NUMERICAL BANK OF ELECTROSTATICS FOR IIT PMT.pdf

A NAME IN CONCEPTS OF PHYSICS

XI &XII (CBSE & ICSE BOARD)

S1

S2

+

–

+Q

IIT-JEE / NEET /AIIMS / JIPMER / uptU

S4 S3

–Q 

E

Based on Charge and Coulomb’s Law 1.

4.

A total charge Q is broken in two parts Q1 and Q2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

(c) 22.50 N

[MP PET 1990]

Q 2Q Q Q , Q1  Q  (b) Q2  , Q1  Q  R R 4 3 3Q Q Q Q (c) Q2  , Q1  (d) Q1  , Q2 

(a) Q2 

4

(d)

2.

2

1 4 0

Q2 (2 L)2

(b) 90o ,

Q2 L2

1 4 0

F  r  F  2  1   2  F 2  r 1  r 1

2



L

L

Thus angle  = 180° and Force 

[AMU 2002]

(a)  (b)  (c)

2Q 4 b

2

Q 4 b

2

, ,

Q 4 c 2

Q a

2 4 c

b

Q

0,

c

2 4 c

(d) None of the above (a)

Surface charge density (  ) 

4 0



between q1 and q3 is F 13 , the ratio of magnitudes (a) 1 / 2

(b) 2

(c) 1 / 2

(d)

F 12 

1 4 0



So

F 12 F 13

is

6.

 inner 

2Q 4 b

2

a

b

+2Q

c

and

 Outer 

Q 4 c 2

Three charges each of magnitude q are placed at the corners of an equilateral triangle, the electrostatic force on the charge placed at the center is (each side s ide of triangle is L) [DPMT 2009]

(a) Zero

2

1 q2 q2   and F 13 4 0 (a 2 )2 a2

NUMERICAL BANK FOR IIT - PMT

– 2Q

Q2 (2 L)2

Three equal charges are placed on the three corners of a square. If the force between q1 and q 2 is F 12 and that

Charge Surface area – Q + 2Q = Q

+Q

1

2

 0.04     F 2  11.25 N F 2  0.06  5

A solid conducting sphere of radius a has a net positive charge 2Q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge – Q. The surface charge density on the inner and outer surfaces of the spherical shell will be

1 Q2 Q2 o (d) 180 , 4 0 2 L2 4 0 L2 The position of the balls in the satellite will become as shown below 180o +Q

(b)

5.

(d) 45.00 N

1

(c) 180o ,

3.

2

(b)

QQ  k 1 2 2 .....(ii) Q1  Q2  Q ..... (i) and F  r kQ (Q  Q ) From (i) and (ii) F  1 2 1 r Q dF For F to to be maximum  0  Q1  Q2  2 dQ1 Two small spheres each having the charge Q are suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two [IIT 2007] suspensions and the tension in each will be (a) 180o ,

(a)

4

The force between two charges 0.06 m apart is 5 N . If each charge is moved towards the other by 0.01m , then [SCRA 1994] the force between them will become (a) 7.20 N (b) 11.25 N



F 12 2 F 13

(c)

1

(b) 3q2

4 0 L2

(d)

1

q

2

4 0 L2

1 q2 12 0 L2

ELECTROSTATICS

P.L . SHARMA ROAD, ROAD, center SHASTRI NAGAR NAGAR center CENTRAL MARKET, Opp. Sagar Complex Meerut

OPP. SUMIT NURSING HOME, 1 ST FLOOR AIM INTERNATIONAL Page 1


XI &XII (CBSE & ICSE BOARD) (a)


 1  2 2   (c)  

In the following figure since | F A || F B || F C | and they are equally inclined with each other, so their resultant will be zero. q



2

q2

 4 0 a

2

 1  (d)  2  

q2

2  4 0 a



2

A

(c)

After following the guidelines mentioned above F C F D

Q F C

F B

+Q

F A

q

q

Three charges are placed at the vertices of an equilateral triangle of side ‘a ‘a’ as shown in the following figure. The force experienced by the charge placed at the vertex A in a [AIIMS 2003] direction normal to BC is +Q

F net 

(c) Zero (d) Q /(2 0a ) 2

| F B | | F C | k.

– Q Q B

C

Q2 a2

F C cos 60o F B cos 60o

F B

– Q Q B

A

60o

F B sin 60o

60o

60o

+Q

a

between the two is (Where K 

(c) 9.

(d)

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r . The coulomb force F 1 4 0

(a)  K

e r r 3

(b) K

(c)  K

e2  r r 3

(d) K

F   k

2

ˆ

)

ˆ

3q 2 4 0 a

2

2

e

3

r

e2



4q 2 4 0 a


q2 1  kq2   2  2  2   2  4 0 a 2 a  2a

 1  2 2     2   

(d) 3 F / 8

Q2 (fig. A). Finally when a third spherical r 2 conductor comes in contact alternately with B and C then removed, so charges on B and C are Q / 2 and 3Q / 4 respectively (fig. B)

Initially F   k.

Q

Q /2

3Q /4

r  r 

C

C r

(A)

(B)

11.

The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5  1011m, will wil l be (Charge on electron electron = 1.6  10–19 C, mass of electron = 9.1  10–31 kg , mass of proton = 1.6  1027 kg , G  6.7  1011 Nm2 / kg 2)

2

B

r

 Q   3Q     2   4  3   F Now force F '  k. 8 r 2

r ˆ

r 2

kq2

Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of [AIEEE 2004] repulsion between B and C is (a) F / 4 (b) 3 F / 4

Q

r

ˆ

(b)

a2

B

Equal charges q are placed at the four corners A, B, C, D of a square of length a . The magnitude of the force on the [MP PMT 1994; DPMT 2001] charge at B will be (a)

2

[CBSE PMT 2003]

  r   

e2 e2 r k r .   r 2 r 3

2kq

kq2 kq2 and  F D a2 (a 2 )2

(c) F / 8

C

Hence force experienced by the charge at A in the direction normal to BC is zero. 8.

10.

F C sin 60o

60o 60o

C

+Q

a

F C

+Q

Since F A  F C 

(b)  Q /(4 0a )

(c)

F A

F net  F AC  F D  F A2  F C2  F D

2

2

B

D

A

(a) Q2 /(4 0a 2 ) 

F AC

B

C

7.

