The force between two small charged spheres having charges of 0.2 μC and 0.6 μC placed 0.6 m apart in the air will be

0.003 N

The correct answer is Option (4) → 0.003 N

Given:

$q_1 = 0.2\ \mu C = 0.2 \times 10^{-6}\ C$

$q_2 = 0.6\ \mu C = 0.6 \times 10^{-6}\ C$

$r = 0.6\ m$

Coulomb's law:

$F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$

$F = 9 \times 10^9 \cdot \frac{(0.2 \times 10^{-6}) (0.6 \times 10^{-6})}{(0.6)^2}$

$F = 9 \times 10^9 \cdot \frac{0.12 \times 10^{-12}}{0.36}$

$F = 9 \times 10^9 \cdot 0.333 \times 10^{-12}$

$F \approx 3 \times 10^{-3}\ N$