A charge of 23.0 nC is placed in a uniform electric field that is directed vertically...

Question

Question

Answer 1

Answer #1

part C)

q = 23 nC = 23 *10^-9 C

E = 4 *10^4 N/C

d = 2.4 m

theta = 45 degree

work done by the electric force = -q *E * d * cos(45 degree)

work done by the electric force = -23 * 10^-9 * 4 *10^4 * 2.4 * cos(45 degree)

work done by the electric force = -0.00156 J

the work done by the electric force is -0.00156 J

Answer 2

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