+Q

A

(a) 2.36  1039 (c) 2.34  1041

[RPET 1997; Pb PMT 2003]

(b) 2.36  1040 (d) 2.34  1042 ELECTROSTATICS




XI &XII (CBSE & ICSE BOARD) (a)

Gravitational force F G  F G 

Gmemp r 2

Based on Electric Field and Potential

6.7  1011  9.1  1031  1.6  1027 = 3.9  10 10–47 N 11 2 (5  10 )

Electrostatic force F e 

1 4 0

1.

e2 r 2

9  109  1.6  1019  1.6  1019

F e 


(5  1011)2

A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium, if q is equal to [IIT 1987; CBSE PMT 1995; CPMT 1999; AIEEE 2002; AFMC 2002]

(a)  –8

= 9.22  10 N

F e 9.22 108   2.36  1039  47 F G 3.9  10 An infinite number of charges, each of charge 1  C, are placed on the x -axis -axis with co-ordinates x = = 1, 2, 4, 8, .... . If a charge of 1 C is kept at the origin, then what is the net [DCE 2004] force acting on 1 C charge (a) 9000 N (b) 12000 N (c) 24000 N (d) 36000 N The schematic diagram of distribution of charges on x -axis -axis is shown in figure below :

(b)

(b)

1C

1 C

1 C

1 C

1 C

O

x =1 =1

x =2 =2

x =4 =4

x =8 =8



 1  1  106 1  1  106   4 0  (1)2 (2)2

1

1

(c) 

4

Q

(d) 

4

2.

2

4 0

Q2 1 qQ  2 4 0 x 2 4 x



q

F C

C

x 1

q

Q 4 Q B = Q

x 2

F A B

x , y, z all in metres) The electric potential V at any point O ( x in space is given by V  4 x 2 volt . The electric field at the [IIT 1992] point (1m, 0, 2m) in volt / / metre is

(a) (b) (c) (d) (a)

 1  1  106 1  1  106    ....  (4)2 (8)2 

8 along negative X  axis 8 along positive X  axis 16 along negative X  axis 16 along positive Z  axis

The electric potential V ( x , y, z)  4 x 2 volt  V V V    j k y  z    x V V V 0  8 x ,  0 and Now  x y  z

Now E    i

ˆ

1

ˆ

ˆ

Hence E   8 x i , so at point (1 m, 0, 2m) ˆ

E   8i volt/metre or 8 along negative X -axis. -axis. ˆ

3.

(b)

4.

A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre [IIT 1983; MNR 1990; DPMT 2004] of the sphere is (a) 0 V (b) 10 V (c) Same as at point 5 cm away from from the surface (d) Same as at point point 25 cm away from the surface Since potential inside the hollow sphere sphere is same as that on the surface. 10  10 9 C are placed at each of the four 3 corners of a square of side 8 cm . The potential at the

Charges of 

intersection of the diagonals is


Q

x

   1       ... ...    9  109  10 6     1  4 0  1 4 16 64  1   4  4 4  9  109  106   9  103  = 12000 N 3 3 10 6  1

1

A

1



2

Q

Q A = Q

Total force acting on 1 C charge is given by F 

(b) 

Suppose in the following figure, equilibrium of charge B is considered. Hence for it's equilibrium | F A || F C |

So,

12.

Q

[BIT 1993]

(a) 150 2 volt

(b) 1500 2 volt

(c) 900 2 volt

(d) 900volt

ELECTROSTATICS




XI &XII (CBSE & ICSE BOARD) (b)

Potential at the centre O, where Q 

V  4 

1


Q

.

(c) EC = E

4 0 a / 2

10  10 9 C and a  8 cm  8  10 2 m 3 +q

EC

E B

120o

E A

+q B

A



120o

E BC = E

E B = E 

120o E A = E

E A = E

Enet = 0

8.

O

On rotating a point charge having a charge q around a charge Q in a circle of radius r . The work done will be [CPMT 1990, 97; AIIMS 1997; DCE 2003]

D

10  10 9 So V  5  9  109  3 2  1500 2 volt 8  10

A uniform electric field having a magnitude E 0 and direction along the positive X  axis exists. If the potential V is zero at x  0 , then its value at X   x will be

(c)

Since charge Q moving on equipotential surface so work done is zero.

9.

There is an electric field E in X-direction. If the work done on moving a charge 0.2 C through a distance of 2 m along a line making an angle 60 with the X-axis is 4.0, what is [CBSE PMT 1995] the value of E (a) 3 N / / C (b) 4 N / / C (c) 5 N / / C (d) None of these

(d)

W  qV  qE.d

[MP PMT 2009]

(b) 6.

(a) V ( x x )   xE0

(b) V x   xE0

(c) V x   x 2 E 0

(d) V x   x 2 E 0

E  



dV dX



(b)

(c) Zero

2 5.

q  2 Q r Q (d) 2 0 r

(a) q  2 r

+q C

+q

V x   xE0

4 = 0.2  E  (2 cos 60 o) = 0.2 E  (2  0.5)



Figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are E A and E B respectively and if the displacement between A and B is [CPMT 1986, 88] r then



E 

2m 

4  20 NC1 0.2

E

60°

X

d

10. A

r

B

Four equal charges Q are placed at the four corners of a square of each side is ' a' . Work done in removing a charge [AIIMS 1995] – Q from its centre to infinity is (a) 0

(a) E A  E B

(c)

(b) E A  E B

E B E (d) E A  2B r r In non-uniform electric field. Intensity is more, where the lines are more denser. ABC is an equilateral triangle. Charges  q are placed at each corner. The electric intensity at O will be [AIEEE 2002] 1 q +q (a) 2 A 4 0 r 1 q (b) 4 0 r r (c) Zero r r O 1 3q (d) +q +q 4 0 r 2 (c) E A 

(a) 7.

B


(c)

(b) 2Q 2  0 a

(d)

2Q 2 4 0 a

Q2 2 0 a

Potential at centre O of the square   Q  V O  4   4 ( / 2 )   a 0  

Q

Q

O

Work done in shifting ( – Q) charge from centre to infinity W   Q(V   V O)  QV 0 

4 2 Q2 4 0a



2Q2

a

Q

Q

 0a

C

ELECTROSTATICS




XI &XII (CBSE & ICSE BOARD) 11.


