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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 16F \)
Coulomb's law gives the force between two point charges:
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^{2}} \)
So for a given pair \( F \propto q_1q_2/r^{2} \). Only the ratio is needed, so the constants cancel.
\( \displaystyle \frac{F'}{F} = \frac{(2q)(2q)}{q\cdot q} \times \frac{r^{2}}{(r/2)^{2}} = 4 \times 4 = 16 \)
So the new force is \( 16F \).