Question

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 50 V. What is the change in potential energy of the charge associated with the change in position?

Answer 1

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J