A particle A has charge  q and a particle B has charge  4q with each of them having the same mass m . When allowed to fall from rest through the same electric potential v difference, the ratio of their speed A will become v B

15.

Two point charges 100  C and 5  C are placed at points A and B respectively with AB  40 cm . The work done by external force in displacing the charge 5  C from B to C,

[BHU 1995; MNR 1991; UPSEAT 2000]

(a) 2 : 1 (c) 1 : 4 (b)

1

(b) 1 : 2 (d) 4 : 1 2QV

Using v 



m

v Q 

v A v B

Q B



q 4q



(a) 9 J

1 2

(c) 12.

BC  30 cm ,

Deutron and   particle are put 1 Å apart in air. Magnitude of intensity of electric field due to deutron at   [MP PET 2010] particle is (a) Zero (b) 2.88  1011 newton / coulomb

(d)

angle

 9  10 9 Nm2 / C 2

4 0 Q A



where

9 25

ABC ABC 



2

and

[MP PMT 1997]

(b)

81 20

J

9

J

(d)  J 4

Work done in displacing charge of 5  C C from B to C is W  5  106 (V C  V B ) where C 100  C

A

40 cm

(c) 1.44  1011 newton / coulomb

50 cm

(d) 5.76  1011 newton / coulomb (a)

13.

V  E  r 

B +5  C C

V 3000  6m r   E 500

(a) (c)

q 2

4 0 a

3q 4 0 a

2

(b)

(d)

| E A || EB | k.

(c)

16.

2 0 a

2

q a2



C

(b)

60° a

a

17.

2

3q

q

4 0a 2

a

A

q B

Two equal charges q are placed at a distance of 2a and a third charge 2q is placed at the midpoint. The potential [MP PMT 1997] energy of the system is (a)

q2 8 0 a

(c)  (c)

Enet 

a

(b)

7q 2 8 0 a

U system system 

(d)

1

(q) (2q)

4 0

a

U system em  



An alpha particle is accelerated through a potential difference of 10 6 volt . Its kinetic energy will be (a) 1 MeV MeV (c) 4 MeV MeV

E B Enet E A

3 k. q



(a)

6q 2

18.

8 0 a 9q 2

K  qV  2e  106 J 

(2q) (q)

4 0

a

7q 8 0a 2




(b) 2 MeV MeV (d) 8 MeV MeV 2e  106

eV  2 MeV e The electric potential V is given as a function of distance x (metre) by V  (5 x 2  10 x  9) volt . Value of electric field at x  1 is [MP PET 2010] (a) 20 V / (b) 6 V / / m / m (c) 11 V / / m (d) 23 V / / m

dV d (5 x 2  10 x  9)  10x  10  dx dx  ( E) x 1  10  1  10   20 V / m A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is E  

8 0 a

1

100  106 9   106 V 0.4 4

100  106 9   106 V 0.5 5 9 9  9  So W  5  10 6    106   106    J 4 4  5 

4 0 a 2 q

C

and V C  9  109 

2q

So, Enet  E A2  E B2  2 E A E B cos 0o

14.

V B  9  109 

Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a . The magnitude of [MP PMT 1997] electric field at the point C is

30 cm

[CBSE PMT 1998; Kerala PMT 2005]

(q) (q) 4 0 2a

2

(a) qEy

1

(c)

(b) qE 2 y

(c) qEy (d) q 2 Ey Kinetic energy = Force  Displacement = qEy

ELECTROSTATICS





A hollow insulated conducting sphere is given a positive charge of 10  C . What will be the electric field at the centre [CBSE PMT 1998] of the sphere if its radius is 2 meters (a) Zero (b) 5  Cm 2 (c)

(a)

20  Cm 

2

(d)


24.

[AMU 1999]

8  Cm 

2

The intensity of electric field inside a hollow conducting conducting sphere is zero. (b)

20.

An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t 1 . A proton of mass m p also initially at rest takes time t 2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t 2 / t 1 is nearly [IIT 1997 Cancelled] equal to (a) 1 (b) (m p / me )1 / 2 (c) (me / mp )1 / 2

(b)

For electron s 

An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg /m3, the radius of the drop will be (a) 2.0  10 6 m

(b) 1.7  106 m

(c) 1.4  10 6 m In equilibrium QE = mg

(d) 1.1  106 m

Q.



12000 4   r 3  900  10 2 3 2  10 –6 = 1.7  10 m  r = 2  1.6  1019 



25.

(d) 1836

eE 2 eE 2  t 2  t 1 , For proton s  m p me

The ratio of momenta of an electron and an  -particle -particle which whic h are accelerat acce lerated ed from rest by a potential poten tial difference diffe rence [UPSEAT 1999] of 100 volt is is (a) 1

1 / 2

m p  m p  t 22 m p t    2     2 t 1 me  me  t 1 me 21. A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of [KCET 2007, 2009] this cube will be 2 (a) q / b (b) q / 2b 2

(d) 22.

(c)

(d)

(c) 32q / b 2 (d) Zero Due to symmetric charge distribution. Point charges 4q,  q and 4 q are kept on the x  axis at points x  0, x  a and x  2a respectively, then

26.

[CBSE PMT 1992]

(c)

23.

(a) Only q is in stable equilibrium (b) None of the charges are in equilibrium (c) All the charges are in unstable equilibrium (d) All the charges are in stable equilibrium Force on each charge charge isis zero. But ifif any of the the charge charge isis displaced, the net force starts acting on all of them. If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be [SCRA 1998; JIPMER 2001, 02]

(b)

(a)

1

(c)

3

3 2

2

V

(b)

V

(d) 3 V

3



V out V



2 3

V  V surface  

V out 

2 3

q 10

stat volt , V out 

V


me m

(b)

2me

(d)

me

m

2m

Momentum p  2mK ; where K = = kinetic energy = Q.V 

p  2mQV



p  mQ 

pe meQe   p m Q

me 2m

A proton is accelerated through 50,000 V . Its energy will [JIPMER 1999] increase by (a) 5000 eV (c) 5000 J

(b) 8  1015 J (d) 50,000 J

(b)

Kinetic energy K = = Q.V  K = = (+e) (50000 V ) = 50000 eV = = 50000  1.6  10–19 J = = 8  10–15 J

27.

