(a) What work was done by the electric force? (b) What is the potential of the s
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Question
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the end point? (b) What is the potential of the starting point with respect to the end point?
(c) What is the magnitude of the electric field?
1/3 points I Previous Answers YF14 23 P053. A particle with charge +6.70 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 5.00 cm, the additional force has done 6.7S 10's of work and the particle has 4.35 x 10.5 J of kinetic energy. My Notes Ask Your Tea a) What work was done by the electric force? 2.4.10-os x (b) What is the potential of the starting point with respect to the end point? 358208 xV (c) What is the magnitude of the electric field? 1641.79 V/m
Explanation / Answer
A) 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
So, Welec + Wappl = dKE
=> Welec + (6.75e-5 J) = (4.35e-5 J)
=> Welec = 2.4e-5 J
B) The difference in electric potential is the work done by the particle on the field normalized by its charge q.
Hense,V2 - V1 = (-Welec) / q
=> V1 - V2 = (+ Welec) / q = (2.4e-5 J) / (6.7e-9 C) = 3.58e+4 Volt
C) Electric field is:
E = F/q = Welec / (dq)
=> E = 2.4e-5 / (6.7e-9 * 5e-2) = 7.164e+4 N/C
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