Asked by mageda

A particle has a charge of +2.2 μC and moves from point A to point B, a distance of 0.16 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 = +7.8 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
Answered by jjk
4.8
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