(c)

When a proton is accelerated through 1 V, then its kinetic [CBSE PMT 1999] energy will be (a) 1840 eV (b) 13.6 eV (c) 1 eV (d) 0.54 Ev  KE  qV  eV  e  1  1eV

28.

Electric charges of 10 C,  5 C,  3 C and 8  C are

V

Potential inside the sphere will be same as that on its surface i.e.

 4   mg    r 3   g d  3 

V

q 15

stat volt

placed at the corners of a square of side 2 m. the potential [KCET (Engg./Med.) 1999] at the centre of the square is (a) 1.8 V

(b) 1.8  106 V

(c) 1.8  105 V

(d) 1.8  104 V ELECTROSTATICS




XI &XII (CBSE & ICSE BOARD) (c)

Length of each side of square is 2

centre from each corner is

2


2 m so distance of it’s

32.

 1m.

In the rectangle, shown below, the two corners have charges q1  5 C and q2  2.0 C . The work done in moving a 3.0 C from B charge to A is (take 10 2 2 [AMU 2008] 1 / 4 0  10 N - m / C ) q1

A

O

5 cm

1m B

Potential at the centre  10  106 5  106 3  106 8  106  V  9  10      1 1 1 1  

(a) 2.8 J (c) 4.5 J

9

5

= 1.8  10 V 29.

(b) 3.5 J (d) 5.5 J

Work done W  3  10 6 (V A  V B ); where

(a)

 (5  106 )

V A  1010 

Ten electrons are equally spaced and fixed around a circle of radius R. Relative to V = 0 at infinity, the electrostatic potential V and and the electric field E at the centre C are

and



(a) V  0 and E  0

(b) V  0 and E  0



2

 15  10 10



(d) V  0 and E  0

33.

6 2  10  2 

5  10 



1 15

 106 volt

 (2  106 ) 5  106  13     106 volt 2 2  15 5  10  15  10

1  13   10 6     10 6  = 2.8 J  15   15

W  3  10 6 



(c) V  0 and E  0 At centre E = 0, V  0



V B  10 

[AMU 2000] 

(b)

q2

15 cm

Potential at a point x -distance -distance from the centre inside the conducting sphere of radius R and charged with charge Q is [MP PMT 2001]

30.

The displacement of a charge Q in the electric field

(a)

E  e1i  e 2 j  e 3k is r  ai  b j . The work done is ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

(c)

[EAMCET (Engg.) 2000]

(a)

(a) Q(ae1  be2 )

(b) Q (ae1 )2  (be2 )2

(c) Q(e1  e 2 ) a 2  b 2

(d) Q( e12  e22 ) (a  b)

31.

ˆ

ˆ

ˆ

(b)

q 1 2

(d)

(b)

Q

35.

2 2

+q

+q a

kQq kq kQq   0 a a a 2  kq   Q  q  Q   0  Q   2q    a  2 2 2 

Net electrostatic energy U 


2

(d) xQ

(a) 3 2

(b) 4 2

(c) 5 2

(d) 7

dV dV  3  (5)  5; E y   dy dx dV   15 and E z   dz

E x  

Enet  E x 2  Ey2  E z2

2q

(c) 2q (d)  q

x

Electric potential at any point is V  5 x  3y  15 z , then [MP PET 2002] the magnitude of the electric field is

[IIT-JEE (Screening) 2000]

(a)

x 2

Q

34.

ˆ

Three charges Q,  q and  q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to

Q

(b)

Potential at any point inside the charged spherical conductor equals to the potential at the surface of the conductor i.e. Q/R.

W  Q [(e1 i  e 2 j  e 3 k).( ).(ai  b j)]  Q (e1a  e 2 b) ˆ

R

(a)

By using W  Q ( E.r ) 

Q



(5)2  (3)2  ( 15 )2



7

A drop of 106 kg water carries 10 6 C charge. What electric field should be applied to balance its weight (assume  10m / s 2 ) g 

[MP PET 2002]

(a) 10 V /m upward

(b) 10 V /m downward

(c) 0.1 V /m downward

(d) 0.1 V /m upward

mg 106  10   10 V /m; QE  mg  E  (a) By using Q 10 6 upward because charge is positive. ELECTROSTATICS






If 3 charges are placed at the vertices of equilateral triangle

40.

of charge ‘q ‘ q’ each. What is the net potential energy, if the

side of equilateral  is l cm (a) (c) (c)

4 0

q2 l

1

3q2

4 0

l

1

(d)

1

2q2

Field is responsible for the retardation of motion of electron.

4 0

l

Now evaluate the distance travelled by the electron before

1

4q 2

4 0

l

coming to rest for an instant (mass of e  9  1031 Kg . charge  1.6  1019 C)

QQ U  . 1 2 ; net potential energy r 4 0 1

U net  3  37.

shooted parallel to the electric field of intensity 1  10 3 N /C .

[AIEEE 2002]

(b)

1 4 0

.

An electron moving with the speed 5  106 per sec is

q2 l

If identical charges (q) are placed at each corner of a cube of side b, then electric potential energy of charge ( q) which is placed at centre of the cube will be

(c)

(a) 7 m

(b) 0.7 mm

(c) 7 cm

(d) 0.7 cm

Electric force qE  ma  a 

QE

(c) (d)

8 2q 2 4 0b

 4 2q 2  0b

From v 2  u 2  2as



 0b

 4q2

(d)

Distance s 



3 0b

Length of the diagonal of a cube cube having having each side b is 3 b. So distance of centre of cube from each vertex is 3 b 2

u  5  10 6 and v  0

 8 2q 2

(b)

41.

 1 (q) (q)   4q  .   4 0 3 b / 2  3 0b 2

U  8  

A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV , its [AIIMS 2003; DCE 2001] kinetic energy will be (a) 1840 keV (b) 1/1840 keV (c) 1 keV (d) 920 keV KE  QV = e  103 V = (c) = 1 KeV . 39. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The 38.

electrostatic potential at a point p a distance centre of the shell is (q  Q) 2 (a) 4 0 R

(d)



s 

(5  10 6 ) 2  9 2  1.6  1015

u2 2a  7 cm.(approx)

The electric potential at the surface of an atomic nucleus Z = ( Z = 50) of radius 9.0× 10 13 cm is [CPMT 1990; Pb. PMT 2002; BVP 2003; MP PET 2008]

.

