The force between two small charged spheres having charges of $2 × 10^{-7} C$ and $3 × 10^{-7} C$, respectively placed 30 cm apart in air would be

$6 × 10^{-3} N$

The correct answer is Option (3) → $6 × 10^{-3} N$

Given:

$q_1 = 2 \times 10^{-7}\ \text{C}$, $q_2 = 3 \times 10^{-7}\ \text{C}$, $r = 30\ \text{cm} = 0.3\ \text{m}$

Using Coulomb’s law:

$F = \frac{1}{4πϵ_0} \frac{q_1 q_2}{r^2}$

Substitute values:

$F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.3)^2}$

$F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{0.09}$

$F = 9 \times 10^9 \times 6.67 \times 10^{-13}$

$F = 6.0 \times 10^{-3}\ \text{N}$

Answer: $F = 6.0 \times 10^{-3}\ \text{N}$