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 \(\theta\).
Step 1: Balance torques about the suspension point instead of resolving forces along the thread. Each ball hangs from the same fixed point by a thread of length $L$, under its weight $mg$, the electrostatic repulsion $F_e$, and the thread tension. Taking torques about the suspension point removes tension from the picture entirely, since tension acts along the thread and passes straight through that pivot.
Step 2: Write the torque balance. The weight acts at a perpendicular distance $L\sin\theta$ from the vertical through the pivot, and $F_e$ acts at a perpendicular distance $L\cos\theta$. Balancing these turning effects, \[ mg(L\sin\theta) = F_e(L\cos\theta) \] \[ \tan\theta = \frac{F_e}{mg} \] This is the same relation force-resolution gives, reached without ever writing $T\cos\theta$ or $T\sin\theta$.
Step 3: Use the small-angle limit and the geometry of the triangle. Since $\theta$ is small, $\tan\theta \approx \theta$, and the horizontal separation is $x \approx 2L\theta$, so $\theta \approx \frac{x}{2L}$.
Step 4: Bring in Coulomb's law. With identical charge $q$ on each ball, $F_e = \frac{kq^2}{x^2}$, so $F_e \propto \frac{1}{x^2}$.
Step 5: Combine the two proportionalities. From Step 3, $\theta \propto \frac{x}{L}$; from Steps 2 and 4, $\theta \propto F_e \propto \frac{1}{x^2}$. Setting these equal, \[ \frac{x}{L} \propto \frac{1}{x^2} \quad\Rightarrow\quad x^3 \propto L \quad\Rightarrow\quad x \propto L^{1/3} \] Final Answer: Comparing with $x \propto L^{\beta}$, \[ \boxed{\beta = \frac{1}{3}} \]
