x

A

Fig. : Gaussian Surface for a Uniformly Charged Infinite Plane Sheet

Let be the uniform surface charge density of an infinite plane sheet. We take the x-axis normal to the given

plane. By symmetry, we assume that the electric field will not depend on y and z co-ordinates and its direction

at every point must be parallel to the x-direction.

We have taken the Gaussian surface as a cylinder (a rectangular parallolepiped will also do) of cross-sectional

area A as shown. As seen from the figure, only the two faces 1 and 2 will contribute to the flux (as between E�

and A�

is zero on these faces). Electric field lines are parallel to other curved surface ( = 90º) and they,

therefore, do not contribute to the total flux.

The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x

direction. Therefore, flux ·E S��

through both the surfaces are equal and add up. Hence, the net flux through the

Gaussian surface is 2EA. Now the charge enclosed by the closed surface is A. Therefore by Gauss’s law

2EA = 0

A

or, E = 0

2

vectorially, 0

ˆ2

E n

�

where ˆn is a unit vector normal to the plane and going away from it.



(ii) Field at O, 1 2

E E E � � �

+

+

+ +

1

2

OE

1=1

02

i

E2=1

02

jE

= 1 2

0 0

ˆ ˆ

2 2i j

Magnitude of electric field

2 2

1 2E E E

2 2

1 2

0

1

2E

.

��



Objective Types Questions

(Electric Charges, Conductors and Insulators, Charging by Induction, Properties of Electric Charge)

1. If a body has positive charge on it, then it means it has

(1) Gained some protons (2) Lost some protons

(3) Gained some electrons (4) Lost some electrons

Sol. Answer (4)

Due to lack of electron body get positive charge.

2. Sure check for presence of electric charge is

(1) Process of induction (2) Repulsion between bodies

(3) Attraction between bodies (4) Frictional force between bodies

Sol. Answer (2)

Due to similar (like charge), repulsion force is possible but attraction force may be due to uncharged body.

3. If a solid and a hollow conducting sphere have same radius then

(1) Hollow sphere will hold more maximum charge

(2) Solid sphere will hold more maximum charge

(3) Both the spheres will hold same maximum charge

(4) Both the sphere can’t hold charge

Sol. Answer (3)

Excess charge spread on outer surface only from their property.

4. Five balls marked a to e are suspended using separate threads. Pairs (b, c) and (d, e) show electrostatic

repulsion while pairs (a, b), (c, e) and (a, e) show electrostatic attraction. The ball marked a must be

(1) Negatively charged (2) Positively charged

(3) Uncharged (4) Any of the above is possible

Sol. Answer (3)

5. When a plastic rod rubbed with wool is brought near the knob of a negatively charged gold leaf electroscope,

the gold leaves

(1) Contract (2) Dilate

(3) Start oscillating (4) Collapse completely

Sol. Answer (2)

6. Which of the following is not true about electric charge?

(1) Charge on a body is always integral multiple of certain charge known as charge of electron

(2) Charge is a scalar quantity

(3) Net charge on an isolated system is always conserved

(4) Charge can be converted into energy and energy can be converted into charge

Sol. Answer (4)

A rest charge cannot be converted into energy.

Solutions (Set-2)



7. What is the amount of charge possessed by 1 kg of electrons?

(1) 1.76 × 1011 C (2) 1.76 × 10–9 C

(3) 1.76 × 10–7 C (4) 1.76 × 10–5 C

Sol. Answer (1)

319.1 10 kge

m ∵

191.6 10 C

eq

So charge due to 1 kg electron

19

11

31

1.6 101.76 10

9.1 10Q C

8. Which of the following processes involves the principle of electrostatic induction?

(1) Pollination (2) Chocolate making (3) Xerox copying (4) All of these

Sol. Answer (4)

These are properties of electrostatic induction.

