Parabola Charge Stasis

Since all the electrostatic forces on the outermost two particles will act to push these particles away from the centre, the outermost particles must be at the ends of the wires, and will therefore have coordinates $(-1,1)$ and $(1,1)$ . By symmetry, the inner two particles will have coordinates $(x,x^2)$ and $(-x,x^2)$ for some $0 < x < 1$ .

We must minimize the electrostatic potential energy of the system, and hence must minimize $V(x) \; = \; \frac{2}{(1-x)\sqrt{x^2 + 2x + 2}} + \frac{2}{(1+x)\sqrt{x^2 - 2x + 2}} + \frac{1}{2x} + \frac12$ (over $0 < x < 1$ ), where $V(x)$ is the sum of the reciprocals of the distances between the six pairs of particles. Handling this problem numerically, the minimum of $V$ occurs when $x = \boxed{0.450162}$ .