1. Forces on each ball

Each ball is in equilibrium under three forces:

• Weight: $mg$ downward
• Tension: $T$ along the thread
• Electrostatic repulsion: $F_e = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{x^2}$ acting horizontally

Here, $x$ is the separation between the two balls.

2. Resolve forces

If each thread makes a small angle $\theta$ with the vertical, then for one ball:

• Vertical equilibrium: $T\cos\theta = mg$

• Horizontal equilibrium: $T\sin\theta = F_e = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{x^2}$

Dividing, $\tan\theta = \frac{F_e}{mg} = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{mgx^2}$

3. Use small angle geometry

Since the balls are suspended from the same point, the separation is $x = 2l\sin\theta$

For small $\theta$, $\sin\theta \approx \tan\theta \approx \theta$

So, $x \approx 2l\tan\theta$

Hence, $\tan\theta = \frac{x}{2l}$

4. Substitute into equilibrium condition

Using $\tan\theta = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{mgx^2}$ and also $\tan\theta = \frac{x}{2l}$

Equate them: $\frac{x}{2l} = \frac{1}{4\pi \varepsilon_0}\frac{q^2}{mgx^2}$

Multiply both sides by $2lmgx^2$: $mgx^3 = \frac{2lq^2}{4\pi \varepsilon_0}$

So, $x^3 = \frac{q^2 l}{2\pi \varepsilon_0 mg}$

Therefore, $x = \left(\frac{q^2 l}{2\pi \varepsilon_0 mg}\right)^{1/3}$

5. Match with options

This matches Option B.

$\boxed{x = \left(\frac{q^2 l}{2\pi \varepsilon_0 mg}\right)^{1/3}}$