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(D) The force between two point charges $q_1$ and $q_2$ in air at a distance $R$ is given by Coulomb's Law: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_

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$R_1 = \frac{R}{\sqrt{K}}$

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(D) The force between two point charges $q_1$ and $q_2$ in air at a distance $R$ is given by Coulomb's Law: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{R^2}$.When the space between the charges is filled with a medium of dielectric constant $K$, the force $F'$ at a new distance $R_1$ is given by: $F' = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{R_1^2}$.Given that the force remains the same $(F' = F)$, we equate the two expressions:$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{R^2} = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{R_1^2}$.Canceling the common terms $\frac{q_1 q_2}{4\pi\epsilon_0}$ from both sides, we get: $\frac{1}{R^2} = \frac{1}{K R_1^2}$.Rearranging for $R_1^2$, we get: $R_1^2 = \frac{R^2}{K}$.Taking the square root on both sides: $R_1 = \frac{R}{\sqrt{K}}$.

Two point charges placed in air separated by distance $R$ exert a force $F$ on each other. Now the space between those charges is filled with a medium of dielectric constant $K$. At distance $R_1$ between the charges, the same force is exerted as $F$. Then:

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