Q. If the electric field inside a conductor in electrostatic equilibrium is zero, what can be said about the electric field just outside the conductor?
A.It is zero
B.It is uniform
C.It is perpendicular to the surface
D.It is parallel to the surface
Solution
The electric field just outside a conductor in electrostatic equilibrium is perpendicular to the surface.
Correct Answer: C — It is perpendicular to the surface
Q. If the electric potential at a point is 10 V and the electric field is uniform, what is the work done in moving a charge of 2 C from that point to a point where the potential is 0 V?
A.20 J
B.10 J
C.5 J
D.0 J
Solution
Work done W = q(V1 - V2) = 2 C (10 V - 0 V) = 20 J.
Q. If the electric potential at a point is 150 V and the electric field is directed towards the point, what can be said about the charge creating the field?
A.It is positive
B.It is negative
C.It is neutral
D.Cannot be determined
Solution
If the electric field is directed towards the point, it indicates that the charge creating the field is negative.
Q. If the electric potential at a point is 200 V and the electric field is uniform, what is the work done in moving a charge of 0.5 C to a point where the potential is 100 V?
A.50 J
B.100 J
C.200 J
D.0 J
Solution
Work done W = q * (V1 - V2) = 0.5 C * (200 V - 100 V) = 50 J.
Q. If the electric potential at a point is 300 V and the electric field is uniform, what is the work done in moving a charge of 3 μC to a point where the potential is 600 V?
A.0.9 mJ
B.0.6 mJ
C.0.3 mJ
D.1.2 mJ
Solution
Work done W = q * ΔV = 3 × 10^-6 C * (600 V - 300 V) = 3 × 10^-6 * 300 = 0.9 mJ.
Q. If the electric potential at a point is 50 V and the electric field strength is 5 N/C, what is the distance from the point to the reference point where the potential is zero?
Q. If the electric potential at a point is increased, what happens to the electric field at that point?
A.Increases
B.Decreases
C.Remains the same
D.Cannot be determined
Solution
An increase in electric potential at a point generally indicates a stronger electric field, as the electric field is related to the rate of change of potential.
Q. If the electric potential at point A is 5 V and at point B is 15 V, what is the work done by the electric field in moving a charge of 2 C from A to B?
A.-20 J
B.20 J
C.10 J
D.30 J
Solution
Work done W = Q(V_B - V_A) = 2 C * (15 V - 5 V) = 20 J.
Q. If the radius of a spherical Gaussian surface is doubled while keeping the charge inside constant, how does the electric field change?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
Q. If the radius of a spherical Gaussian surface is doubled, how does the electric field change if the enclosed charge remains constant?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
