A particle with a charge of 5nC has a distance of 0.5m away from a charge of 9.5nC. What is its electric potential energy?
Answer
electric potential energy is given by
U=kqq′r=9∗109∗5∗10−9∗9.5∗10−90.5=85.5∗10−8JU =\frac{kqq'}{r}\\=\frac{9*10^9*5*10^{-9}*9.5*10^{-9}}{0.5}\\=85.5*10^{-8}JU=rkqq′=0.59∗109∗5∗10−9∗9.5∗10−9=85.5∗10−8J
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