Two point charges exert a force $F$ on each other. If each charge is halved and the distance is also halved, the new force is
A$F/4$
B$F/2$
C$F$
D$4F$
Answer & Solution
Correct answer: C. $F$
Start with $F = k\dfrac{q_1 q_2}{r^2}$. After halving each charge and the distance, $$F' = k\frac{(q_1/2)(q_2/2)}{(r/2)^2} = k\frac{q_1 q_2/4}{r^2/4} = k\frac{q_1 q_2}{r^2} = F.$$ The reduction in the product of charges is exactly compensated by the reduction in $r^2$. Options A and B ignore the inverse-square change in distance, while D overestimates the effect of reducing $r$.
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