Two positive point charges are 3 m apart in the air and the sum of th…

The correct answer is Option (3) → \( 15\,\mu\mathrm{C} \) and \( 5\,\mu\mathrm{C} \)

By Coulomb's law the force between two point charges in air is

    \( \displaystyle F = \frac{k\,q_1 q_2}{r^{2}}, \qquad k = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \)

So the product of the charges is

    \( \displaystyle q_1 q_2 = \frac{F r^{2}}{k} = \frac{0.075\,\mathrm{N} \times (3\,\mathrm{m})^{2}}{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}} = 7.5\times10^{-11}\,\mathrm{C^{2}} \)

Since \( 1\,(\mu\mathrm{C})^{2} = 10^{-12}\,\mathrm{C^{2}} \), this is \( q_1 q_2 = 75\,(\mu\mathrm{C})^{2} \). With \( q_1 + q_2 = 20\,\mu\mathrm{C} \) (charges in \( \mu\mathrm{C} \)):

    \( q_1(20 - q_1) = 75 \)
    \( q_1^{2} - 20q_1 + 75 = 0 \)
    \( (q_1 - 15)(q_1 - 5) = 0 \)

So the charges are \( 15\,\mu\mathrm{C} \) and \( 5\,\mu\mathrm{C} \).

Shortcut: Every option adds up to \( 20\,\mu\mathrm{C} \), so only the product decides. The products are 84, 64, 75 and 96 \( (\mu\mathrm{C})^{2} \); only \( 15 \times 5 = 75 \) matches.