Question

The force between two electrical charges is 0.4 N. What would be the force if the charges are doubled, and the distance between them increased by 100%? 

Answer 1

$$F=k\frac{q_1q_2}{r^2}$$

$$F'=k\frac{2q_1*2q_2}{(2r)^2}=k\frac{q_1q_2}{r^2}=F=0.4\:\rm N$$