Hence potential energy of the given system of charge is

(c)

m

1.6  1019  1  103 1.6   1015  31 9 9  10

a 

[CBSE PMT 2002]

(a)

[MP PMT 2003]

2Q 4 0 R



2q 4 0 R

R 2

from the

(b)

42.

(a) 80 volts

(b) 8 ×

10 volts

(c) 9 volts

(d) 9 ×

10 volts

V  9  109 

6 5

50  1.6  1019  8  106 V 9  1015

A charge of o f 10 e.s.u. is placed at a distance of 2 cm from a charge of 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs) [CPMT 1976; MP PET 1989]

(a) 87.5 (c) 150

(b) 112.5 (d) 250 10  40 10  20   250 erg 2 4

(d)

Energy 

43.

If an insulated non-conducting sphere of radius R has charge density  . The electric field at a distance r from the  R) will be [BHU 2003] centre of sphere (r 

[AIEEE 2007]

2Q

(b)

4 0 R

2Q

(d)

Electric potential at P k.Q k.q V   R / 2 R q 2Q   4 0 R 4 0 R

4 0 R

q




R Q

q 4 0 R

(a) (c)

P R/ 2

(c)

 R R

3 0  r

3 0

(b)

(d)

For non-conducting sphere Ein 

 r r  0

R 3  R  0

k.Qr  r r  3 3 0 R ELECTROSTATICS






Point charge q1  2 C and q2  1  C are kept at points x  0 and x  6 respectively. Electrical potential will be [MP PMT 2004] zero at points (a) x  2 and x  9 (b) x  1 and x  5 (c) x  4 and x  12 (d) x  2 and x  2

(a)

q3

40 cm

q1 = 2 C

M

O x =0 =0

q2 = –1 C

N

x =6 =6

x =12 =12

x =4 =4

l'

l



l = 2

So distance of M from from origin; x = = 6 – 2 = 4 N ) : At exterio exteriorr point point ( N 



 2  10 (1  10 )    0 4 0  (6  l ' ) l'  6

1

U 

47.

As per this diagram a point charge q is placed at the origin O . Work done in taking another point charge Q from the point A [co-ordinates (0, a) ] to another point B [co-ordinates (a, 0)] along the straight path AB is

(c)

(d)

L 4

– 2q

x = = 0

A

P

x = = L L

 qQ 1  O B X  2a (d)  2 4   0 a   Since A and B are at equal potential so potential difference between A and B is zero. Hence W = = Q.V = = 0

q3 4 0

q3

48.

k , where k is [CBSE PMT 2005]

(c) 6q 2

(b)

 Q 1 1    4 0  R R 2  d 2 

(c) QR / 4 0d 2

(d)

 Q 1 1    2 0  R R 2  d 2 

Potential at the centre of rings are –q

R

30 cm

O2

O1

q2


(a) Zero

R

40 cm

A

(d)

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are q and  q . The potential difference [AIEEE 2005] between the centres of the two rings is

+q

C

q1

l

Suppose E.F. is zero at P as as shown. k.(2q) 8q Hence at P ; k.  l = L.  2 2 l ( L  l)

To charges q1 and q 2 are placed 30 cm apart, shown in the figure. A third charge q 3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the

(d) 6q1

(8q 2 )

The net field will be zero at a point point outside the charges and near the charge which is smaller in magnitude.

Y

 qQ 1  a  (c)  2  4   a 0   2

(b) 8 q1

4 0

Two point charges +8 q and 2q are located at x  0 and x   L respectively. The location of a point on the x axis at which the net electric field due to these two point [AIEEE 2005] charges is zero is (a) 8 L (b) 4 L

+ 8q

potential energy of the system is

q3

k = 8q2

(a) Zero

(a) 8 q 2

4 0

[8q 2 q 3 ] 

(c) 2 L

[CBSE PMT 2005]

46.

1

6

l '  6 , So distance of N from from origin, x = = 6 + 6 = 12

   qQ 1   2a (b)  2  4 0 a 

D

30 cm 10 cm 40 cm  q q q q   q q q q  1  1 3 U   2 3    1 3  2 3   4 0  0.4 0.1   0.4 0.5 



 2  106 (1  106 )  1   At intern internal al point point ( M ) : M  0 l 4 0  (6  l )  

q2

q1

6

(a)

50 cm

Potential will be zero at two points

(c)

45.

Change in potential energy ( U ) = U f – U i

D B

d

ELECTROSTATICS




XI &XII (CBSE & ICSE BOARD) k.q k(q)  R R 2  d 2

V O1 

V O2 

,

k(q) kq  R R 2  d 2

1  1 q V O1  V O2  2kq     R R2  d 2  2 0




in an electric field of 2  105 N /C. The maximum torque on [MP PMT 1987] the dipole will be

1  1     R R2  d 2  (b)

49.

Three infinitely long charge sheets are placed as shown in [IIT-JEE 2005] figure. The electric field at point P is is (a)

2  o

(b)  (c)

2  o

 o

50.

(d)

E  

3.

k

Z = 3a

2

Z = a

k

ˆ



Z = a

k

ˆ



k ˆ

2 o

2 2  k k k 2 o 2 o  o ˆ

ˆ

Charges 4Q, q and Q and placed along x -axis -axis at positions x  0, x  l / 2 and x   l , respectively. Find the value of q [DPMT 2005] so that force on charge Q is zero

An electric dipole is placed along the x  axis at the origin O . A point P is at a distance of 20 cm from this origin

(a) Q

(b) Q / 2

(c) – Q / 2

(d) – Q

The total force on Q Qq 4Q 2  0 2 4 0l 2  l 

(a)



(c)

2

3

3

3

with the x-axis. If the

Q

x = = 0

x = = l / 2

x = = l

Based on Electric Dipole

An electric dipole is kept in non-uniform electric field. It [AIIMS 2003; DCE 2001] experiences (a) A force and a torque (b) A force but not not a torque (c) A torque but not not a force (d) Neither a force nor a torque As the dipole will feel two forces which which are although opposite but not equal.  A net force will be there and as these forces act at different points of a body. A torque is also there. An electric dipole consisting of two opposite charges of 2  10 6 C each separated by a distance of 3 cm is placed

[MP PMT 1997]

 3    tan tan 1   2  3  

(b)



(d)

tan tan 1 

 3    2   



 / 3

O

q

the x -axis, -axis, the

P

Q – Q

4Q

 with

E

Y

 










3

+Q



X



P

tan     where tan 

1  tan tan 2 3

tan 1 3 / 2 so,     tan



3

tan 1 3 / 2  tan

4.

