The electric potentials V are measured at distances x from P along a line PQ. The results are

Question 29

The electric potentials V are measured at distances x from P along a line PQ. The results are

V / V               13                    15                    18                    21                    23

x / m                0.020               0.030               0.040               0.050               0.060

The component along PQ of the electric field for x = 0.040 m is approximately

A 75 Vm^-1 towards P.

B 300 Vm^-1 towards Q.

C 300 Vm^-1 towards P.

D 450 Vm^-1 towards Q.

E 450 Vm^-1 towards P.

Reference: Past Exam Paper – J78 / II / 22

Solution:

Answer: C.

The electric potential at a point is defined as the work done per unit positive charge in moving a small test charge from infinity to the point. At infinity, the electric potential is defined to be zero.

The electric field strength E is related to the potential gradient ΔV/Δx as follows:

Electric field strength: E = – ΔV / Δx


We want to find the electric field strength at x = 0.040 m. So, we need to consider values of potential around this point.

Concerned points (V, x): (15, 0.030), (18, 0.040) and (21, 0.050)

The distance are being measured from point P. So, the smaller the value of x, the closer it is to P.


EITHER
E = – ΔV / Δx                             
E = – (18-15) / (0.040-0.030) = – 300 V m^-1

OR
E = – ΔV / Δx
E = – (21-18) / (0.050-0.040) = – 300 V m^-1


The negative sign in the value of the electric field strength E indicates that E is in the direction of decreasing x.

The values of x are smaller closer to P. So, the direction is towards P.