Answer to Question #77087 in Electric Circuits for Mys

Question

A charge of 40 µC is pushed by a force of 15 µN a distance of 1.3 mm in an electric field.

What is the electric potential difference?

2.0 × 10-8 V
4.6 × 10-7 V
4.9 × 10-4 V
3.8 × 10-1 V

Answer 1

The work done
W=FS
W=qV
So
V=FS/q=(15×〖10〗^(-9)×1.3×〖10〗^(-3))/(40×〖10〗^(-9) )=4.9×〖10〗^(-4) V