The distance between H  and Cl  ions in HCl molecule is 1.28 Å. What will be the potential due to this dipole at a [MP PMT 2002] distance of 12 Å on the axis of dipole (a) 0.13 V (b) 1.3 V (c) 13 V (d) 130 V

(a)

V  9  10 9 .

4 0    4 

2.

Maximum torque = pE = 2  10–6  3  10–2  2  105 = 12  10–3 N-m.

(b)

Qq 4Q2    q  Q. 2 4 0l 2  l 

(a)

(d) 24  10 3 N m

ˆ

4 0    2 

1.

(c) 24  10 1 N m

electric field at P makes an angle value of  would be

x

 o

(b) 12  10 3 N m

such that OP makes an angle

P

ˆ

4



(b)

ˆ



4

(d) 

Z

k

(a) 12  10 1 N m

 9  109  5.

p r 2

(1.6  1019 )  1.28  10 10 (12  10 10 ) 2

= 0.13V

Two charges  3.2  10 19 C and  3.2  10 9 C kept 2.4 Å apart forms a dipole. If it is kept in uniform electric field of intensity 4  10 5 volt /m then what will be its electrical energy in equilibrium (a)  3  10 23 J

(b)

[MP PMT 2003]

(b)  3  10 23 J

(c)  6  10 23 J (d)  2  10 23 J Potential energy of electric dipole U   pE cos   (q  2l)E cos U   (3.2  1019  2.4  1010 )4  105 cos  U   3  1023 (approx.)

ELECTROSTATICS






Based on Electric Flux and Gauss's Law 1.

(d)

4.

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

A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for [CPMT 1975] the surface of the cylinder is given by (a) 2 R 2 E

(b)

(c) ( R 2   R) / E

(d) Zero

Flux through surface A

S

[AIIMS 2009]

+q

+q

 R 2 / E

+q

(a) 3q /  0 (c) q /  0

2 2  A  E   R and  B   E   R

(b) 2q /  0 (d) Zero



ds

(b)

C

A

5.

B

o cos 90 = 0 Flux through curved surface C   E.ds   E ds cos

Total flux through cylinder   A   B   C = 0



2.

Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical [MP PET 2001] surface is Q + (a) +  0

(b)

(b) (c)

10Q

 0

 0

 Qenc 

1  0

(2q)

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1 = –14 nC, q2 = 78.85 nC, [KCET 2005] q3 = – 56 nC) (a) 103 Nm2 C–1 Gaussian surface A 3 -1 –2 (b) 10 CN m q1 q3 Gaussian q2 (c) 6.32  103 Nm2 C–1 surface B 3 -1 –2 (d) 6.32  10 CN m (a) Flux is due to charges enclosed enclosed per  0

Total flux =



 8.85  10 9 C 

(14  78.85  56)nC /  0

4  8.85  109  9  109  4 4 0

 1000.4 Nm 2 / C i..e. 1000 Nm2C 1

1m

( 0 ) + + +

100Q ( 0 )

50cm

Charge enclosed by cylindrical surface (length 100 cm) is

Qenc  100Q. By applying Gauss's law 3.

1

+

100Q

(d)

 

 

1  0

(Qenc. ) 

1  0

(100Q)

The inward and outward electric flux for a closed surface in units of N - m2 / C are respectively 8  10 3 and 4  103. Then the total charge inside the surface is [where  0  permittivity constant] [KCET 2003; MP PMT 2002]

(a) 4  10 C

(b)  4  103 C

3

(c) (d)

(4  103 ) 

By Gauss’s law    

(d)  4  10 3  0 C

C

1  0

(Qenclosed)

Qenclosed   0  (8  10 3  4  10 3 ) 0

 4  10 3  0 Coulomb NUMERICAL BANK FOR IIT - PMT

ELECTROSTATICS






4.

TARGET QUESTIONS 1.

A solid metallic sphere has a charge  3Q . Concentric with this sphere is a conducting spherical shell having charge Q . The radius of the sphere is a and that of the spherical shell is b(b  a) . What is the electric field at a distance R(a  R  b) from the centre [MP PMT 1995] (a) (c)

(c)

2.

Q 2 0 R

3Q 4 0 R

2

(b)

(d)

(a)

Q( R  r ) 4 0 ( R  r ) 2

(b)

(c) Zero (d)

q1  q 2  Q and q1 

3.

(d)

Qr 2 R 2  r 2

(c) (d)

(a)

4 0

q 6 0

q

4 0 R 2

So flux from given square q (i.e. one face)   5.

(c) K

R  r

(b)

4 0 ( R 2  r 2 )

2 1  2

R

 1  2

F 2 l

Also



p


(d) K

2 1 2

R  1  2

R



 1

 2

2k 1 2

Q

l

R

F 1 F 2 F 2k 1 2    l l l R

R

Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge qfrom the centre of one ring to that of other is [MP PMT 1999; AMU (Engg.) 1999]

(a) Zero (c)

V 

2a

R 2

R 2  r 2

p cos 4 0 r p cos p sin (c) V  (d) V  4 0 r 2 0 r 2 For the given situation, diagram can be drawn as follows As shown in figure component of dipole moment  along the line OP will be r p p'  p cos  . Hence electric potential at point P will be  1 p cos V  . q q – 4 0 r 2 +q

(b) K

2

Force on l length of the wire 2 is 2k 1 F 2  QE1  ( 2l) R 

6.

a

Two infinitely long parallel wires having linear charge densities  1 and  2 respectively are placed at a distance of R metres. The force per unit length on either wire will be (a) K

QR Q( R  r )

a/ 2

6 0

QR 2

(b)

(d)

An imaginary cube can be made by considering charge q at the centre and given square is one of it's face.

