Electrostatic force between two charges placed in vacuum is F. If the same charges are placed at the same separation in a medium of dielectric constant K. force between them will be

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K^{2} F

\frac{F}{K}

answer is C.

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Detailed Solution

In vacuum electro static force

F_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}

In medium $F_{2} = \frac{1}{4 π ε} \frac{q_{1} q_{2}}{r^{2}}$

$Since ε = k ε_{0}$

$= \frac{1}{4 π K ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

$F_{2} = \frac{F_{1}}{K}$

$F_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}$

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