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Two charges of equal magnitude at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled, then the new force acting on each charge is:

1. $F/8$
2. $F/4$
3. $4F$
4. $F/16$

Correct answer: $F/16$

Solution

By Coulomb’s law, $F\propto \dfrac{q_1q_2}{r^2}$. Halving both charges makes the product one-fourth, and doubling the distance makes the force one-fourth again. So the new force is $F/16$.

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