1. Given data

• Mass of each particle: $m = 10\,\text{g} = 0.01\,\text{kg}$
• Charge on each particle: $q = 2.0 \times 10^{-7}\,\text{C}$
• Coefficient of friction: $\mu = 0.25$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

Since the particles are on a horizontal table and are in limiting equilibrium, the electrostatic repulsion is balanced by the maximum static friction.

2. Maximum friction on each particle

Normal reaction: $N = mg = 0.01 \times 10 = 0.1\,\text{N}$

Maximum static friction: $f_{\max} = \mu N = 0.25 \times 0.1 = 0.025\,\text{N}$

3. Electrostatic force between the charges

By Coulomb's law, $F = \frac{kq^2}{L^2}$ where $k = 9 \times 10^9\,\text{N m}^2\text{/C}^2$.

At limiting equilibrium, $\frac{kq^2}{L^2} = f_{\max}$

So, $\frac{9 \times 10^9 \times (2 \times 10^{-7})^2}{L^2} = 0.025$

4. Simplify

First, $(2 \times 10^{-7})^2 = 4 \times 10^{-14}$

Thus, $9 \times 10^9 \times 4 \times 10^{-14} = 36 \times 10^{-5} = 3.6 \times 10^{-4}$

Hence, $\frac{3.6 \times 10^{-4}}{L^2} = 0.025$

Therefore, $L^2 = \frac{3.6 \times 10^{-4}}{0.025}$

$L^2 = 1.44 \times 10^{-2}$

$L = \sqrt{1.44 \times 10^{-2}} = 0.12\,\text{m}$

$L = 12\,\text{cm}$

5. Option check

• A: $12\,\text{cm}$ ✅
• B: $10\,\text{cm}$ ❌
• C: $8\,\text{cm}$ ❌
• D: $5\,\text{cm}$ ❌

So the correct option is A.