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Standard XII

Physics

Introducing Epsilon

Two charges q...

Question

Two charges $q_{1}$ and $q_{2}$ are placed in vacuum at a distance d and the force acting between them is F. If a medium of dielectric constant 4 is introduced around them, the force now will be

4 F

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2 F

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\frac{F}{2}

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\frac{F}{4}

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Solution

The correct option is D $\frac{F}{4}$
$g i v e n : F = \frac{K q_{1} q_{2}}{d^{2}}$
$F^{'} = \frac{F}{k} = \frac{K q_{1} q_{2}}{k d^{2}} ⟶ \frac{F}{k}$

In the presence of medium force becomes $\frac{1}{K}$ times

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