Question

A point charge q1 = +2.80 µC is at origin. How far should the second point charge of +5.20 µC be placed to have electric potential energy of 0.600 J?

Answer 1

Answer

The distance is

$r=\frac{kQQ'}{U}\\=\frac{9*10^9*2.8*5.2*10^{-12}}{0.6}\\=6.5cm$