Two point charges of 1 C each are separated from each other by a distance of 1 m in vacuum. The force of interaction between them is

$9 × 10^9 N$

The correct answer is Option (2) → $9 × 10^9 N$

Given:

Charges: $q_1 = q_2 = 1\ \text{C}$

Distance: $r = 1\ \text{m}$

Vacuum permittivity: $\epsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}$

Coulomb's law: $F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$

Substitute values:

$F = \frac{1}{4 \pi \cdot 8.854 \times 10^{-12}} \cdot \frac{1 \cdot 1}{1^2} = \frac{1}{1.112 \times 10^{-10}} \approx 8.99 \times 10^9\ \text{N}$

∴ Force of interaction = 8.99 × 10⁹ N