Two balls of same mass and carrying equal charge are hung from a fixed support of length $$l$$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to :

Consider two identical balls, each with mass $$m$$ and charge $$q$$, suspended from a fixed support by strings of length $$l$$. At electrostatic equilibrium, the balls repel each other, causing the strings to make small angles with the vertical. We need to find how the separation $$x$$ between the balls depends on $$l$$.

For each ball, three forces act: the gravitational force $$mg$$ downward, the tension $$T$$ along the string, and the electrostatic repulsive force $$F_e$$ horizontally away from the other ball. The electrostatic force is given by Coulomb's law:

$$ F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

where $$x$$ is the separation between the balls and $$\epsilon_0$$ is the permittivity of free space.

Since the setup is symmetric and the angles are small, let $$\theta$$ be the small angle each string makes with the vertical. The separation $$x$$ can be expressed in terms of $$\theta$$. Each ball is displaced horizontally by $$l \sin\theta$$ from the vertical, so the total separation is:

$$ x = 2 l \sin\theta $$

Using the small-angle approximation, $$\sin\theta \approx \theta$$ (where $$\theta$$ is in radians), we get:

$$ x \approx 2 l \theta $$

Resolving the forces for one ball at equilibrium:

• Vertically: The vertical component of tension balances the weight.

$$ T \cos\theta = mg $$

Since $$\theta$$ is small, $$\cos\theta \approx 1$$, so:

$$ T \cdot 1 = mg \quad \Rightarrow \quad T = mg $$

• Horizontally: The horizontal component of tension balances the electrostatic force.

$$ T \sin\theta = F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Using $$\sin\theta \approx \theta$$ and substituting $$T = mg$$:

$$ mg \theta = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

From the separation equation $$x \approx 2 l \theta$$, solve for $$\theta$$:

$$ \theta \approx \frac{x}{2l} $$

Substitute this expression for $$\theta$$ into the horizontal force equation:

$$ mg \left( \frac{x}{2l} \right) = \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Multiply both sides by $$2l$$ to simplify:

$$ mg x = 2l \cdot \frac{1}{4\pi\epsilon_0} \frac{q^2}{x^2} $$

Simplify the right side:

$$ mg x = \frac{2l}{4\pi\epsilon_0} \frac{q^2}{x^2} = \frac{l}{2\pi\epsilon_0} \frac{q^2}{x^2} $$

Multiply both sides by $$x^2$$:

$$ mg x \cdot x^2 = \frac{l}{2\pi\epsilon_0} q^2 $$

$$ mg x^3 = \frac{l}{2\pi\epsilon_0} q^2 $$

Solve for $$x^3$$:

$$ x^3 = \frac{l}{2\pi\epsilon_0} \frac{q^2}{mg} $$

The terms $$\frac{q^2}{2\pi\epsilon_0 mg}$$ are constant since $$q$$, $$m$$, $$g$$, and $$\epsilon_0$$ are constants. Let $$k = \frac{q^2}{2\pi\epsilon_0 mg}$$, so:

$$ x^3 = k l $$

Taking the cube root of both sides:

$$ x = (k l)^{1/3} = k^{1/3} l^{1/3} $$

Thus, $$x$$ is proportional to $$l^{1/3}$$.

Comparing with the options:

A. $$l$$

B. $$l^2$$

C. $$l^{2/3}$$

D. $$l^{1/3}$$

Hence, the correct answer is Option D.