Question

What happens to the electrostatic force when both charges are doubled?

Answer 1

Initial case:

Charges are $q_1,q_2$

DIstance =$r$

$\therefore F=K\frac{q_1{q}_2}{r^2}$

Final case:

$q_1^{\prime }=2q_1$

$q_2^{\prime }=2q_2$

$r^{\prime }=r$

$\therefore F^{\prime }=K\frac{2q_1\times 2q_2}{r^2}$

$F^{\prime }=4F$

Hence electrostatic force increases $4$ times of its initial value.