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 emf of a cell is 12 V and the potentiometer wire is 20 m long, what is the potential gradient if the wire is calibrated to give a reading of 0.6 V/m?
A.12 V
B.0.6 V/m
C.0.3 V/m
D.0.5 V/m
Solution
The potential gradient is given as 0.6 V/m, which is the calibration value for the potentiometer.
