A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.164

Question

A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.164 m in the presence of an external electric field of magnitude 265 N/C directed toward the right as in the following figure.

------->E------>
A B
(+) ------d------ (+)
q

(a) Find the electric force exerted on the charge.
magnitude ________??? N

(b) Find the work done by the electric force.
________???? J

(c) Find the change in the electric potential energy of the charge.
______???? J

(d) Find the potential difference between A and B.
VB - VA = ________??? V

Explanation / Answer

a) F = qE = 41x10^-6 x 265 = 0.010865 N b) W= Fxd = 0.010865x0.164 = 1.78x10^-3 Joules c) Change in potential energy = W = 1.78x10^-3 Joules d) Vb- Va = W/q = 1.78x10^-3 / (41x10^-6) = 43.41 Volts

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