The magnitude of the force of a 400-N/C electric field on a 0.02-C point charge is: A)8 N B)5 * 10^-5 N C)8 *10^-3 N D)0.08 N E)2 *10³ N

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

Quizplus · Accepted Answer

The force on a charge in an electric field is given by F = qE, where q is the charge (0.02 C) and E is the electric field strength (400 N/C). Thus, F = 0.02 C * 400 N/C = 8 N.

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