Question

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

Answer 1

Force between two charges is given by

$F = \frac{kq_1q_2}{r^2}$

here force is inversely depends on the square of the distance

so here if distance is decreased the force will increase'

now if force is tripled

$\frac{3F}{F} = \frac{r^2}{r'^2}$

initially distance between charges was 8.45 cm

$3 = \frac{8.45^2}{r'^2}$

$r' = \frac{8.45}{\sqrt3}$

$r' = 4.88 cm$

so the distance between charges is 4.88 cm now