Two charged particles exert a force of magnitude F on one another. If the distance between them is doubled and the charge of one of the particles is doubled, what is the new force acting between them?

1/4 F

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1/2 F

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Solution

The correct option is B 1/2 F
By Coulomb's law, $F = \frac{k q_{1} q_{2}}{r^{2}}$
If the distance between them is doubled and the charge of one of the particles is doubled, the force becomes, $F^{'} = \frac{k q_{1} (2 q_{2})}{(2 r)^{2}} = F / 2$

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Two charged particles exert a force of magnitude F on one another. If the distance between them is doubled and the charge of one of the particles is doubled, what is the new force acting between them?

The correct option is B 1/2 FBy Coulomb's law, F=kq1q2r2If the distance between them is doubled and the charge of one of the particles is doubled, the force becomes, F′=kq1(2q2)(2r)2=F/2

The correct option is B 1/2 F
By Coulomb's law, $F = \frac{k q_{1} q_{2}}{r^{2}}$
If the distance between them is doubled and the charge of one of the particles is doubled, the force becomes, $F^{'} = \frac{k q_{1} (2 q_{2})}{(2 r)^{2}} = F / 2$