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

Answer

The distance must

$R=\frac{Kqq'}{U}\\=\frac{9*10^9*3.9*6.30}{0.7}\\=315.9m$