Electrostatics for JEE

ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI

IIT – JEE-2014 TIME: 1 HR

TW TEST TOPIC: ELECTROSTATISTICS

MARKS: 60 DATE: 29/04/13

SINGLE CHOICE QUESTIONS (+3, –1) 1.

2.

A spherical charged conductor has surface density of charge as  . The electric field intensity on its surface is E. If the radius of surface is doubled, keeping  unchanged , what will be the electric field intensity on the new sphere? (a) E 2 (b) E 4 (c) 2E (d) E

  Charge Q is given a displacement r  aiˆ  b ˆj in an electric field E  E1ˆi  E 2ˆj . The work done is (a) Q  E1a  E 2b 

(b) Q

 E1a 2   E 2b 2

(c) Q  E1  E2  a 2  b 2

(d) Q

 E12  E22 

2

a 2  b2

3.

An insulated sphere of radius R has a uniform volume charge density  . The electric field at a point P inside the sphere at a distance r from the centre is R r 2  r  (a) (b) (c) zero (d)   30 30 3  0 

4.

Consider two concentric spherical surfaces S1 with radius a and S2 with radius 2a , both centred on the origin. There is a charge q at the origin, and no other charges. Compare the flux 1 , through S1 with the flux 2 through S2 . (a) 1  42 (b) 1  22

(c) 1  2

(d) 1  2 2

5.

If the flux of the electric field through a closed surface is zero, then (a) the electric field must be zero everywhere on the surface . (b) the total charge inside the surface must be zero. (c) the electric field must be uniform throughout the closed surface. (d) the charge outside the surface must be zero.

6.

Two particles A and B (B is right of A) having charges 8 106 C and 2  106 C , respectively, are held fixed with separation of 20 cm . Where should a third charged particle be placed so that it does not experience a net electric force. (a) 5 cm right of B (b) 5 cm left of A (c) 20 cm left of A (d) 20 cm right of B

7.

A point charge of 100 C is placed at 3iˆ  4ˆj m . Find the electric field intensity due to this charge









at a point located at 9iˆ  12ˆj m . (a) 8000 Vm 1 8.

(b) 9000 Vm1

(c) 2250 Vm1

(d) 4500 V m 1

It is required to hold equal charges q in equilibrium at the corners of a square. What charge when placed at the center of the square will do this? q q q q (a)  1  2 2 (b) 1  2 2 (c) 1  2 2 (d)  1  2 2 2 2 4 4

















CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

9.

Four identical charges Q are fixed at the four corners of a square of side a . The electric field at a point P located symmetrically at a distance a 2 from the centre of the square is (a)

Q 2 2 0a

2

(b)

Q 20a

(c)

2

2 2Q 0 a

2

(d)

2Q 0 a 2

10.

A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. Find the electric potential at the centre of the shell 1 q 1 4q 1 5q 1 2q (a) (b) (c) (d) 40 2R 40 3R 40 6R 40 3R

11.

We have three identical metallic spheres A,B and C.A is given a charge Q and B and C are uncharged. The following processes of touching of two spheres one carried out in succession. Each process is carried out with sufficient time: (i) A and B (ii) B and C (iii) C and A (iv) A and B (v) B and C The final charges on the sphere are 11Q 5Q 11Q 11Q 11Q 5Q 8Q 5Q 5Q 5Q 11Q 11Q (a) (b) (c) (d) , , , , , , 32 16 32 32 32 16 8 16 16 16 32 32

12.

A non-conducting sphere with a cavity has volume charge density .O1 and O2 represent the two centres as shown. The electric field inside the cavity is

E0 . Now, an equal and opposite charge is given uniformly to the sphere on its outer surface. The magnitude of electric field inside the cavity becomes (a) zero (b) E0 (c) 2E 0 (d) 3E 0 13.

A conducting spherical shell is earthed. A positive charge  q1 is placed at the centre and another small positive charge q2 is placed at a distance

r from q1 (see figure). Ignore the effect of induced charge due to q 2 on the sphere. Then the coulomb force on q 2 is qq (a) zero (b) 1 2 4r 2 q1q 2 q1q 2 (c) (d) 2 4 0  r  R  40 r 2  R 2



14.



Consider a system of three charges q 3, q 3 and  2q 3 placed at points A,B and C, respectively, as shown in Figure. Take O to be the centre of the circle of radius R and angle CAB  60 . Then (a) the electric field at point O is q 80 R 2 directed along the negative x  axis (b) the potential energy of the system is zero (c) the magnitude of the force between the charges at C and B is q 2 54 0 R 2 (d) the potential at point O is q 12 0R

15.

