A particle with charge +7.90 nC is in a uniform electric field directed to the l

Question

A particle with charge +7.90 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 cm, the additional force has done 6.95 Times 10^-5 J of work and the particle has 4.35 Times 10^-5 J of kinetic energy. What work was done by the electric force? What is the potential of the starting point with respect to the end point? What is the magnitude of the electric field?

Explanation / Answer

a)

Work done by electric field = (4.35-6.95)x10^-5

Work = -2.6 x 10^-5 J

b)

W = -qV

V = W/-q

V = 3291.14 Volts

c)

Ed = V

E = 41139.2 N/C

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