Suppose two equipotential surfaces are the planes x = 3 m and x = 5 m.If the potential at the x = 3 m surface is 10 V and the potential at the x = 5 m surface is 50 V,what is the magnitude of the electric field between the two surfaces?
A)40 N/C
B)20 N/C
C)15 N/C
D)10 N/C

Question

Quizplus · Accepted Answer

The potential difference between two equipotential surfaces is given by the formula:

ΔV = V2 - V1

where V2 is the potential of the second surface and V1 is the potential of the first surface.

In this case, we have:

ΔV = 50 V - 10 V = 40 V

The electric field between the two surfaces is given by the formula:

E = ΔV/d

where d is the distance between the two surfaces.

In this case, we have:

d = 5 m - 3 m = 2 m

So:

E = 40 V/2 m = 20 N/C

Therefore, the answer is B.

Suppose Two Equipotential Surfaces Are the Planes X = 3