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Two positive charges separated by a distance of 2m repel each other with a force of 0.36 N, if the combined charge is $26 μC$ , the charges are

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$13 μ C, 13 μ C$

$16 μ C, 10 μ C$

$20 μ C, 6 μ C$

$18 μ C, 8 μ C$

answer is B.

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Detailed Solution

$F = 9 \times 10^{9} \frac{q_{1} q_{2}}{r^{2}}$

$\Rightarrow 0 .36 = 9 \times 10^{9} \frac{q_{1} q_{2}}{4} \Rightarrow q_{1} q_{2} = 160 (μC)^{2}$
${(q_{1} - q_{2})}^{2} = {(q_{1} + q_{2})}^{2} - 4 q_{1} q_{2} = 36 (μC)^{2}$

$\Rightarrow q_{1} - q_{2} = 6 μC$

$since, q_{1} + q_{2} = 26 μC$

$q_{1} = 16 μC and q_{2} = 10 μC$

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