Electrostatic Potential Test

Result

Electrostatic Potential Test - 2

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Weekly Quiz Competition

Question 1
1 / -0.25

The minimum velocity v with which charge q should be projected so that it manages to reach the centre of the ring starting from the position shown in figure is

A

B

C

D

Solution
Question 2
1 / -0.25

The work done in moving a positive test charge qo from infinity to a point P at a distance r from the charge q is

A

B

C

D

Solution

Definition based
Question 3
1 / -0.25

Electric field intensity is equal to

A
time rate of change of potential

B
minimum rate of change of potential with distance

C
maximum rate of change of potential with distance

D
none of the above

Solution

Definition based
Question 4
1 / -0.25

Electric potential at a point located far away from the charge is taken to be

A
Unity

B
May be zero or infinite

C
Zero

D
Infinite

Solution

Electric potential at a point located far away from the charge is taken to be zero Because the distance is much larger so in other words there is no effect of other charges on that charge.
Question 5
1 / -0.25

A charge of 6 mC is located at the origin. The work done in taking a small charge of -2 x 10^-9 C from a point P (0, 3 cm, 0) to a Q (0,4 cm, 0) is

A
0.5 J

B
1.12 J

C
1.2 J

D
0.9 J

Solution

q=6x10^-10 c
Q=-2x10^-9c
r₁=3x10^-2m
r₂=4x10^-2m
△v_Q=w/q
(1/4πε_o)-2x10^-9/10^-2[(1/4)-(1/3)]=W/6x10^-3
9x10⁹x2x10^-9x6x10^-3/12x10^-2=W
W=9x10^-3x10²
W=0.9J
Question 6
1 / -0.25

A particle of mass 1 Kg and charge 1/3 μC is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. The minimum intial velocity V₀ of projection of particle required if the particle just grazes the shell is

A

B

C

D
1 ms^-1

Solution

From conservation of angular momentum,
mr₀ = r/2 = mvr
⇒ v = v₀/2
From conservation of energy,
Question 7
1 / -0.25

On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is

A
8 V

B
0.5 V

C
0.1 V

D
2 V

Solution

Potential difference between two points is given by

V_a - V_b = W/q₀

Work, W = 2 J

Charge, q₀ = 20 C

Potential difference = 2/20 = 0.1 V

The correct option is C.
Question 8
1 / -0.25

The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is

A
Zero

B
Infinity

C
One

D
Two

Solution

Since Potential difference between two points in equipotential surfaces is zero, the work done between two points in equipotential surface is also zero.
Question 9
1 / -0.25

A charge is uniformly distributed inside a spherical body of radius r₁ = 2r₀ having a concentric cavity of radius r₂ = r₀ (ρ is charge density inside the sphere). The potential of a point P or a distance 3r₀/2 from the centre is

A

B

C

D
none of these

Solution

Electric field at a distance r,
(r₀ < r < 2r₀) from the center is given by
Question 10
1 / -0.25

Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC^-1 and the electric potential at a point ‘B’ due to same charge is 6 JC^-1. The distance between AB is

A
2 m

B
1.5 m

C
1 m

D
0.5 m

Solution

E.l = V where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B,
l = (6 / 12) m or (1 / 2) m = 0.5 m