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A charge of 1 $μ C$ is divided into two parts such that their charges are in the ratio of 2 :3. These two charges are kept at a distance 1 m apart in vacuum. Then the electric force between them ( in Newton ) is

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0.216

0.0216

0.00216

2.16

answer is B.

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Detailed Solution

$q_{1} = \frac{2}{5} \times Q = \frac{2}{5} \times 10^{- 6}$

$q_{2} = \frac{3}{5} Q = \frac{3}{5} \times 10^{- 6}$

$F = \frac{9 \times 10^{9} \times \frac{2}{5} \times 10^{- 6} \times \frac{3}{5} \times 10^{- 6}}{1^{2}}$

$F = 2.16 \times 10^{- 3}$

$F = 0.00216 N$

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