Coulomb's choreography

We begin by noting that resulting force exerted on any of the charges is always directed toward the center center of square. Let $l$ be the side of the square at a specific moment in time. Then the total force acting on a charge is given by $\vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\vec{F}_{4},$ as shown in the figure below.

It is relatively simple to find its magnitude. Using some trigonometry we find that $F=k \frac{q ( \frac{(3-2\sqrt{2})}{2}q)}{l^{2}}=k \frac{q ( \frac{(3-2\sqrt{2})}{4}q)}{r^{2}}$ where $r=\frac{l}{\sqrt{2}}$ is the distance from the charge to the center of the square. This is a very important observation because it suggests that if we want to describe the motion of one charge, the rest of the charges can be replaced by an effective charge $Q_{eff}=- \frac{(3-2\sqrt{2})}{4}q$ located at the center of the square. In addition, since both Coulomb and Gravitational forces are of the form $F= \gamma/r^{2}$ the orbit of any or the charges must be elliptical.

In the figure below, we show the positions of the charges corresponding to $r_{max}=\frac{ l_{max}}{\sqrt{2}}$ and $r_{min}=\frac{l_{min}}{\sqrt{2}}.$

The semi-major axis of the ellipses is $a=\frac{r_{min}+r_{max}}{2}=\frac{5}{8\sqrt{2}}L_{0}.$ Kepler's law states that the orbital period depends only on the semi-major axis. Consider a planet describing an elliptical orbit around the sun, then Kepler's law tells you that the period is given by $T=2 \pi \sqrt{\frac{a^{3}}{GM_{s}}}.$ Thus, by exploiting the similarity between Coulomb and Gravitational forces we can write $T= 2 \pi \sqrt{\frac{a^{3}}{k q|Q_{eff}|/m}}= \frac{5\pi}{8}\sqrt{\frac{5}{\sqrt{2}(3-2\sqrt{2})kq^{2}/mL_{0}^{3}}}=0.089~\mbox{s}.$