The work done by an external force to move a -7.75 µC charge from point a to point b is 25.0 * 10-4 J. If the charge was started from rest and had 4.73*10-4 J of kinetic energy when it reached point b, what is the magnitude of the potential difference between a and b?

Answer 1

The charge times the potential difference is the increase in potential energy. That plus the acquired kinetic energy equals the work done.

(Vb - Va)*Q + 4.73*10^-4 J
= (25.0)*10^-4 J

Answer 2

I don't understand how you get the answer. What is the Vb, Va and Q?