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The force between two charges at a certain distance between those in air is unchanged when those charges are each increased by 60% and the distance between them in a particular medium is 20% less than that in air. Then dielectric constant of that medium is

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Detailed Solution

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

In a medium

$F_{m} = \frac{1}{4 π ε_{0} k} \frac{q_{1} q_{2}}{r_{m}^{2}}$

$F_{m} = \frac{1}{4 π ε_{0} k} \frac{1.6 q_{1} 1.6 q_{2}}{{(0.8 r)}^{2}} = F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}}$

solving k=4

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