Question

Two identical charged spheres are suspended from a common point by massless strings of length l, initially at a distance d(d≪ l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with velocity v. Then v varies as a function of the distance x between the spheres as:

• A. \(v\propto x^{1/2}\)
• B. \(v\propto x\)
• C. \(v\propto x^{-1/2}\)
• D. v∝ x⁻1

Hint:

When charge varies uniformly with time, combine F∝ (q²)/(x²) with small-angle approximations to find velocity–distance relations.

Answer 1

The Correct Option is C

Step 1: Electrostatic repulsion: F=(kq²)/(x²)
Step 2: As charge leaks at a constant rate, q∝ t.
Step 3: For small angles, restoring force is proportional to displacement: F ∝ x
Step 4: Equating and differentiating with respect to time gives: v=(dx)/(dt)∝ x⁻1/2