Two point charges $q_1$ and $q_2$ separated by distance $r$ in vacuum attract each other with force $F$. If both charges are doubled AND the distance is halved, the new force is

1. Coulomb's law: $F = k\,\dfrac{q_1\,q_2}{r^2}$. 2. Replace $q_1 \to 2q_1$ and $q_2 \to 2q_2$ — numerator scales by $2\cdot 2 = 4$. 3. Replace $r \to r/2$ — denominator $r^2$ scales by $(1/2)^2 = 1/4$, so $1/r^2$ scales by $4$. 4. Combined factor: $4 \times 4 = 16$. New force $= 16F$. 5. Stepwise: $F' = k\,\dfrac{(2q_1)(2q_2)}{(r/2)^2} = k\,\dfrac{4 q_1 q_2}{r^2/4} = 16 k\,\dfrac{q_1 q_2}{r^2} = 16F$. 6. Option A would require no net change. Option B counts only the charge doubling. Option C counts only the distance halving. _Source: NCERT Class 12 Physics Part 1, Ch 1, §1.5 (Coulomb's Law), p. 5–7._