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Target 2021 DPP - 3 Duration: 45min

Topic : Electrostatics
7. A positively charged disc is placed on horizontal plane. A
1. An electrical charge 2  10 8 C is placed at the point (1, 2, 4)
charged particle is released from a certain height on its axis.
m. At the point (4, 2, 0) m, the electric The particle just reaches the centre of the disc. Select the
(a) potential will be 36 V correct alternative.
(b) field will be along y-axis (a) Particle has negative charge on it
(c) field will increase if the space between the points is filled (b) Total potential energy (gravitational + electrostatic) of the
with a dielectric particle first increases, then decreases
(d) All of the above (c) Total potential energy of the particle first decreases, then
2. Charge Q is given a displacement r  aˆi  bˆj in an electric increases
(d) Total potential energy of the particle continuously
field E  E1iˆ  E2 ˆj. The work done is decreases
2 2
8. A Point charge q1 = q is placed at point P. Another point
(a) Q  E1a  E2b  (b) Q  E1a    E2 b  charge q 2 = – q is placed at point Q. At some point
R  R  P, R  Q  , electric potential due to q1 is V1 and electric
(c) Q  E1  E2  a 2  b 2 (d) Q E12  E22 a 2  b 2
potential due to q2 is V2. Which of the following is correct?
3. Six charges are placed at the vertices of a rectangular hexagon
as shown in the figure. The electric field on the line passing (a) Only for some points V1  V2
through point O and perpendicular to the plane of the figure (b) Only for some points V2  V1
as a function of distance x from point O is  x  a  (c) For all points V1  V2 (d) For all points V2  V1
Qa 9. Intially the sphere A and B are at potentials at potentials VA
(a) 0 (b)  x 3 and VB respectively. Now, sphere B is earthed by closing the
0
switch. The potential of A will now become
2Qa 3Qa
(c)  x 3 (d) (a) 0
0  0 x 3
(b) VA
4. If the electric potential of the inner shell is 10 V and that of
the outer shell is 5 V, then the potential at the centre will be (c) VA  VB
(d) VB
10. A particle of mass m and charge q is fastened to one end of a
string of length l. The other end of the string is fixed to the
point O. The whole system lies on a frictionless horizontal
plane. Initially, the mass is at rest at A, A uniform electric field
in the direction shown is then switched on. Then,
(a) 10 V (b) 5 V (c) 15 V (d) zero
5. There are four concentric shells, A, B, C and D of radii a, 2a,
3a and 4a respectively. Shells B and D are given charges +q
and –q respectively. Shell C is now earthed. The potential
 1 
difference VA  VC is  k  4  
 0 

kq kq kq kq 2qEl
(a) (b) (c) (d) (a) the speed of the particle when it reaches B is
2a 3a 4a 6a m
6. Potential difference between centre and surface of the sphere
qEl
of radius R and uniform volume charge density  within it (b) the speed of the particle when it reaches B is
will be m
(c) the tension in the string when the particle reaches at B is
R 2 R 2 R 2 R 2 qE
(a) (b) (c) (d)
6 0 4 0 30 2 0 (d) the tension in the string when the particle reaches at B is
zero

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11. A charged particle of mass m and charge q is released from
q2 q2
rest from the position  x0 , 0  in a uniform electric field E ˆj . 0
(c)
4 0 a
 
2 1 (d)
4 0 a
 
2 1

The angular momentum of the particle about origin 17. Two identical positive charges are placed at x   a and
(a) is zero (b) is constant x  a. The correct variation of potential V along the x-axis is
(c) increases with time (d) decreases with time given by
12. A charge + Q is uniformly distributed in a spherical volume of
radius R. A particle of charge +q and mass m projected with
velocity 0 from the surface of the spherical volume to its
centre inside a smooth tunnel dug across the sphere. The
minimum value of 0 such that it just reaches the centre
(assume that there is no resistance on the particle except (a) (b)
electrostatic force) of the spherical volume is

Qq Qq
(a) 20 mR (b)  0 mR

2Qq Qq
(c) (d) (c) (d)
 0 mR 4 0 mR
13. Two identical coaxial rings each of radius R are separated by
a distance of 3R. They are uniformly charged with charges 18. Two identical charges are placed at the two corners of an
+Q and –Q respectively. The minimum kinetic energy with equilateral triangle. The potential energy of the system is U.
which a charged particle (charge +q) should be projected
The work done in bringing an identical charge from infinity to
from the centre of the negatively charged ring along the axis
of the rings such that it reaches the centre of the positively the third vertex is
charged ring is (a) U (b) 2 U (c) 3 U (d) 4 U
19. A charged particle q is shot from a large distance towards
Qq Qq Qq 3Qq
(a) 4 R (b) 2 R (c) 8 R (d) 4 R anoter charged particle Q which is fixed, with a speed , It
0 0 0 0 approaches Q up to a closest distance r and then returns. If q
14. A uniform electric field exists in x-y plane. The potential of were given a speed 2 the distance of approach would be
points A (2m, 2m), B (–2m, 2m) and C (2m, 4m) are 4 V , 16 V
and 12 V respectively. The electric field is

