Two insulated charged copper spheres A and B are separated by $50 ext{ cm}$ and experience a mutual

Question

Two insulated charged copper spheres A and B are separated by $50	ext{ cm}$ and experience a mutual repulsive force of $1.52 	imes 10^{-2}	ext{ N}$. What is the force of repulsion if each sphere is charged double the original amount, and the distance between them is halved?

Professor PC Thomas Classes · Answer

New force $F' = k \frac{(2q_1)(2q_2)}{(r/2)^2} = k \frac{4 q_1 q_2}{r^2 / 4} = 16 \left(k \frac{q_1 q_2}{r^2}ight) = 16F$. Thus $F' = 16 	imes 1.52 	imes 10^{-2} = 0.2432	ext{ N}$.