(Coulomb's Law, Forces between Multiple Charges)

9. When a conducting soap bubble is negatively charged then

(1) Its size starts varying arbitrarily (2) It expands

(3) It contracts (4) No change in its size takes place

Sol. Answer (2)

Due to repulsion force between diametrically opposite wall, it expands.

10. Coulomb’s law is analogous to

(1) Charge conservation law (2) Newton’s second law of motion

(3) Law of conservation of energy (4) Newton’s law of gravitation

Sol. Answer (4)

Coulomb’s law and Newton’s law of gravitation are inverse square law.

11. Two point charges Q1 and Q

2 exert a force F on each other when kept certain distance apart. If the charge

on each particle is halved and the distance between the two particles is doubled, then the new force between

the two particles would be

(1)2

F(2)

4

F(3)

8

F(4)

16

F

Sol. Answer (4)

Given

1 2KQQ

Fr …(i)

if, 1

12

QQ

2

22

QQ

& r = 2r

Then = 116

FF



12. Two equally charged identical small balls kept some fixed distance apart exert a repulsive force F on each

other. A similar uncharged ball, after touching one of them is placed at the mid-point of line joining the two

balls. Force experienced by the third ball is

(1) 4F (2) 2F (3) F (4)2

F

Sol. Answer (3)

First case : Q Qr

2

1 2

KQF F

r …(i)

Second case : Q

2

r2

Q

2

r

F1 2

Q

F2

2 1NetF F F

2 2

2 2

4 4

4. 2

K Q K Q

r r

2

2

KQF

r

Force remain’s constant

13. Two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular

bisector at a distance ‘d ’ from the centre will experience maximum electrostatic force when

(1)2 2

Rd (2)

2

Rd (3) 2d R (4) 2 2d R

Sol. Answer (1)

2

1 2 2

2

4

KQF F

Rd

F2

F1

F

R(+)Q (+)Q

d

a

R

2

2

2

4

Rd

FN

= F1 cos + F

2 cos

= 2F1 cos =

2

2 2

2 2

12. .

2

4 4

N

KQ dF

R Rd d

If F = Maximum than 0dF

dd

So we get 2 2

R



14. A charged gold leaf electroscope has its leaves apart by certain amount having enclosed air. When the

electroscope is subjected to X-rays, then the leaves

(1) Further dilate (2) Start oscillating

(3) Collapse (4) Remain unaltered

Sol. Answer (3)

15. Two equal positive charges Q are fixed at points (a, 0) and (–a, 0) on the x-axis. An opposite charge –q at

rest is released from point (0, a) on the y-axis. The charge –q will

(1) Move to infinity

(2) Move to origin and rest there

(3) Undergo SHM about the origin

(4) Execute oscillatory periodic motion but not SHM

Sol. Answer (4)

In question

Net force on q is not proportional to (x)F

2

–q

F1

x

x = a– x = a+

(+)Q (+)Qas F (–x) [For SHM]

but Net force on q is

1

2 2 2

.K q Q xF

x a

This is condition for periodic motion

16. Four charges each equal to Q are placed at the four corners of a square and a charge q is placed at the

centre of the square. If the system is in equilibrium then the value of q is

(1) (1 2 2)2

Q (2)–

(1 2 2)4

Q (3) (1 2 2)4

Q (4)–

(1 2 2)2

Q

Sol. Answer (2)

Net force on Q due to other corner charge is

F123

= 2 2

3 1 2F F F

Q Q

Q

Q

F1

F3

F4

2l

F2

= 3 1

2F F

= 2 2

2 2

2

2

KQ KQ

l l

Force on Q1 due to centre charge –q

4 2.2

K Q qF

l



If net force on corner charge Q is zero

Then

F123

+ F4 = 0

So 1 2 24

Qq

17. According to Coulomb’s Law, which is correct relation for the following diagram?

q1

q2F

12F21

(1) q1 q

2 < 0 (2) q

1 q

2 > 0 (3) q

1 q

2 = 0 (4) q

1 q

2 >> 100 C

Sol. Answer (1)

Both charge should be unlike charge

q1

= +Q , q2

= –Q

So q1

q2

= –Q2

So q1

q2

= Negative

So q1

q2

< 0

18. A charge q is to be distributed on two conducting spheres. What should be the value of the charges on the

spheres so that the repulsive force between them is maximum when they are placed at a fixed distance from

each other in air?

