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Two positive charges separated by a distance 2 m each other with a force of 0.36 N. If the combined charge is $26 μ C$ . The charges are

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$26 μ C, 6 μ C$

$16 μ C, 10 μ C$

$18 μ C, 8 μ C$

$13 μ C, 13 μ C$

answer is B.

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Detailed Solution

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

$0.36 = 9 \times 10^{9} \times \frac{q_{1} q_{2}}{4}$

$q_{1} q_{2} = 160 {(μ C)}^{2}$

${(q_{1} - q_{2})}^{2} = {(q_{1} + q_{2})}^{2} - 4 q_{1} q_{2}$ $= 36 {(μ C)}^{2}$

$q_{1} - q_{2} = 6 μ C$

$\frac{q_{1} + q_{2} = 26 μ C}{q_{1} = 16 μ C}$ and $q_{2} = 10 μ C$

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