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Step-by-Step Solution
Step 1: Understand the Physical Setup
Two identical spheres, each carrying the same charge, are suspended from a common point by two strings of length $l$. Because of their mutual electrostatic repulsion, they stay apart at some small separation compared to $l$. Over time, both spheres leak charge at the same constant rate, causing their charges to decrease identically. As a result, they move closer together, and we want to find how their relative speed $v$ depends on the distance $x$ between them at any given moment.
Step 2: Approximate the Tension and Angles for Small Separations
Since $d \ll l$, the angle each string makes with the vertical is small. Hence, the tension $T$ in each string is nearly equal to the weight $mg$ of each sphere, and its horizontal component balances the electrostatic repulsion between the spheres.
Step 3: Relate Charge and Geometry
Let the instantaneous charge on each sphere be $q(t)$. For small angles, the horizontal component of tension $T \theta$ (with $\theta \approx \frac{x}{2l}$) equals the electrostatic force $F = \frac{k\,q^2}{x^2}$, where $k = \frac{1}{4 \pi \varepsilon_0}$. Thus:
$$
T \theta \;\approx\; mg \cdot \frac{x}{2l} \;=\; \frac{k\,q^2}{x^2}.
$$
Since $T \approx mg$, we get:
$$
mg \cdot \frac{x}{2l} \;=\; \frac{k\,q^2}{x^2}.
$$
Rearranging, this implies
$$
q^2 \;\propto\; x^3.
$$
Therefore,
$$
q \;\propto\; x^{\frac{3}{2}}.
$$
Step 4: Use the Constant Rate of Leakage
Since charge is leaking out of each sphere at a constant rate, we can write
$$
\frac{dq}{dt} = -\alpha \quad (\text{a constant}).
$$
However, from $q \propto x^{\frac{3}{2}}$, we have
$$
q = C\,x^{\frac{3}{2}} \quad \text{(for some constant $C$)},
$$
Differentiating with respect to time $t$,
$$
\frac{dq}{dt} = C \cdot \frac{3}{2} x^{\frac{1}{2}} \frac{dx}{dt}.
$$
But $\frac{dx}{dt} = v$, the speed at which the spheres approach each other. Hence,
$$
\frac{dq}{dt} = C \cdot \frac{3}{2} x^{\frac{1}{2}} \, v.
$$
Since $\frac{dq}{dt}$ is a constant (say $-\alpha$), we see that
$$
v \;\propto\; \frac{1}{x^{\frac{1}{2}}}.
$$
Step 5: Conclude the Proportionality
Therefore, the velocity $v$ at which the spheres approach each other as a function of the instantaneous separation $x$ obeys:
$$
v \propto x^{-\frac{1}{2}}.
$$
This matches Option 1: $v \propto x^{-\frac{1}{2}}.$
Referenced Figure
