Charges On a Ring

We want to minimise the electrostatic potential energy of the system. Thus we want to minimise $V \; = \; \frac{q^2}{8\pi \varepsilon_0 r} \left\{ \frac{1}{\sin\theta} + \frac{6}{\sin\frac12\theta}\right\}$ where $r$ is the radius of the ring, and $\frac{dV}{d\theta} \; = \; -\frac{q^2}{8\pi \varepsilon_0 r}\times \frac{\cos\theta + 12\cos^3\frac12\theta}{\sin^2\theta}$ so we need to solve the equation $12\cos^3\tfrac12\theta + \cos\theta \,=\, 0$ . The numeric solution to this equation is $\theta = \boxed{134.3757917^\circ}$ .

A conventional solution by balancing forces: Denote the top +q charge to be vertex A, the bottom +q charge to be vertex B and the +3q charge to be vertex C. Let the centre be O. By symmetry, we have $\angle AOC = \angle BOC = \theta$ Now consider vertex A. The charge there will experience repulsive forces in the direction $BA$ and $CA$ . By simple geometry, $\angle BAO = \theta - \pi/2$ , $\angle CAO = \pi/2 - \theta/2$ . Now, balancing forces in the tangential direction, we obtain $\frac{(\sin(\theta - π/2))}{(1 - \cos(2 π - 2 \theta))} = \frac{(3 \sin(π/2 - \theta/2))}{(1 - \cos(\theta))}$ Solving this, we get the answer.