A uniformly charged thin spherical shell of radius R carries uniform surface charge density of  per unit area. It is made of two hemispherical shells, held together by pressing them with force F. F is proportional to 1 2 2 1 2 (a)  R (b)  R 0 0 (c)

16.

1 2 0 R

(d)

1 2 0 R 2

Three concentric metallic spherical shells of radii R, 2R,3R are given charges Q1, Q2 , Q3 , respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells Q1 : Q2 : Q3 is (a) 1: 2 : 3 (b) 1:3: 5 (c) 1: 4 : 9 (d) 1: 8:18

B R

2R 3R

COMPREHENSION TYPE (+3, –1) The electric potential varies in space according to the relation V  3x  4y . A particle of mass 0.1kg starts from rest from point  2,3.2  under the influence of this field. The charge on the particle is 1 C . Assume V and  x, y  are in S.I. units

17.

The component of electric field in the x  direction  E x  is (a) 3Vm1

18.

20.

(c) 5Vm1

(d) 8Vm1

 

The component of electric field in the y  direction E y is (a) 3Vm1

19.

(b) 4 Vm1

(b) 4 Vm1

The time taken to cross the x  axis is (a) 20 s (b) 40 s

(c) 5Vm1

(d) 8Vm1

(c) 200 s

(d) 400 s

The velocity of the particle when it crosses the x  axis is (a) 20 103 ms 1

(b) 40  103 ms1

(c) 30 103 ms1

(d) 50 103 ms1

ANSWER KEY 1.

(d)

2.

(a)

3.

(b)

4.

(a)

5.

(b)

6.

(d)

7.

(b)

8.

(d)

9.

(b)

10.

(c)

11.

(d)

12.

(b)

13.

(a)

14.

(c)

15.

(a)

16.

(b)

17.

(a)

18.

(b)

19.

(d)

20.

(a)

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA

ELECTROSTATISTICS SOLUTION 1.

(d) E

 does not depend upon radius if  is constant 0

2.

(a)     W  F.r  qE.r

3.

(b) 3 1  4 3 R r r E   30 R 3 40 R3

kQr

4.

(a)

5.

(b) q q   in  0  in  qin  0 0 0

6.

(d) B

20cm

x Q

A 6

6

2 10 C

810 C

Let the third charge Q be placed at a distance x to the right of B. Then kQ  8 10 6

 20  x 2



 7.



kQ 2 106



x2

x  20 cm

(b)  r   9  3 iˆ  12  4  ˆj  6iˆ  8jˆ  r  62  82  10 m E

9  109  100 106 10

8.

2

 9000 Vm 1

(d) AC  2l  BD 1  BO  2

q

D

Q O



FBO  FBD   FBA  FBC  cos 45 Solving, we get q Q  1  2 2 , Q should be negative of q 4





C

FBO A

q

B

FBA

q FBC

FBD

9.

(b)

Q

E

Q C

D

 P

a

 a

a

O

2 B

A

a

E net  4E cos   10.

11.

Q

Q

O

Q

2

4 Q 1 Q  2 40 a 2 2  0a 2

(c) q q q   R 2R 3R 6q  3q  2q 6R

q 2R

3R

(d) (i) Q 2, Q 2, 0 (iv) 5Q 16,5Q 16,3Q 8

(ii) Q 2, Q 4, Q 4 (iii) 3Q 8, Q 4,3Q 8 (v) 5Q 16,11Q 32,11Q 32

12.

(b) Electric field will remain the same, because electric field due to surface charge distributed uniformly will be zero at any point inside sphere.

13.

(a) There is no charge on the outer surface. Hence, no force on q 2

14.

(c) FBC 

15.

1 4 0

R 3

2

q2 540 R 2

(a) Pressure 

16.

 q  2q     3 3     

(b) Q1

4R

2



2 2 and force  R 2 20 2 0

Q1  Q2 4   2R 

2

Q  Q 2  Q3  1 2 4  3R 

 Q1 : Q2 : Q3 ::1: 3 : 5

17.

(a)

18. (b) 19. V  y Ex    3Vm 1; E y    4Vm1 x y

qE x 1106  3   3 105 ms2 m 0.1 qE y 1106  4 ay    4 105 ms 2 m 0.1 Time taken to cross the x  axis 1 Using s  ut  at 2 2 1 3.2   4  105  t 2 2 t  400 s ax 

x  a x t  3  10 5  400  12  103 ms 1

 y  a y t  4 10 5  400  16  103 ms 1

  2x  2y  20  10 3 ms 1

(d)

20.

(a)