[MP PET 1997]

p cos 4 0 r 2

4Q

Potential at common centre QR 2  Q( R  r ) 1  Qr 2   2  2 2 2 4 0  ( R  r )r ( R  r ) R  4 0 ( R 2  r 2 ) Two equal charges q of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p . If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle  with the axis of the dipole, then the  2a) (Where p  2qa ) potential at P is given by (r  (a) V 

q

2 0 R

q1 q  2 2 (given) 2 4 r 4 R and q 2 

 0

 0

3Q

Electric field at a distance R is only due to sphere because electric field due to shell inside it is always zero. Hence 1 3Q electric field = . 2 4 0 R If on the concentric hollow spheres s pheres of radii r and R(  r ) the charge Q is distributed such that their surface densities are same then the potential at their common centre is [IIT 1981] 2

A point charge q is placed at a distance a /2 directly above the centre of a square of side a. The electric flux through the [AMU 1999] square is q q (a) (b)

(b)

q 2 (Q1  Q2 ) 4 0 R

(b)

q(Q1  Q2 )( 2  1) 2.4 0 R

(d)

q(Q1  Q2 )( 2  1) 2.4 0 R

W  q (V O2  V O1 ) where V O1 

P

and V O2  

Q1 4 0 R

Q2 4 0 R

V O2  V O1 

So, W 

Q2

Q1





Q2 4 0 R 2

R

R

Q1 4 0 R 2

O1

(Q 2  Q1 )  1  1   4 0 R  2 

O2

R

q.(Q2  Q1) ( 2  1) 4 0 R 2 ELECTROSTATICS





A negatively charged plate has charge density of 2  106 C / m2 . The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of 200eV [RPET 1997] (a) 1.77 mm (b) 3.51mm (c) 1.77 cm


10.

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that  is so small that tan tan   sin , then for equilibrium x is is equal to [AMU 2000]

(d) 3.51cm  

(a)

Let an electron is projected towards the plate from the r distance as shown in fig. =  =

2  10–6 C/m2

– – – – – – – – – – – – – – – – –

e r

q KE = 200 eV

= Electric field due to c harge plate 



2 0

1

. Hence minimum value of r is is given by



r  

r 

KE 200 eV 400  8.86  1012    1.77mm eE e   2  10 6

eE

1

 qL2  3  (b)    mg 2 0  

1

1

 q 2 L2  3  (c)    4 0 mg 

)

 q 2 L  3  (d)    4 0 mg 

(a) 

2 0

8.

T sin

9.

x

(b) 2.67  107 C (d) 1.67  10 23 C

mg ....... (i) In equilibrium F e = T sin sin mg = = T cos cos ....... (ii) x / 2 F q2 tan   sin   also tan tan  e  2 L mg 4 o x  mg

Q  ne ; where n = number of moles  6.02  1023  10 

Q

500 18

 6.02  10 23  10  1.6  10 19  2.67  107 C

Hence

Electric potential is given by

V  6 x  8 xy 2  8y  6yz  4 z 2

(d)

11.

E y  

dy

(a) r 2 1

r

(b)

(b) r

 ( 16 xy  8  6z)

E x  

E z  



1 / 3

The electric potential at a point ( x , y) in the x   y plane is

(c)

dV   (6y  8z) dz At origin x = = y = z = 0 so, E x   6, Ey  8 and E z  0

2q 2 L

given by V  kxy . The field intensity at a distance r from [UPSEAT 2002] the origin varies as

dV E x    (6  8y 2 ), dx dV

x q2  2 L 4 o x 2  mg

 q 2 L    x   x     4 o mg  2 o mg  3

Then electric force acting on 2C point charge placed on [RPET 1999] origin will be (a) 2 N (b) 6 N (c) 8 N (d) 20 N



F e

[RPET 1997]

(b)

T cos

T

The charge on 500 cc of water due to protons will be (a) 6.0  10 27 C (c) 6  10 23 C

q

x

 q 2 L  3  (a)     mg 2 0  

E

Er ) (where E It will not strike the plate if and only if KE  e( E

KE

L

L

(d)

1

r 2

dV dV   ky ; Ey     kx dy dx Y

E  E x 2  E y2  10 N /C .

P ( x, x, y) r

y

Hence force F  QE  2  10  20 N O 


X

x

E  E x 2  Ey2  k x 2  y 2  kr  E



r

ELECTROSTATICS





n electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I . If the dipole is displaced slightly from the equilibrium position, then the angular frequency of its oscil lations is


14.

A small sphere carrying a charge ‘q ‘ q’ is hanging in between two parallel plates by a string of length L. Time period of

pendulum is T 0 . When parallel plates are charged, the time period changes to T . The ratio T / T 0 is equal to

[MP PET 2003] 1 / 2

pE  (a)     I 

pE  (b)   

p  (d)     IE 

(a)

3 / 2

d 2   pE (as dt 2

  I   I

–

–

1 / 2

–

–

–

–

1 / 2

qE    g   m  (a)   g     

When dipole is given a small angular displacement  about it's equilibrium position, the restoring torque will be (as sin = =  )    pE sin   pE or I

[UPSEAT 2003]

m

 I 

1 / 2

 I  (c)    pE 

+ + + + + + + + + L

–

–

–

   g   (b)   g  qE    m  

3 / 2

1 / 2

   g   (c)   g  qE    m  

d 2 ) dt 2

(d) None of these

(c)

13.

or

d    2 with 2 dt



 

2

 2 

pE I

+

pE I

An infinite number of electric charges each equal to 5 nano-axis at x  1 cm, coulomb (magnitude) are placed along X -axis x  2 cm, x  4 cm x  8 cm ………. and so on. In the setup if the consecutive charges have opposite sign, then the x  0 electric field in Newton/Coulomb is Newton/Coulomb at  1 9 2 2     4  9  10 N  m / c  0  

(c)

–

Net downward force mg '  mg  QE

(b) 24  10

(c) 36  104

(d) 48  104

Hence time period T  2

4

9 9  5  109 5  10 5  10 .   E  4 0  (1  10 2 )2 (2  10 2 )2 (4  10 2 )2

15.

