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Electrical force between two point charges is 200N. If we increase 10% charge on one of the charges and decrease 10% charge on the other, then electrical force between them for the same distance becomes

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198 N

100 N

200 N

99 N

answer is A.

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Detailed Solution

$F = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}}; q_{1}^{1} = \frac{110}{100} q_{1} and q_{2}^{1} = \frac{90}{100} q_{2}$

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