Q#29 (Past Exam Paper – J78 / II / 22)
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.
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.
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