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A particle has a charge of +2.6 ?C and moves from point A to point B,...

A particle has a charge of +2.6 ?C and moves from point A to point B, a distance of 0.23 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 = +8.1 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.

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Answer #1

a) F = dU/d

= 8.1*10^-4/0.23

= 3.52*10^-3 N

b) E = F/q

= 3.52*10^-3/(2.6*10^-6)

= 1354.5 N/c

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