1

  E 

 

Effect acceleration g '   g 



[EAMCET 2003]

(a) 12  10 4

QE mg

 (5  10 9 )  .....  2 2 (8  10 ) 

9 9  9  10  5  10  1 1 1 ...  1  2  2  2  ... 4 10  (2) (4) (8) 

  1 1 ...   E  45  104 1  2   ... 2  (4) (16)   1  1 1  45  104  2  2   ... ...  2 (8) (32)  (2) 

  4   1  45  10  1 1 4 1 2   E  45  10   ..   2 2 (2) (16) 1  1   4   16 

(c)

QE 



m 

l l  2 QE   g '  g   m  

Three charges q1 ,  q2 and q3 are placed as shown in the figure. The x -component -component of the force on q1 is [AIEEE 2003] proportional to q q Y – q q3 (a) 22  32 sin  b a q q (b) 22  32 cos  b a  a b q2 q3 (c) 2  2 sin  X b a – q q1 +q2 q q (d) 22  32 cos b a – q q3 a

b F 2



q1 – q

F 3 sin  +q2



F 3 F 3 cos 


ELECTROSTATICS






F 2 = Force applied by

q 2 on q1 F 3 = Force applied by (q 3 ) on – q1

kQ2 kQ2 KQq Q2 and F O  F A  k 2 , F C  2 , F D  2 a a (a 2 ) (a 2)2

x-component of Net force on q1 is F x = F 2 + F 3 sin 



16.

F C

qq qq  k 1 22  k. 1 23 sin  b a

F AC A

qq q q  F x  k  1 22  1 2 3 sin   a  b 

F O O

q q q   q  F x  k  q1  22  32 sin    F x   22  32 sin   a a b   b 

D

A piece of cloud having area 25  10 6 m 2 and electric potential of 10 5 volts. If the height of cloud is 0.75 km , then energy of electric field between earth and cloud will be (a) 250 J

(b) 750 J

(c) 1225 J

(d) 1475 J

Energy 

1 2

 0 E 2  ( A  d) 

1  kQ2 kQ2 kQ2   F A  F C  F D  2 2  2  2  2   2  2a a a  2



19.

18.

(c)  (b)

Q 2

4

(1  2 2 )

(d)

Q 2

(1  2 2 )

If all charges are in equilibrium , system is also in equilibrium. Charge at centre : charge q is in equilibrium because no net force acting on it corner charge : If we consider the charge at corner B. This charge will experience following forces


q

Q 4

1  2 2 

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

5

Q

4



s t l o v 4 n i l a 3 i t n e 2 t o P1

r

(d)

KQ2  1  2 KQq  2  2 2  a  a2

[NCERT 1975; MP PMT 2003]

[AIEEE 2004]

(a) r (b) 2r (c) r /2 (d) r /4 Charge q will momentarily come to rest at a distance r from from charge Q when all it's kinetic energy converted to potential 1 1 qQ energy i.e. mv2  . 2 4 0 r Therefore the distance of closest approach is given by 1 qQ 2  2 r  .  r  2 4 0 mv v Hence if v is doubled, r becomes becomes one fourth Four charges equal to – Q are placed at the four corners of a square and a charge q is at its centre. If the system is in [AIEEE 2004] equilibrium the value of q is Q Q (1  2 2 ) (a)  (1  2 2 ) (b)

2kQq

a2 For equilibrium of charge at B, F AC  F D  F O

V 2  1   0  2  Ad d  2  

v

2

Force at B towards the centre  F O 

A charged particle q is shot towards another charged particle Q which is fixed, with a speed  . It approaches Q upto a closest distance r and and then returns. If q were given a speed 2 , the closest distances of approach would be q

C

a

Force at B away from the centre = F AC  F D

1 8.85  1012  (105 )2  25  106    1475 J 2 0.75  103 17.

F A

B

+q

[RPET 1997]

(d)

F D

0

1

2

3

4

5

6

Distance R in metres

(a)

(a) 2.5 volt / / m

(b) 2.5 volt / / m

(c) 2 / 5 volt / / m

(d) 2 / 5 volt / / m

Intensity at 5m is same as at any point between B and C because the slope of BC is same throughout ( i.e., electric field between B and C is uniform). Therefore electric field at R = 5m is equal to the slope of line BC hence by dV (0  5) V E   2.5 E   ; dr 64 m s t l o v n i l a i t n e t o P

A

5 4 3 2 1 O

1

B

2 3 4 5 6 Distance R in meters

At R = 1 m, E  

C

V (5  0)  2.5 m (2  0)

and at R = 3m potential is constant so E = 0. ELECTROSTATICS






The figure gives the electric potential V as a function of distance through five regions on x -axis. Which of the following is true for the electric field E in these regions

V

(c)

(d)

V

[AMU 2000]

r

V

r

V inside 

(b)

1

2

3

4

(a) E1  E2  E3  E4  E5 (b) E1  E3  E5 and E2  E4

4 0 R

and V out 

5

 R ....(i) for r 

Q 4 0r

 R ....(ii) for r 

i.e. potential inside the hollow spherical shell is constant 1 and outside varies according to V  . r

x

(c) E2  E4  E5 and E1  E3 (d) E1  E2  E3  E4  E5 (b)

Q

23.

Electric field in the region 1, 3 and 5 is zero i.e. E i.e. E1 = E3 = E5 Slope of the line BC < Slope of the line DE i.e. E i.e. E2 < E4

The electric field due to a uniformly charged sphere of radius R as a function of the distance from its centre is [AIIMS 2004] represented graphically by E

E

(a)

(b)

D

C

V R

B

A

(c)

r

R

r

R

r

E

E

(d)

E

1 21.

3

2

4

5

Which of the following graphs shows the variation of electric field E due to a hollow spherical conductor of radius R as a function of distance from the centre of the spher [AMU 2001] (a)

E

E

(b)

(b)

Einside 



3 0

R

r

r

(r < < R)

Eoutside  i.e. inside the ( E  r ) with

r

R

(c)

E

R

(a)

E

r

E 

3 0r 2

 R) (r 

uniformly charged sphere field varies linearly distance and outside varies according to

1

r 2

r

R

Electric field due to a hollow spherical conductor is governed by following equation E  0, for r < < R ...(i) Q  R ....(ii) and E  for r  4 0r 2 i.e. inside the conductor field will be zero and outside the conductor will vary according to E 

22.

x

R

(d)

 R3

O

1

r 2

In a hollow spherical shell potential ( V ) changes with respect [DCE 2001, 03] to distance ( r ) from centre V

V

(a)

(b)

r


r

ELECTROSTATICS



NUMERICAL BANK OF ELECTROSTATICS FOR IIT PMT.pdf

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