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

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0.216

0.00216

0.0216

2.16

answer is B.

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Detailed Solution

$\begin{array}{l} F = 9 \times 10^{9} \frac{q_{1} q_{2}}{r^{2}} = 9 \times 10^{9} \frac{(\frac{2}{5} \times \frac{3}{5}) 10^{- 12}}{(1)^{2}} \\ = 0 .00216 N \end{array}$

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