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(D) The force between two charges in air is given by $F_a = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} = 10^{-4} \ N$.When a medium (oil) with dielec

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$4$

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(D) The force between two charges in air is given by $F_a = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} = 10^{-4} \ N$.When a medium (oil) with dielectric constant $K$ is introduced, the force becomes $F_m = \frac{1}{4\pi\epsilon_0 K} \frac{q_1q_2}{r^2} = 2.5 	imes 10^{-5} \ N$.The ratio of the force in air to the force in the medium is given by $K = \frac{F_a}{F_m}$.Substituting the values: $K = \frac{10^{-4}}{2.5 	imes 10^{-5}} = \frac{10}{2.5} = 4$.Therefore, the dielectric constant of the oil is $4$.

Two charges placed in air repel each other with a force of $10^{-4} \ N$. When oil is introduced between the charges, the force becomes $2.5 \times 10^{-5} \ N$. The dielectric constant of the oil is:

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