Question

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 1

Answer

electric potential energy is given by

$U =\frac{kqq'}{r}\\=\frac{9*10^9*5*10^{-9}*9.5*10^{-9}}{0.5}\\=85.5*10^{-8}J$