A particle with charge +8.50 nC is in a uniform electric field directed to the l

Question

A particle with charge +8.50 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 6.00 cm, the additional force has done 6.70 10-5 J of work and the particle has 4.35 10-5 J of kinetic energy.

A)What work was done by the electric force? -2.35e-5

B)What is the potential of the starting point with respect to the end point? 2764.71(Wrong!!!!!!)(2.35e-5/8.5e-9)Need Help!!

C)What is the magnitude of the electric field?

Explanation / Answer

Choose a coordinate system so that values on the x-axis increase to the right; then the particle moves a distance d > 0 in the direction as the applied force Fappl > 0 and opposite to the electric field E < 0. The work done by the field on the particle is
. . 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 + (6. 7e-5 J) = (4.35e-5 J)

A) Work done by electric force = -2.35*10^-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 = (6.7e-5 J - 4.35e-5 J) / (8.5e-9 C)

B) Potential = 0.27x 10^ 4 V

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
. . (8.5e-9 C) E_x (6e-2 m) = (4.35e-5 J) - (6.7e-5 J)

C) Magnitude of electric field = -.046 x 10^6 N/C

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