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

The correct answer is Option (3) → 15 μC and 5 μC

Given:

Distance between charges, $r = 3\ \text{m}$

Sum of charges, $q_1 + q_2 = 20\ \mu\text{C} = 20 \times 10^{-6}\ \text{C}$

Force between charges, $F = 0.075\ \text{N}$

Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$

Permittivity of free space: $k = 9 \times 10^9\ \text{Nm²/C²}$

Substitute values:

$0.075 = 9 \times 10^9 \frac{q_1 q_2}{3^2} \Rightarrow q_1 q_2 = \frac{0.075 \cdot 9}{9 \times 10^9} = 0.075 \times 10^{-9} = 7.5 \times 10^{-11}\ \text{C²}$

System of equations:

$q_1 + q_2 = 20 \times 10^{-6}$

$q_1 q_2 = 7.5 \times 10^{-11}$

Let $q_1 = x$, $q_2 = 20 \times 10^{-6} - x$

Then $x (20 \times 10^{-6} - x) = 7.5 \times 10^{-11}$

$20 \times 10^{-6} x - x^2 = 7.5 \times 10^{-11}$

$x^2 - 20 \times 10^{-6} x + 7.5 \times 10^{-11} = 0$

Solving quadratic:

$x = \frac{20 \times 10^{-6} \pm \sqrt{(20 \times 10^{-6})^2 - 4 \cdot 7.5 \times 10^{-11}}}{2}$

So, $q_1 = \frac{20 \times 10^{-6} \pm 10 \times 10^{-6}}{2}$

⇒ $q_1 = \frac{30 \times 10^{-6}}{2} = 15 \times 10^{-6}\ \text{C}$

or $q_1 = \frac{10 \times 10^{-6}}{2} = 5 \times 10^{-6}\ \text{C}$

Thus, the charges are $q_1 = 15\ \mu\text{C}$, $q_2 = 5\ \mu\text{C}$