Electrostatic Potential Test

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Electrostatic Potential Test - 3

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Question 1
1 / -0.25

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 2
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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 3
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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
[M² L T^-2 A^-1 ]

Solution
Question 4
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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/(4 πεr), where r = 1m
V = (2 X 10^-9 )/(4 πεx 1) = 18 volts.
Question 5
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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=q₀ x[v_final -v_initial ]
100=4x[v-(-5)]
v+5=25
v=20V
Question 6
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A particle of charge q₁ = 3 μC is located on x-axis at the point x₁ = 6 cm. A second point charge q₂ = 2 μC 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[(2 µc/4) + (3 µc/6)]
=9c10⁹ x10^-6 [1/2+1/2]x10²
=9x10⁵
Question 7
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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 [ L² M¹ T^-3 I^-1 ]
Question 8
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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 9
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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 10
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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,