Question

If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them will decrease by a factor of of how much?

Answer 1

Coulomb's Law:
$F=\frac{kq_1q_2}{r^2}$
k is a constant
q1 and q2 are charges
r is the distance


$F_1=\frac{kq_1q_2}{r^2}$

the distance is tripled:
$F_2=\frac{kq_1q_2}{(3r)^2}=\frac{kq_1q_2}{9r^2}$


$\frac{\frac{kq_1q_2}{9r^2}}{\frac{kq_1q_2}{r^2}}= \frac{r^2}{9r^2}=\frac{1}{9}$

The strength of the electrostatic repulsion will decrease by a factor of 9.