(1) and2 2

q q(2)

3and

4 4

q q(3)

2and

3 3

q q(4)

4and

5 5

q q

Sol. Answer (1)

Force between both is

2

K Q q QF

r

If F = maximum then, 0dF

dQ

So 2

qQ

So both charge be ,2 2

q q

19. A point charge q1 exerts an electric force on a second point charge q

2. If third charge q

3 is brought near, the

electric force of q1 exerted on q

2

(1) Decreases

(2) Increases

(3) Remains unchanged

(4) Increases if q3 is of same sign as q

1 and decreases if q

3 is of opposite sign

Sol. Answer (3)

Electric force between ‘2’ charge do not depend on the ‘3’rd charge.



20. Three charges +4q, Q and q are placed in a straight line of length l at points 0, 2

� and � distance away from

one end respectively. What should be Q in order to make the net force on q to be zero?

(1) –q (2) 4q (3) –2

q(4) –2q

Sol. Answer (1)

FNet

on q is

x = 0 x = l

+4q Q q

2

lx

3

2 2

.4 .4K q KQqF

l l

If F = 0 then

Q = –q

21. A particle of mass m and carrying charge –q1 is moving around a charge +q

2 along a circular path of radius

r. Find period of revolution of the charge –q1

(1)

3 3

0

1 2

16 mr

q q

(2)

3 3

0

1 2

8 mr

q q

(3)

1 2

3 3

016

q q

mr (4) Zero

Sol. Answer (1)

2

1 2

2

0

1.

4

q qmv

r r

1

2

1 2

0

1

4

q qv

rm

q2

r

m

(– )q

For 1 trip,

1

20 1 2

22 4

rT r mr q q

v

3 3

0

1 2

16 mrT

q q

22. Consider three point objects P, Q and R. P and Q repel each other, while P and R attract. What is the nature

of force between Q and R?

(1) Repulsive force (2) Attractive force (3) No force (4) None of these

Sol. Answer (2)

(Electric Field and Electric Field Lines)

23. The electric field intensity at a point in vacuum is equal to

(1) Zero

(2) Force a proton would experience there

(3) Force an electron would experience there

(4) Force a unit positive charge would experience there

Sol. Answer (4)

This is defination of electric field.



24. A sphere of radius r has electric charge uniformly distributed in its entire volume. At a distance d from the

centre inside the sphere (d < r) the electric field intensity is directly proportional to

(1)1

d(2) 2

1

d(3) d (4) d2

Sol. Answer (3)

Electric field inside volume charge is given by

3

0

1

4

q dE

r

d

r

q

E d

25. The electric field at 2R from the centre of a uniformly charged non-conducting sphere of radius R is E. The

electric field at a distance 2

R from the centre will be

(1) Zero (2) 2E (3) 4E (4) 16E

Sol. Answer (2)

Given 22

KqE

R …(i)

Then 3

.2

'

RKq

ER

…(ii)

Find E' = 2E

26. In a uniform electric field if a charge is fired in a direction different from the line of electric field then the

trajectory of the charge will be a

(1) Straight line (2) Circle (3) Parabola (4) Ellipse

Sol. Answer (3)

F = qE = m ax

x

qEa

m

(+ )qy

y

x

E( )x

Then, 21

02

xx a t …(i)

But y = uxt then

yt

ux

So, 2

2

1.

2

qE yx

m u , so x y2 for parabola

27. A positively charged pendulum is oscillating in a uniform electric field pointing upwards. Its time period as

compared to that when it oscillates without electric field

(1) Is less (2) Is more

(3) Remains unchanged (4) Starts fluctuating

Sol. Answer (2)

Effective g decreases.



