Question

What happens to the force between two charged metal spheres in a vacuum if the charge on each is doubled and the distance between them in multiplied by three?

Answer 1

$F_1=k\frac{q_1q_2}{x^2}$

$F_2=k\frac{2q_12q_2}{(3x)^2}$

$F_1/F_2=9/4=2.25\to F_2=F_1/2.25=0.44F_1$ . So, the force of the interaction will decrease .