Two identical point charges placed at a distance 'r' apart exert a force F on each other. If each charge is halved and the distance between them is doubled, then the new force acting on each charge would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → F/16
Initial force:
$F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$
New charges: $\frac{q}{2}$ and $\frac{q}{2}$
New distance: $2r$
New force:
$F' = \frac{1}{4\pi\epsilon_0} \frac{(\frac{q}{2})^2}{(2r)^2}$
$F' = \frac{1}{4\pi\epsilon_0} \frac{q^2/4}{4r^2}$
$F' = \frac{1}{16}\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$
$F' = \frac{F}{16}$
New force = F/16