Question

A proton moves in a constant electric field from point A to point B. The magnitude of the electric field is 6.4 × 10^4 N/C; and it is directed as shown in the drawing, the direction opposite to the motion of the proton. If the distance from point A to point B is 0.50 m, what is the change in the proton's electric potential energy, EPEB – EPEA?

Answer 1

6.4×10^4 N/C x 0.50 m * 1.60*10^-19 C
(Joules)

EPE increases going from A to B (against the field force)

Answer 2

Thanks for the help, but I don't quite understand why you chose to multiply everything together.