Question:
Two identical charged spheres suspended from a common point by two massless strings of length l are initially at a distance 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 a velocity v. Then v varies as a function of distance x between the spheres as:
Two identical charged spheres suspended from a common point by two massless strings of length l are initially at a distance 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 a velocity v. Then v varies as a function of distance x between the spheres as:
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Use small-angle approximation and equilibrium conditions for charged pendulums.
Updated On: Mar 23, 2026
- \(v \propto x^{1/2}\)
- \(v \propto x\)
- \(v \propto x^{-1/2}\)
- v ∝ x⁻1
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The Correct Option is C
Solution and Explanation
Step 1: For small angles, equilibrium gives:
\( \dfrac{k q^2}{x^2} \propto x \)
\( \Rightarrow q^2 \propto x^3 \)
Step 2: Since charge leaks at a constant rate:
\( q \propto t \)
\( \Rightarrow x^{3/2} \propto t \)
Step 3: Velocity:
\( v = \dfrac{dx}{dt} \propto x^{-1/2} \)
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