Question:
As shown in the figure, two tiny conducting balls of identical mass and identical charge are suspended by two non-conducting threads of equal length \(L\). The separation between the balls in equilibrium is \(x\) and is related to \(L\) as \(x \propto L^{\beta}\). If \(\theta\) is assumed to be very small then the value of \(\beta\) is:
As shown in the figure, two tiny conducting balls of identical mass and identical charge are suspended by two non-conducting threads of equal length \(L\). The separation between the balls in equilibrium is \(x\) and is related to \(L\) as \(x \propto L^{\beta}\). If \(\theta\) is assumed to be very small then the value of \(\beta\) is:
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In small-angle equilibrium problems, always combine geometry relation \(x \approx L\theta\) with force law to eliminate \(\theta\).
Updated On: Jul 18, 2026
- 3
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding equilibrium configuration of charged balls.
Two identical charged spheres suspended from a common point experience electrostatic repulsion due to Coulomb force. In equilibrium, this repulsive force is balanced by horizontal component of tension in the string, while vertical component balances weight. The geometry forms a small angle \(\theta\) with the vertical.
Step 2: Writing force balance equations.
For each ball: Vertical balance: \[ T\cos\theta = mg \] Horizontal balance: \[ T\sin\theta = F_e \] Dividing both equations: \[ \tan\theta = \frac{F_e}{mg} \] For small \(\theta\), we use: \[ \tan\theta \approx \theta \]
Step 3: Express electrostatic force in terms of separation.
The separation between balls is: \[ x \approx 2L\sin\theta \approx 2L\theta \] So, \[ \theta \propto \frac{x}{L} \] Coulomb force: \[ F_e \propto \frac{1}{x^2} \]
Step 4: Combine relations to eliminate \(\theta\).
From equilibrium: \[ \theta \propto F_e \propto \frac{1}{x^2} \] But also: \[ \theta \propto \frac{x}{L} \] Equating: \[ \frac{x}{L} \propto \frac{1}{x^2} \]
Step 5: Solve proportionality for \(x\).
\[ x^3 \propto L \] \[ x \propto L^{1/3} \]
Step 6: Final conclusion.
Thus, \[ \beta = \frac{1}{3} \] \[ \boxed{\frac{1}{3}} \]
Two identical charged spheres suspended from a common point experience electrostatic repulsion due to Coulomb force. In equilibrium, this repulsive force is balanced by horizontal component of tension in the string, while vertical component balances weight. The geometry forms a small angle \(\theta\) with the vertical.
Step 2: Writing force balance equations.
For each ball: Vertical balance: \[ T\cos\theta = mg \] Horizontal balance: \[ T\sin\theta = F_e \] Dividing both equations: \[ \tan\theta = \frac{F_e}{mg} \] For small \(\theta\), we use: \[ \tan\theta \approx \theta \]
Step 3: Express electrostatic force in terms of separation.
The separation between balls is: \[ x \approx 2L\sin\theta \approx 2L\theta \] So, \[ \theta \propto \frac{x}{L} \] Coulomb force: \[ F_e \propto \frac{1}{x^2} \]
Step 4: Combine relations to eliminate \(\theta\).
From equilibrium: \[ \theta \propto F_e \propto \frac{1}{x^2} \] But also: \[ \theta \propto \frac{x}{L} \] Equating: \[ \frac{x}{L} \propto \frac{1}{x^2} \]
Step 5: Solve proportionality for \(x\).
\[ x^3 \propto L \] \[ x \propto L^{1/3} \]
Step 6: Final conclusion.
Thus, \[ \beta = \frac{1}{3} \] \[ \boxed{\frac{1}{3}} \]
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