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Two particles having charges $Q_{1}$ and $Q_{2}$ , when kept at a certain distance exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is tripled, the force between the particles would be

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36F

16F

answer is C.

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Detailed Solution

From Coulomb's law we know

$F = \frac{1}{4 π ε_{0}} \frac{Q_{1} Q_{2}}{r^{2}}$

If charges are tripled and distance is reduced to half, the force becomes

$F^{'} = \frac{1}{4 π ε_{0}} \frac{(3 Q_{1}) (3 Q_{2})}{{(r / 2)}^{2}} = 36 \frac{1}{4 π ε_{0}} \frac{Q_{1} Q_{2}}{r^{2}} = 36 F$

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