A particle with a charge of 4.50 nC is in a uniform electric field \"E\" directe

Question

A particle with a charge of 4.50 nC is in a uniform electric field "E" directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm , its kinetic energy is found to be 1.50

Explanation / Answer

Welec = E dot d < 0 and the work done by the applied force is . . Wappl = Fappl dot d > 0 Energy is conserved, so net work on the particle must equal the change in internal energy dKE . . Welec + Wappl = dKE . . Welec + (6e-5 J) = (3.9e-5 J) The difference in electric potential is the work done by the particle on the field normalized by its charge q. . . V2 - V1 = (- Welec) / q . . V1 - V2 = (+ Welec) / q = (6e-5 J - 3.9e-5 J) / (8.6e-9 C) Electric force on the particle is . . Felec = qE where E is the (vector) electric field. Work is force x distance (force dot distance, really), where here force and distance are in opposite directions. . . Felec dot d = qE dot d = Welec

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