Question

Two charged particles are 8.45 cm apart. They are moved and the force on each of them is found to have tripled. How far are they now?

Answer 1

$3F/F=(q_{1}q_2/x^2)\div (q_1q_2/8.45^2)$

$3=8.45^2/x^2$

$x=8.45/ \sqrt3=4.88cm$