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?

Coulomb's Law:[tex]F=\frac{kq_1q_2}{r^2}[/tex]k is a constantq1 and q2 are chargesr is the distance[tex]F_1=\frac{kq_1q_2}{r^2}[/tex]the distance is tripled:[tex]F_2=\frac{kq_1q_2}{(3r)^2}=\frac{kq_1q_2}{9r^2}[/tex][tex]\frac{\frac{kq_1q_2}{9r^2}}{\frac{kq_1q_2}{r^2}}= \frac{r^2}{9r^2}=\frac{1}{9}[/tex]The strength of the electrostatic repulsion will decrease by a factor of 9.