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Q.
The electric force between two point charges separated by a certain distance d in air is F.The distance at which they should be placed in a medium of relative permittivity k so that the force remain the same is
Electric force between two point charges $q_1$ and $q_2$ separated by a distance d in air is given by
$F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} \, \, \, \, \, \, $ ...(i)
Electric force between two point charges $q_1$ and $q_2$ separated by a distance $d'$ in a medium of relative permittivity $k$ is given by
$F' = \frac{1}{4\pi E_{0}k } . \frac{q_{1}q_{2}}{d'^{2}}$
According to question $, F = F' $
$\therefore \, \, \frac{1}{4\pi e_{0}} \frac{q_{1}q_{2}}{d^{2}} = \frac{ 1}{4\pi e_{0}k} \frac{q_{1}q_{2}}{d'^{2}} $
$ \frac{1}{d^{2} } = \frac{1}{kd'^{2}} $ or $kd'^{2} = d^{2} $
or $d' = \frac{d}{\sqrt{k}} . $