Two charges of equal magnitude and at distance r exert a force F on each other. If each charge is doubled and the distance between them is halved, then the new force acting on each charge will be

16F

The correct answer is Option (2) → 16F

$F = k \frac{q_1 q_2}{r^2}$

Original force: $F = k \frac{q \cdot q}{r^2} = k \frac{q^2}{r^2}$

New charges: $q_1 = 2q$, $q_2 = 2q$, new distance: $r' = \frac{r}{2}$

New force: $F' = k \frac{(2q)(2q)}{(r/2)^2} = k \frac{4q^2}{r^2/4} = 16 \frac{k q^2}{r^2} = 16F$

Answer: $F' = 16F$