28. How many electrons should be removed from a coin of mas 1.6 g, so that it may float in an electric field of

intensity 109 N/C directed upward?

(1) 9.8 × 107 (2) 9.8 × 105

(3) 9.8 × 103 (4) 9.8 × 101

Sol. Answer (1)

qE = mg

neE = mg

Use mg

neE

29. ABC is an equilateral triangle. Charges +q are placed at each corner. The electric field intensity at the centroid

of triangle will beA

CB

O

+q

+q+q

r

(1)2

0

1

4

q

r

(2)

2

0

1 3

4

q

r

(3)0

1

4

q

r

(4) Zero

Sol. Answer (4)

FN = 0

30. A charge Q is placed at the centre of a square. If electric field intensity due to the charge at the corners of

the square is E1 and the intensity at the mid point of the side of square is E

2, then the ratio of

1

2

E

E will be

(1)1

2 2(2) 2

(3)1

2(4) 2

Sol. Answer (3)

2

1 2

0

1

4

QE

l

…(i)

4

2 2

0

1

4

QE

l

E2 = 2E

1

1

2

1

2

E

E



31. Point charges each of magnitude Q are placed at three corners of a square as shown in the diagram. What

is the direction of the resultant electric field at the fourth corner?

A

B C

D

E

+ Q

+ Q – Q

O

(1) OC (2) OE (3) OD (4) OB

Sol. Answer (4)

Resultant force act along OB

32. Two charges e and 3e are placed at a distance r. The distance of the point where the electric field intensity

will be zero is

(1)(1 3)

r

from 3e charge (2)

(1 3)

r

from e charge

(3)(1– 3)

r

from 3e charge (4)1

13

r

from e charge

Sol. Answer (2)

Net electric field at P is zero then

rx

Pe 3e

O = E1

– E2

E1 = E

2

22

3k e k e

x r x

so,1 3

x r x

3r x x

1 3r x

1 3

rx

33. If electric lines of force in a region are represented as shown in the figure, then one can conclude that, electric

field is

(1) Non-uniform (2) Uniform

(3) Both uniform and non-uniform (4) Zero everywhere

Sol. Answer (1)

Diverging electric line of force denote non-uniform electric field.



34. An uncharged sphere of metal is placed in a uniform electric field produced by two oppositely charged plates.

The lines of force will appear as

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

+

Sol. Answer (3)

35. An electron released on the axis of a positively charged ring at a large distance from the centre will

(1) Not move (2) Do oscillatory motion (3) Do SHM (4) Do non periodic motion

Sol. Answer (2)

3

2 2 2

.

N

k Q xF q

x r

For SHM, F (–x)

36. Figure shows electric lines of forces due to charges Q1 and Q

2. Hence

Q1

Q2

(1) Q1 and Q

2 both are negative (2) Q

1 and Q

2 both are positive

(3) Q1 > Q

2(4) Both (2) & (3)

Sol. Answer (4)

37. Figure shows electric lines of force. If Ex and E

y are the magnitudes of electric field at points x and y

respectively, then

x

y

(1) Ex > E

y(2) E

x = E

y

(3) Ex < E

y(4) Any of these

Sol. Answer (1)

(Electric Flux, Electric Dipole, Dipole in a Uniform External Field)

38. Electric charge Q, Q and –2Q respectively are placed at the three corners of an equilateral triangle of side

a. Magnitude of the electric dipole moment of the system is

(1) 2Qa (2) 3Qa (3) Qa (4) 2Qa

Sol. Answer (2)



P1 = Q.a

P2 = Q.a

2 2

1 2 1 22 cosP P P PP

Q

Q – Q2

P1

q

q

P23 .

