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?

asked by Danielle

12 years ago

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1 answer

ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} =
=-7.05 J

answered by Elena

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Question

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?

1 answer

ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} =
=-7.05 J

Elena · Answer

ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} = =-7.05 J