Two identical tennis balls each having mass $$m$$ and charge $$q$$ are suspended from a fixed point by threads of length $$l$$. What is the equilibrium separation when each thread makes a small angle $$\theta$$ with the vertical?

Let us consider one of the two identical tennis balls. The ball has mass $$m$$, charge $$q$$ and is attached to a thread of length $$l$$ which makes a small angle $$\theta$$ with the vertical. Because both balls carry the same charge they repel each other and settle at a horizontal separation $$x$$. Geometrically the two equal threads form an isosceles triangle, so the horizontal distance between the balls is

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

We now analyse the forces acting on a single ball. Three forces are present:

(i) its weight $$\;mg$$ acting vertically downward,

(ii) the tension $$T$$ in the thread acting along the thread, and

(iii) the electrostatic repulsive force $$F$$ exerted horizontally by the other ball.

According to Coulomb’s law, the magnitude of the electrostatic force between two point charges is

$$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2},$$

where $$r$$ is the distance between the charges. In the present situation

$$r = 2l\sin\theta,$$

so

$$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2l\sin\theta)^2} = \dfrac{q^2}{16\pi\varepsilon_0 l^2\sin^2\theta}.$$

For equilibrium we resolve the tension into components. The vertical component balances the weight and the horizontal component balances the electrostatic repulsion:

$$T\cos\theta = mg,$$

$$T\sin\theta = F.$$

Dividing the second equation by the first gives

$$\tan\theta = \dfrac{F}{mg}.$$

Because the angle is small we may use the small-angle approximations

$$\sin\theta \approx \theta,\qquad \tan\theta \approx \theta,\qquad \cos\theta \approx 1.$$

Thus the previous relation becomes

$$\theta = \dfrac{F}{mg} = \dfrac{1}{mg}\left(\dfrac{q^2}{16\pi\varepsilon_0 l^2\theta^2}\right).$$

Rearranging, we collect all powers of $$\theta$$ on one side:

$$\theta^3 = \dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}.$$

Taking the cube root,

$$\theta = \left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right)^{1/3}.$$

The required separation $$x$$ is related to $$\theta$$ by $$x = 2l\sin\theta \approx 2l\theta,$$ so

$$x = 2l \left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right)^{1/3}.$$

To simplify, we cube both sides temporarily:

$$x^3 = 8l^3\left(\dfrac{q^2}{16\pi\varepsilon_0 l^2 mg}\right) = \dfrac{8}{16}\,\dfrac{q^2 l}{\pi\varepsilon_0 mg} = \dfrac{q^2 l}{2\pi\varepsilon_0 mg}.$$

Finally taking the cube root once more we obtain

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

This expression matches Option 2 (Option B).

Hence, the correct answer is Option 2.