NP Qa

39. An electric dipole placed in a uniform electric field experiences maximum moment of couple when the dipole

is placed

(1) Against the direction of the field (2) Towards the electric field

(3) Perpendicular to the direction of the field (4) At 135° to the direction of the field

Sol. Answer (3)

= PE sinFor = 90°

= Max.

40. Force of interaction between two co-axial short electric dipoles whose centres are R distance apart varies as

(1)1

R(2) 2

1

R(3) 3

1

R(4) 4

1

R

Sol. Answer (4)

1 2

4

6K P PF

r

r

+Q–Q +Q–Q

41. Two charges of +25 × 10–9 coulomb and –25 × 10–9 coulomb are placed 6 m apart. Find the electric field

intensity ratio at points 4 m from the centre of the electric dipole (i) on axial line (ii) on equatorial line

(1)1000

49(2)

49

1000(3)

500

49(4)

49

500

Sol. Answer (1)

axial 22 2

.2k PrE

r l

…(i)

eq. 3

2 2 2

.k pE

r l

…(ii)

Find axial

1000

49eq

E

E

42. The electric force on a point charge situated on the axis of a short dipole is F. If the charge is shifted along

the axis to double the distance, the electric force acting will be

(1) 4F (2)2

F(3)

4

F(4)

8

F

Sol. Answer (4)

3 3

1 1F F

r r

If r1 = 2r

Then force become 8

F



43. An electric dipole is placed at an angle 60° with an electric field of strength 4 × 105 N/C. It experiences a

torque equal to 8 3 Nm . Calculate the charge on the dipole, if dipole is of length 4 cm

(1) 10–1 C (2) 10–2 C (3) 10–3 C (4) 10–4 C

Sol. Answer (3)

= PE sin

2 58 3 4 10 . 4 10 sin60q

find q = 10–3

44. A charge q is situated at the centre of a cube. Electric flux through one of the faces of the cube is

(1)0

q

(2)0

3

q

(3)0

6

q

(4) Zero

Sol. Answer (3)

Total

0

of 6 surfaceq

One surface

06

q

45. A charge Q is placed at the centre of the open end of a cylindrical vessel. Electric flux through the surface

of the vessel is

(1)0

2

q

(2)0

q

(3)0

2q

(4) Zero

Sol. Answer (1)

If charge Q is surrounded with two cylinder then flux of ‘2’ cylinder is 0

q

Flux of one cylinder = 0

2

q

46. A hemispherical surface of radius R is kept in a uniform electric field E as shown in figure. The flux through

the curved surface is

E

R

(1) E2R2 (2) ER2 (3) E4R2 (4) Zero

Sol. Answer (2)

qNet

= 0 So Net

= 0

in +

out = 0

out

= –in = –[E.R2cos180°]

= E.R2



47. Total electric flux associated with unit positive charge in vacuum is

(1) 40

(2)0

1

4 (3)0

1

(4) 0

Sol. Answer (3)

0

q for q = 1

0

1

48. A charged body has an electric flux F associated with it. Now if the body is placed inside a conducting shell

then the electric flux outside the shell is

(1) Zero (2) Greater than F (3) Less than F (4) Equal to F

Sol. Answer (4)

Charge remains constant so flux remains constant.

49. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The

outward flux over the surface of the cylinder is given by

(1) 2R2E (2)2

2

R E(3) 2RLE (4) R2E

Sol. Answer (4)

Not net flux only outward flux

φ = ER2

50. A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 V/m, such

that the surface makes an angle of 30° with the direction of electric field. Find the flux of the electric field

through the rectangular surface

(1) 0.1675 N/m2C (2) 0.1875 Nm2/C (3) Zero (4) 0.1075 Nm2/C

Sol. Answer (2)

15 cm

10 cm

25 V/m

A

30°

φ = EAcos30° = 0.1875 Nm2/C

51. If an electric field is given by ˆ ˆ ˆ10 3 4i j k , calculate the electric flux through a surface of area 10 units lying

in yz plane

(1) 100 units (2) 10 units (3) 30 units (4) 40 units

Sol. Answer (1)

ˆ ˆ ˆ10 3 4E l j k �

ˆ10A l�

So, . 100E A � �

unit



52. There is uniform electric field of 3 ˆ8 10 N/C.i What is the net flux (in SI units) of the uniform electric field

through a cube of side 0.3 m oriented so that its faces are parallel to the coordinate plane?

