Asked by Danielle
A charge of - 0.15 C is moved from a position where the electric potential is 13 V to a position where the electric potential is 60 V. What is the change in potential energy of the charge associated with this change in position?
Answers
Answered by
Elena
ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} =
=-7.05 J
=-7.05 J