Two point charges $Q_1$ and $Q_2$ exerts force $F$ on each other, when kept at certain distance apart. If the charge on each particle is halved and the distance between them is doubled, then the new force between the two particles would be

The correct option is D $\frac{F}{16}$
From Coulomb's law:

$F=\frac{k{Q}_1{Q}_2}{r^2}\dots \dots \left(1\right)$

Since
$Q_1^{{}^{\prime }}=\frac{Q_1}{2},Q_2^{{}^{\prime }}=\frac{Q_2}{2}$ and $r^{{}^{\prime }}=2r$

From Eq. (1), we get

$F^{{}^{\prime }}=\frac{k\left(\frac{Q_1}{2}\right)\left(\frac{Q_2}{2}\right)}{(2r{)}^2}$

$\Rightarrow F^{{}^{\prime }}=\frac{k{Q}_1{Q}_2}{16r^2}......\left(2\right)$

from Eq.(1) and Eq.(2), we get

$F^{{}^{\prime }}=\frac{F}{16}$

Hence, option $\left(d\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question ?}\\ \text{It intends to test your understanding of direct and}\\ \text{inverse proportionality of parameters in Coulomb's law.}\end{array}$