Question

A point charge q₁ = +3.90 is at origin. How far should the second point charge of +6.30  be placed to have electric potential energy of 0.700 J?

Answer 1

Potential energy

$U=\frac{kq_1q_2}{r}$

$r=\frac{kq_1q_2}{U}$

$$r=\frac{9\times10^9\times3.90\times10^{-6}\times6.30\times10^{-6}}{0.700}=0.3159m$$