Electrostatics Test

Result

Electrostatics Test - 12

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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

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

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

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

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

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

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

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

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

Potential at the outer surface,
∴ If V is the required potential at r = 3r₀/2,
Question 10
1 / -0

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
Question 11
1 / -0

E = -dV/dr, here the negative sign signifies that

A
E increases when V decreases

B
E is opposite to V

C
E is negative

D
E is directed in the direction of decreasing V

Solution

Answer :- d
Solution :- The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.
Question 12
1 / -0

Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is

A
6.67J

B
60J

C
Zero

D
15J

Solution

The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q₀=0.
Question 13
1 / -0

Dimensional formula for potential difference is

A
[ML² T^-2 A^-1]

B
[ML² T^-3 A^-1]

C
[ML T^-1 A^-1]

D
[M²L T^-2 A^-1]

Solution
Question 14
1 / -0

A point charge 2nC is located at origin. What is the potential at (1,0,0)?

A
12

B
14

C
16

D
18

Solution

V = Q/(4πεr), where r = 1m
V = (2 X 10^-9)/(4πε x 1) = 18 volts.
Question 15
1 / -0

If 100 J of work has to be done in moving an electric charge of 4C from a place where potential is -5 V to another place, where potential is V volt. The value of V is

A
15 V

B
20 V

C
25 V

D
10 V

Solution

From the definition, the work done to a test charge ‘q0’ from one place to another place in an electric field is given by the formula
W=q₀x[v_final-v_initial ]
100=4x[v-(-5)]
v+5=25
v=20V
Question 16
1 / -0

A particle of charge q₁ = 3μC is located on x-axis at the point x₁ = 6 cm. A second point charge q₂ = 2μC is placed on the x-axis at x₂ = -4 cm. The absolute electric potential at the origin is

A
9 x 10^-6 V

B
9 x 10⁵ V

C
9 x 10^-4 V

D
9 x 10^-2 V

Solution

V=kq/r
Therefore, V_net=Kx[(2µc/4) + (3µc/6)]
=9c10⁹x10^-6[1/2+1/2]x10²
=9x10⁵
Question 17
1 / -0

Electric potential is

A
scalar and dimensionless

B
vector and dimensionless

C
scalar with dimension

D
vector with dimension

Solution

Electric potential is the electric potential energy per unit charge. In equation form, V=U/q, where U is the potential energy, q is the charge, and V is the electric potential. Since both the potential energy and charge are scalar quantities, so does the potential.
And it's dimension [ L²M¹T^-3I^-1]
Question 18
1 / -0

A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral.

A
No potential difference appears between the two cylinders when same charge density is given to both the cylinders.

B
No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders.

C
A potential difference appears between the two cylinders when a charge density is given to the outer cylinder.

D
A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.

Solution

When the charge is given to inner cylinder than there is an electric field is produced between cylinders which is given by
and due to this a potential difference is developed between two cylinders.
Question 19
1 / -0

Equal charges are given to two spheres of different radii. The potential will

A
be equal on both the spheres

B
be more on the smaller sphere

C
be more on the bigger sphere

D
depend on the material of the sphere

Solution

When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.
Question 20
1 / -0

Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be

A
1:1

B
1:2

C
1:4

D
1:8

Solution

By dimensional analysis
but by superposition

Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
therefore,