(1) 2 × 8 × 103 (2) 0.3 × 8 × 103 (3) Zero (4) 8 × 106 × 6

Sol. Answer (3)

3 ˆ8 10E i �

N/C (Uniform) here q = 0

So = 0

53. A charge Q is kept at the corner of a cube. Electric flux passing through one of those faces not touching

that charge is

(1)0

24

Q

(2)0

3

Q

(3)0

8

Q

(4)0

6

Q

Sol. Answer (1)

Net

08

q (Of 3 surface)

One surface

024

q

(Continuous charge distribution, Gauss's law, Applications)

54. The electric field in a region is radially outward and at a point is given by E = 250 r V/m (where r is the

distance of the point from origin). Calculate the charge contained in a sphere of radius 20 cm centred at the

origin

(1) 2.22 × 10–6 C (2) 2.22 × 10–8 C (3) 2.22 × 10–10 C (4) Zero

Sol. Answer (3)

2

0

1.

4

QE

r

9

2

9 10 .250

Qr

r

Use r = 20 × 10–2 m

Find Q = ?

Q = 2.22 × 10–10 C

55. An isolated solid metal sphere of radius R is given an electric charge. Which of the graphs below best shows

the way in which the electric field E varies with distance x from the centre of the sphere?

(1)

x

E

O R

(2)

x

E

O R

(3)

x

E

O R

(4)

x

E

O

Sol. Answer (3)

outside 2

k qE

r

Einside

= 0



56. The electric field intensity at P and Q, in the shown arrangement, are in the ratio

a

b

r

2r

Q

P

Fig.: Hollow concentric shell

q

3q

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

Sol. Answer (3)

2P

k qE

r …(i)

2 2

.3

2 2Q

k q k qE

r r

2 2

.3

4 4

k q k q

r r

2

k q

r …(ii)

EP

: EQ

= 1 : 1

57. Consider an atom with atomic number Z as consisting of a positive point charge at the centre and surrounded

by a distribution of negative electricity uniformly distributed within a sphere of radius R. The electric field at

a point inside the atom at a distance r from the centre is

(1)2 3

0

1–

4

Ze r

r R

(2) 2 3

0

1 1

4

Ze

r R

(3) 2

0

2

4

Ze

r(4) Zero

Sol. Answer (1)

E = E1 – E

2

E = 2 3

.k ze k zer

r R

58. An electron is rotating around an infinite positive linear charge in a circle of radius 0.1 m, if the linear charge

density is 1 C/m, then the velocity of electron in m/s will be

(1) 0.562 × 107 (2) 5.62 × 107 (3) 562 × 107 (4) 0.0562 × 107

Sol. Answer (2)

2mvq E

r

2

0

.2

mve

r r

7

5.62 10 m/sv



59. A dipole with an electric moment p�

is located at a distance r from a long thread charged uniformly with a

linear charge density . Find the force F acting on the dipole if the vector p�

is oriented along the thread

(1) 2

02

p

r

(2)0

2

p

r

(3)

02

p

r (4) Zero

Sol. Answer (4)

F1 = F

2

+

+

+ +

+

+

+ F1

–

F2

So, FN = 0

60. For two infinitely long charged parallel sheets, the electric field at P will be

(1) –2 2( – )x r x

(2)

0 02 2 ( – )x r x

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

r

P

x

(3)0

(4) Zero

Sol. Answer (4)

EN = E

1 – E

2

0 02 2

= 0

��

Documents

Electric Charges and Fields - 1 File Download