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 becomes four times, Find out the new force acting between them?

1 / 4 F

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1 / 2 F

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F

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2 F

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4 F

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Solution

The correct option is A $F$
Let the charge of the particles be $Q_{1}$ and $Q_{2}$ and distance between them be $r$ .
Force between them $F = \frac{K Q_{1} Q_{2}}{r^{2}}$

Now $Q_{1}^{'} = 4 Q_{1}$ and $r^{'} = 2 r$
New force between them $F^{'} = \frac{K Q_{1}^{'} Q_{2}}{(r^{'})^{2}} = \frac{K (4 Q_{1}) Q_{2}}{4 r^{2}} = F$

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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 becomes four times, Find out the new force acting between them?

The correct option is A FLet the charge of the particles be Q1 and Q2 and distance between them be r.Force between them F=KQ1Q2r2Now Q′1=4Q1 and r′=2rNew force between them F′=KQ′1Q2(r′)2=K(4Q1)Q24r2=F

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 becomes four times, Find out the new force acting between them?