Equilibrium constellations of classical point charges in hollow conductors

As far as I know, in free space there can be constellations of $$N$$ point charges in electrostatic equilibrium (albeit unstable equilibrium) for all $$N$$ except $$N=2$$. Some discussions here already addressed this and also equilibrium situations inside a sphere[ref1, ref2, ref3]. It seems that in a hollow conductive sphere $$N$$ point charges can be in (unstable) equilibrium for all $$N$$, including $$N=2$$ (see below).

My question: is this also known for other shapes, e.g. inside a hollow conductive cube? And then how to compute the required positions in a cube? (How to do the computation for cases in empty space and in a sphere is given in my explanation below.)

In empty space we have, for instance:

1. Two or more point charges $$+q\$$ regular polygon corners, around a properly chosen negative balancing charge $$q_c$$ in the center.
2. Positive charges $$+q$$ arranged as the vertices of a platonic solid, again with a balancing negative $$q_c$$ in the center.
3. Less regular point constellations, like the socker ball or bucky ball vertices, etc.

In case 1), because of symmetry, the central charge will float freely, albeit in unstable equilibrium. To see how to keep the $$N$$ positive charges also freely floating, we put them on a centered polygon in the $$xy$$-plane, each at distance $$r$$ from the origin. Then we look at the vector $${\bf r_{12}}$$ from a first one, which is on the $$x$$-axis, to another one with position angle difference $$\varphi$$. We can ignore scale factors like $$1/(4\pi\epsilon_0)$$ since we only compute charge ratios, so the outward force $${\bf F_x}$$ is found as: $${\bf r_{12}}=(r,0,0)-(r\ \cos\varphi, r\ \sin\varphi,0), \ \ \Rightarrow\ \ {\bf E} \sim \frac{q\ {\bf r_{12}}}{r_{12}^3}$$ $$\Rightarrow\ F_x \sim \frac{q^2\ r_{12,x}}{r^3} = \frac{q^2\ (r-r\ \cos\varphi)}{[(r-r\ \cos\varphi)^2+(r\ \sin\varphi)^2]^{3/2}} = \frac{q^2}{4 r^2 \sin(\varphi/2)}$$

Summing this force on the charge from the $$N-1$$ other positive charges, and balancing against the force $$q_c q/r^2$$ from a center charge $$q_c$$, will require, as the reader may verify: $$q_c=-q\ \sum_{n=1}^{N-1} \frac1{4\sin(n\pi/N)}$$ $$N=2 \ \ \Rightarrow \ \ q_c=-q/4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ $$N=3 \ \ \Rightarrow \ \ q_c=-q/\sqrt3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ $$N=4 \ \ \Rightarrow \ \ q_c=-q\ (2\sqrt2 + 1)/4$$ $$N=5 \ \ \Rightarrow \ \ q_c=-q \ \sqrt{1+2/\sqrt5}$$ The balancing requires at least two positive charges, so together with the central charge this requires at least 3 charges.

When placed inside a hollow sphere, we can try the same arrangement, but there will be influence on the charge distribution on the sphere's surface, which means that the fields will change a little inside the sphere. This can be described by mirror images, see Wikipedia, images. If our charges $$q$$ are at radius $$r$$ from the origin and the sphere has radius $$R$$, We get $$N$$ image charges $$q_{\small\ \mbox{im}}=-(R/r) q$$ at radius $$R^2/r$$, as shown in the figure.

The case $$N=3$$:

Now we have to redo the calculation for the balancing $$q_c$$ value. In free space the central $$q_c$$ only had to cancel the force that each $$q$$ feels from the other $$N-1$$ positive particles, now we need to add the force from the image charges as well. If the sphere has radius $$R$$, then the images with charge $$q_{\small\ \mbox{im}}=-(R/r)q$$ are at radius $$r_{\small \mbox{im}}=R^2/r$$ and we get, in the same way as before, the force on one of the charges $$q$$, from an image charge at a position angle differing by $$\varphi$$. It is: $$F(\varphi)=q \ q_{\small\ \mbox{im}} \cdot \ \frac{r_{\small \mbox{im}} \cos\varphi -r} { [r_{\small \mbox{im}}^2 - 2\ r_{\small \mbox{im}} r \cos\varphi + r^2]^{3/2}} =\frac{q^2}{r^2} \ \frac{(R/r)^3 \cos\varphi -R/r} {[(R/r)^4 - 2 (R/r)^2 \cos\varphi +1]^{3/2}}$$ So we have to sum over angles for $$N$$ of those image charges and add that to the earlier expression for the force of the $$N-1$$ other real charges, giving: $$q_c=-q\ \sum_{n=0}^{N-1} \frac{(R/r)^3 \cos(2\pi n/N) -R/r} {[(R/r)^4 - 2 (R/r)^2 \cos(2\pi n/N) +1]^{3/2}} -q\ \sum_{n=1}^{N-1} \frac{ 1}{ 4\sin(n\pi/N)}$$ which for increasing $$N$$ quickly leads to quite complicated expressions. For $$N=3$$ as in the drawing, it gives: $$q_c=-q\ \left( \frac1{\sqrt3} + \frac{(R/r)^3 +2R/r} {[(R/r)^4 +(R/r)^2+1]^{3/2}} + \frac{R/r} {[(R/r)^2 -1]^2} \right)$$ and if we further choose values with $$r/R=\frac12$$, we get: $$q_c=-q\ \left( \frac29 + \frac1{\sqrt3} + \frac4{7\sqrt{21}} \right) \approx -0.674877 \ q,$$ so only a 17% stronger $$q_c$$ is needed than the $$-q/\sqrt3\$$ that is needed in free space.

One final thing we have to address: how to use only 2 charges in a hollow sphere (since the above case always needs at least three). This can now be done with charges $$q$$ and $$-q$$ both at radius $$r$$ and at opposite sides of the origin. This gives image charges $$-(R/r)q$$ and $$+(R/r)q$$ both at radius $$R^2/r$$. The four charges are all on one line (say, the $$x$$-axis). To find a radius $$r$$ where this is in (unstable) equilibrium, we look at the charge $$q$$ on the positive $$x$$-axis, which has to its left the charge $$-q$$ and further away an image $$-q_{\small\ \mbox{im}}$$ and to its right an image $$+q_{\small\ \mbox{im}}$$. So it feels a force: $$F= + \frac{-q_{\small\ \mbox{im}} q}{(r_{\small \mbox{im}}+r)^2} + \frac{-q \ q}{(2r)^2} - \frac{q_{\small\ \mbox{im}} q}{(r_{\small \mbox{im}}-r)^2}$$ For the sphere with radius $$R$$ we had $$q_{\small\ \mbox{im}} = -(R/r)q$$ and $$r_{\small \mbox{im}} = R^2/r$$, so by defining $$\rho=r/R$$ we can express this as: $$F= \frac{-q^2}{4R^2}\ \left(\frac{-1}{\rho^2} + \frac{8(\rho+\rho^5)}{(\rho^4-1)^2}\right) = \frac{-q^2}{4R^2}\ \cdot\ \frac{\rho^8- 8\rho^7- 2\rho^4 - 8\rho^3 +1}{\rho^2(\rho^4-1)^2}$$ where we have defined $$\rho=r/R$$. The quotient has no cancelling common factors so we have to solve an eighth order(!) polynomial to find the solution for $$F=0$$: $$\rho^8- 8\rho^7- 2\rho^4 - 8\rho^3 +1 = 0.$$ So unfortunately, unlike the earlier cases, this simplest case with only two charges can now only be solved numerically. The only solution with $$0 is: $$\frac rR = 0.4749646535483733526...$$