A particle has a charge of +1.8 ?C and moves from point A to point B, a distance

Question

A particle has a charge of +1.8 ?C and moves from point A to point B, a distance of 0.19 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +9.1 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences. I found part a, but couldn't figure out part b.

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU q = 1.7 x 10^ - 6 d = 0.3 ?PE = 8.7 x 10^ - 4 ?V = ?PE/q = Ed ?PE/q = Ed E = ?PE/qd E = 1706 N/C F = qE F = 0.0029 N

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