This problem involves two identical charged spheres suspended by strings, repelling each other due to their charges. As charge leaks at a constant rate, the spheres approach each other. We need to find how their approach velocity v depends on the distance x between them.

Step 1: Understand the Forces
Each sphere experiences two forces:

• Gravitational force: $mg$ downward
• Electrostatic repulsion: $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{x^2}$ horizontally
• Tension in the string: along the string direction

Since d << l, the strings are almost vertical, and the small angle approximation applies. The electrostatic force balances the horizontal component of tension.

Step 2: Equilibrium Condition
For small angles, the horizontal force balance gives: $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{x^2}=T\sin \theta \approx T\theta$
The vertical force balance gives: $T\cos \theta \approx T=mg$
Also, from geometry: $\sin \theta \approx \theta \approx \frac{x}{2l}$ (since x is the distance between spheres, each is at x/2 from the vertical).
Combining these, we get: $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{x^2}=mg·\frac{x}{2l}$
Simplifying: $q^2\propto x^3$ or $q\propto x^{3/2}$

Step 3: Relate Charge Leakage to Velocity
Charge leaks at a constant rate: $\frac{dq}{dt}=-k$ (constant).
Differentiate $q\propto x^{3/2}$ with respect to time: $\frac{dq}{dt}\propto \frac{d}{dt}\left(x^{3/2}\right)=\frac{3}{2}{x}^{1/2}\frac{dx}{dt}$
Since the spheres are approaching, $\frac{dx}{dt}=-v$ (v is speed of approach).
So, $\frac{dq}{dt}\propto \frac{3}{2}{x}^{1/2}(-v)\propto -x^{1/2}v$
But $\frac{dq}{dt}$ is constant, so $x^{1/2}v=\mathrm{constant}$
Therefore, $v\propto \frac{1}{x^{1/2}}=x^{-1/2}$

Final Answer: v ∝ x^-1/2

Related Topics

• Electrostatics (Coulomb's Law): Deals with forces between charged particles. The repulsive force here is given by Coulomb's law.
• Equilibrium of Forces: The spheres are in equilibrium under tension, gravity, and electrostatic force.
• Related Rates: Using calculus to relate the rate of charge change to the rate of distance change.

Formulae

• Coulomb's Law: $F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$
• Small Angle Approximation: $\sin \theta \approx \theta ,\cos \theta \approx 1$
• Related Rates: $\frac{dq}{dt}=\frac{dq}{dx}·\frac{dx}{dt}$