Asked by Mikie
A charge of 0.25 C moved from a position where the electric potential is 20 V to a position where the potential is 50 V. What is the change in potential energy?
Answers
Answered by
bobpursley
work= PE1-PE= q1V1-q2V2=.25 (-30)=-7.5J
PE change = - work done
PE change = - work done
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