Two equal point charges ' \(q\) ' each exert a force ' \(F\) ' on each other, when they are placed distance ' \(x\) ' apart in air. When the same charges are placed distance ' \(y\) ' apart in a medium of dielectric constant ' k ', they exert the same force. The ratio of distance ' \(y\) ' to ' \(x\) ' is equal to

Force between the two charges in air,
\(\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} \cdot \mathrm{q}}{\mathrm{x}^2}\)
Force between the two charges when in a medium of dielectric constant k,
\(\begin{aligned}
& \mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0 \mathrm{k}} \cdot \frac{\mathrm{q} \cdot \mathrm{q}}{\mathrm{y}^2} \\
& \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{x}^2} \times \frac{4 \pi \varepsilon_0 \mathrm{k} \cdot \mathrm{y}^2}{\mathrm{q}^2}
\end{aligned}\)
\(\begin{array}{ll}\quad 1 & =k \frac{y^2}{x^2} \quad \ldots\left(\because F_1=F_2\right) \\ \therefore \quad & \frac{y}{x}=\frac{1}{\sqrt{k}}\end{array}\)