Question

A charge of +0.08 C moves to the right due to a 4 N force exerted by an electric field. What is the magnitude and direction of the electric field?

0.32 N/C left
0.32 N/C right
50 N/C left
50 N/C right

Answer 1

E = F/q = 4 N / 0.08 C = 50 N/C. For a positive charge the field points in the same direction as the force, so 50 N/C to the right.