Question

The attractive force between two equal but opposite charges is F newtons when the charges are 6 cm apart. What should be their distance apart to increase the attractive force to 6F newtons?

Answer 1

Attractive force

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

$F=-\frac{kqq}{r^2}$

$\frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}$

$\frac{F}{6F}=\frac{r_2^2}{0.06^2}$

$r_2^2=\frac{3.6\times10^{-3}}{6}$

$r=0.0244m$

$r=2.44cm$