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A charge of 1μC 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 N) is:

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0.00216

0.0216

2.16

0.216

answer is B.

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Detailed Solution

$Q = q_{1} + q_{2} = 1 mC, q_{1} : q_{2} :: 2 : 3$

let

$q_{1} = 2 q, q_{2} = 3 q, Q = 2 q + 3 q = 5 q$

$q = \frac{1}{5} μC = 0 .2 μC$

$F = \frac{{kq}_{1} q_{2}}{r^{2}}$

$= \frac{9 \times 10^{9} \times (2 \times 0.2 \times 10^{- 6}) (3 \times 0.2 \times 10^{- 6})}{1^{2}}$

F = 0.00216 N

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