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Two point charges repel each other with a force of 100 N. One of the charges is increased by 10% and the other is reduced by 10%.The new force of repulsion at the same distance would be

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100 N

121 N

99 N

89 N

answer is C.

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Detailed Solution

100 = $\frac{K q_{1} q_{2}}{r^{2}}$ and F=K $\frac{(1.1 q_{1}) (0.9 q_{2})}{r^{2}}$

$\frac{F}{100} = 0.99$ $∴ F = 99 N$

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