Question

The electrostatic force between two point charges is 4.7 x 10-5 N. If the distance between them is tripled, the charge of one of the points triples and the charge of the other point halves. What will be the force between them?

Answer 1

$$F=\frac{kq_1 q_2}{r^2}$$

$$F'=\frac{k(3q_1)(0.5 q_2)}{(3r)^2}=\frac{kq_1 q_2}{6r^2}$$

$$F'=\frac{F}{6}=\frac{4.7 \cdot 10^{-5}}{6}=7.8 \cdot 10^{-6} N$$