Question

Two point charges exert a 5 N force on each other. What will the resulting force be if the distance between them is increases by a factor of three?

Answer 1

$$F=\frac{kq_1q_1}{r^2}\\F'=\frac{kq_1q_1}{(3r)^2}=\frac{F}{9}=\frac{5}{9}=0.556\ N$$