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(C) The force between two point charges $Q_1$ and $Q_2$ in air at a distance $r$ is given by $F = \frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2}$.Whe

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$r/\sqrt{k}$

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(C) The force between two point charges $Q_1$ and $Q_2$ in air at a distance $r$ is given by $F = \frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2}$.When the same charges are placed in a medium of dielectric constant $k$ at a distance $r'$,the force $F'$ is given by $F' = \frac{1}{4\pi\varepsilon_0 k} \frac{Q_1 Q_2}{r'^2}$.Given that the force remains the same,$F = F'$,so $\frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2} = \frac{1}{4\pi\varepsilon_0 k} \frac{Q_1 Q_2}{r'^2}$.Simplifying the equation,we get $\frac{1}{r^2} = \frac{1}{k r'^2}$.Rearranging for $r'$,we get $r'^2 = \frac{r^2}{k}$,which implies $r' = \frac{r}{\sqrt{k}}$.

Two point charges placed at a certain distance $r$ in air exert a force $F$ on each other. Then the distance $r'$ at which these charges will exert the same force in a medium of dielectric constant $k$ is given by

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