The magnitude of the force of a 400-N/C electric field on a 0.02-C point charge is: A) 8.0 N B) 8 * 10^-⁵ N C) 8* 10^-³ N D) 0.08 N E) 2 *10¹¹ N

Question

Quizplus · Accepted Answer

The magnitude of the force on a point charge in an electric field is given by F = qE, where q is the charge and E is the electric field strength. In this case, q = 0.02 C and E = 400 N/C, so F = (0.02 C)(400 N/C) = 8.0 N.

The Magnitude of the Force of a 400-N/C Electric Field