13. 13 points YF14 23P053. My Notes Ask Your A particle with charge +8.00 nC is

Question

13. 13 points YF14 23P053. My Notes Ask Your A particle with charge +8.00 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 9,00 cm, the additional force has done 6.15 x 10-5 J of work e has 4.35 x 10-5 of kinetic energy. electric t is released from rest, it t has and the partide (a) What work was done by the electric force? b) What is the potential of the starting point with respect to the end poin? (c) What is the magnitude of the electric field? V/m

Explanation / Answer

(A) Net work done = change in KE

(6.15 x 10^-5) + W = 4.35 x 10^-5

W = - 1.8 x 10^-5 J

(B) Work done = q deltaV

-1.8 x 10^-5 = (8 x10^-9) V

V = - 2250 Volt  

(C) E = V/d = 2250 / 0.09

E = 25000 N/C

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