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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?

  • physics -

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

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

  • physics -

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

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