1. Set up the equilibrium condition

Let the distance between the two spheres at any instant be $x$. Since the spheres are identical and suspended from a common point, each string makes a small angle $\theta$ with the vertical.

For small angles, $\tan\theta \approx \sin\theta \approx \frac{x/2}{l} = \frac{x}{2l}$

If the charge on each sphere at that instant is $q$, then the electrostatic repulsion is $F_e = \frac{kq^2}{x^2}$

In equilibrium, $T\sin\theta = F_e, \qquad T\cos\theta = mg$ So, $\tan\theta = \frac{F_e}{mg}$

Thus, $\frac{x}{2l} = \frac{1}{mg}\cdot \frac{kq^2}{x^2}$

Rearranging, $x^3 = \frac{2lk}{mg}q^2$

Hence, $x \propto q^{2/3}$

2. Relate charge leakage to motion

The charge leaks at a constant rate, so $\frac{dq}{dt} = -\lambda$ where $\lambda$ is constant.

From $x^3 = Cq^2$ where $C=\frac{2lk}{mg}$ is constant.

Differentiate both sides with respect to time: $3x^2\frac{dx}{dt} = 2Cq\frac{dq}{dt}$

Now from $x^3=Cq^2$, $q = \left(\frac{x^3}{C}\right)^{1/2} \propto x^{3/2}$

Substitute into the differentiated equation: $3x^2\frac{dx}{dt} \propto q\frac{dq}{dt}$ Since $\dfrac{dq}{dt}$ is constant, $x^2\frac{dx}{dt} \propto q \propto x^{3/2}$ Therefore, $\frac{dx}{dt} \propto x^{-1/2}$

Since the spheres approach each other, the speed of approach is $v = -\frac{dx}{dt}$ so in magnitude, $v \propto x^{-1/2}$

3. Check options

• A: $v \propto x^{-1}$ ❌
• B: $y \propto x^{1/2}$ ❌ (irrelevant variable and wrong relation)
• C: $v \propto x$ ❌
• D: $v \propto x^{-1/2}$ ✅

Thus the correct option is D.