 
(a) 4ˆi  5ˆj V / m  
(b) 3ˆi  4ˆj V / m (a) r (b) 2r (c) r/2 (d) r/4
20. Figure shows a closed dotted surface which intersects a con-
 
(c)  3ˆi  4 ˆj V / m (d)  3ˆi  4ˆj V / m ducting uncharged sphere. If a positive charge is placed at
the point P, the flux of the electric field through the closed
15. A small ball of mass m and charge +q tied with a string of surface
length l, rotating in a vertical circle under gravity and a uniform
electric field E as shown. The tension in the string will be
minimum for

1  qE 
(a)   tan  mg 
  (a) will remain zero (b) will become positive
(b)    (c) will become negatve (d) data insufficient

1  qE 
(c)   0o (d)     tan  mg 
 
16. Four point charges A, B, C and D are placed at the four corners
of a side a. The energy required to take the charges C and D
to infinity (they are also infinitely separated from each other)
is

q2
(a)
4  0 a

2q 2
(b)
 0 a

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Target 2021 DPP - 3 Solutions

Topic : Electrostatics
1. (A) q
2 2 2
Q
r  4  1   2  2  0  4 4
= 5m kq kq / 4 kq kq
Now, VA    
1 q 2a 3a 4 a 6a
V .
4 0 r and VC  0
kq

 9 10  2 10   36V
9 8
 VA  VC 
6a
5
6. (A)
Field is in the direction of r  rp  rq
q
2. (A) 
  R3
4 / 3
W  Fr
  QE   r  Q  E  r  4
 q R 3
3
 Q  E1a + E2 b 
3 1 q  1 q
3. (B) VC  VS   .  
2  4 0 R  4 0 R

q

80 R
Substituting the value of q, we have
R 3
VC  VS 
6 0
7. (C)
 1  P 
E  E1  E2  E3    3 
 40  x 

 1  2Qa
  3
 4 0  x
Resultant of E1 and E3 is also equal to E along E2
 Enet  2 E (along E2 ) Hence, in between A and C there is a point B, where speed of

Qa the particle should be maximum. F1  mg = constant


 0 x 3 F2 = electrostatic repulsion (which increases as the particle
4. (A) moves down)
From centre to the surface of inner shell, potential will remain From A to B kinetic energy of the particle increases the poten-
constant = 10 V (given). tial energy decreases . Then from B to C kinetic energy de-
5. (D) creases and potential energy increases.
Let Q charge comes on shell-C from earth. Then, 8. (C)
VC  0 V1 is positive and V2 is negative. Hence at all points
V1 > V2
kq kQ kq
   0 9. (C)
3a 3a 4a
qA will remain unchanged.
Solving, we get Hence, according to principal of generator potential differ-
ence will remain unchanged.

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V ' A  V 'B  VA  VB 15. (D)

T sin   qE
or V ' A  VA  VB (as V 'B  0 )
T cos   mg
10. (B)
WT  0  qE 
   tan  1  
W F   Fe  (displacement in the direction of force)  mg 
e
Minimum tension will be obtained at    .
= kinetic energy of the particle
16. (C)
1 2 l l  Energy required = U  U f  U i
 mv  qE   cos 60o 
2 2 2 
1  q2   q2 q q2 q2 q 2 q 2  
qEl           
 v 40  a   a 2a a a 2a a 
m
11. (C) q2
  2  1
L  mv r  m  at  x0  4 0 a  
17. (C)
 qE  On both sides of the positive charge V    just over the
 m 0  t  x0  or L  t
 m  charge.
12. (D) 18. (B)
U i  Ki  U f  k f 1 q2
U . (a = side of triangle)
1 2 4 0 a
or qVi  mvmin  qV f  0
2
 1 q2 
W  U f  U i  3  U
 1  Q  1 2 3 Q   40 a 
or q     mvmin  q   
 4  0  R  2  2 4  0R 3U  U  2U
From here, we can find vmin . 19. (D)
13. (A) U i  Ki  U f  K f
kC1  U C1  kC2  U C2
1 1 Qq 1
0  mv 2   0 or r 
kmin  qVC1  0  qVC2 ...(i) 2 40 r v2

1  Q Q 1
VC1     If v is doubled, the minimum distance r will remain th.
4 0  2R R  4
20. (C)
1 Q Q 
VC2    
4 0  R 2R 
Substituting these values in Eq. (i), we can find K min
14. (D)
E  E ˆi  E ˆj
x y
The induced charges on conducting sphere due to +q charge
at P are as shown in figure.
Now we can use,  dV    E  dr Now, net charge nside the closed dotted surface is negative.
two times and can find values of Ex and E y . Hence, according to Gauss’ theorem net flux is zero.

ANSWER KEY
1. (A) 6. (A) 11. (C) 16. (C)
2. (A) 7. (C) 12. (D) 17. (C)
3. (B) 8. (C) 13. (A) 18. (B)
4. (A) 9. (C) 14. (D) 19. (D)
5. (D) 10. (B) 15. (D) 20. (C)

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