Tennis Balls Equilibrium Separation
Two identical tennis balls each having mass \( m \) and charge \( q \) are suspended from a fixed point by threads of length \( l \). What is the equilibrium separation between the balls when each thread makes a small angle with the vertical?
Select the correct option:
Detailed Explanation
Each ball is in equilibrium under three forces: the gravitational force \( mg \) downward, the tension \( T \) in the thread, and the horizontal repulsive electrical force given by Coulomb's law: \[ F_e = \frac{1}{4\pi\epsilon_{0}}\frac{q^2}{x^2}. \]
When the balls are displaced by a small angle \( \theta \) from the vertical, the horizontal separation is approximately: \[ x \approx 2l\sin\theta \quad \text{or} \quad \theta \approx \frac{x}{2l}. \]
The components of the forces in the horizontal direction yield: \[ T\sin\theta = F_e. \] Meanwhile, in the vertical direction, the tension balances the gravitational force: \[ T\cos\theta \approx mg. \] For small angles, \( \cos\theta \approx 1 \), hence \( T \approx mg \).
Substituting and using the small-angle approximation \( \sin\theta \approx \theta \), we have: \[ mg \cdot \theta = \frac{1}{4\pi\epsilon_{0}}\frac{q^2}{x^2}. \] Replace \( \theta \) with \( \frac{x}{2l} \): \[ mg \cdot \frac{x}{2l} = \frac{1}{4\pi\epsilon_{0}}\frac{q^2}{x^2}. \]
Rearranging the equation: \[ mg \cdot \frac{x^3}{2l} = \frac{q^2}{4\pi\epsilon_{0}}. \] Solving for \( x^3 \): \[ x^3 = \frac{q^2 \cdot 2l}{4\pi\epsilon_{0} mg} = \frac{q^2 l}{2\pi\epsilon_{0} mg}. \] Hence, the equilibrium separation is: \[ x = \left( \frac{q^2 l}{2\pi\epsilon_{0} mg} \right)^{\frac{1}{3}}. \]
This matches Option B.
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