When the distance between two charged particles is halved, then the force between them becomes :

One-fourth

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Half

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Double

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Four times

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Solution

The correct option is D Four times

Earlier force between two particles is

$F = \frac{k q_{1} q_{2}}{r^{2}}$

When distance is halved force becomes

$F^{'} = \frac{k q_{1} q_{2}}{(r / 2)^{2}} = 4 \frac{k q_{1} q_{2}}{r^{2}} = 4 F$

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