Question

A uniform electric field with a magnitude of 6100 V m−1 points in the positive x direction.
Find the change in electric potential energy when a +14 μC
charge is moved 5 cm in the positive x direction.

Answer in joules.

Answer 1

PE= -Eqd=61OO V/m * 12E-6C*.05, so the energy is reduced.

Electric field potential is defined as the negative of the work done by the electric field moving a charge in an electric field.