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The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is

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F/4

4 F

16 F

answer is D.

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Detailed Solution

$F = \frac{q_{1} q_{2}}{4 π \in_{0} r^{2}}$ ----(1)

$F' = \frac{q_{1} q_{2}}{4 π \in {(r / 2)}^{2}} = \frac{4 q_{1} q_{2}}{4 π K \in_{0} r^{2}} = \frac{4 F}{4} = F$

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