Thursday, July 30, 2009

A positive test charge of 6.68 x 10^-5 C is placed in an electric field of 54.52 N/C?

intensity. What is the strength of the force exerted on the test charge? Answer in N.

A positive test charge of 6.68 x 10^-5 C is placed in an electric field of 54.52 N/C?
F = qE = (6.68 x 10^-5)(54.52) = 3.64 x 10^-3 N


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