Metal sphere and charged ring I think this is an interesting question, to which I don't really know the answer to. (Also, not a homework question.)
Say you have an uncharged metal sphere constrained to move in the z-axis. There is a charged ring lying in the x-y plane centered at the origin. Two cases: 1) the diameter of the ring is larger than the sphere, so that the sphere can pass through the ring, and 2) the diameter of the ring is smaller than the sphere, so the sphere cannot pass through the ring but can touch it.
What are the equilibrium position(s) of the ball?
For case 1) it is obvious that the center, by symmetry, is an equilibrium point. But are there more? The complication arises because of the finite size of the sphere. As the sphere starts to pass through the ring, the charge of opposite sign to the ring is induced near the ring, but the angle is very shallow, so there is not much attractive force. On the other hand the induced charge of the same sign gets pushed to the far end of the sphere, which causes a strong repulsive force, so the sphere might get repelled as it enters the ring. I'm not sure whether this is correct though. It might very well turn out that there is only 1 equilibrium point... 
So aside from solving the system exactly to find the equilibrium points, is there a way to argue how many there are, for the 2 cases, and where they are?
 A: Using the method of images, you can calculate the force between the ring of charge and the sphere.
Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere.
To calculate the force, you can replace the sphere with a charged ring (method of images) with a charge density $\lambda'$ placed at a distance $d$ below the sphere's center and a point charge $Q'$ at the center of the sphere to make the sphere electrically neutral:
$$\lambda'=-\lambda\frac{ \sqrt{R_r^2+z^2}}{R_s}$$
$$d=\frac{zR_s^2}{z^2+R_r^2}$$
$$Q'=Q\frac{R_s}{\sqrt{z^2+R_r^2}}$$
Now the problem reduces to finding the force between the main ring and the induced ring and point charge.
The image below shows the choose of parameters:

I assume that $R_{\text{ring}}>R_{\text{sphere}}$. The equations for the case of $R_{\text{ring}}<R_{\text{sphere}}$ are the same, except that the motion is restricted to $z>\sqrt{R_s^2-R_r^2}$.
The field of the main ring differs for points with $r>R_r$ and $r<R_r$:($Q$ is the ring's total charge)
$$\Phi(r,\theta)=\frac{Q}{4\pi\epsilon_0}\sum_{l=0}^\infty \mathrm{P}_{2l}(\cos \theta)\cases{\frac{R_r^{2l}}{r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r>R_r$}\\\frac{r^{2l}}{R_r^{2l+1}}\mathrm{P}_{2l}(0)\,\,\,\,\,\,\,\,\,\,\text{$r<R_r$}}$$
and 
$$\mathbf{E}=-\frac{\partial \Phi}{\partial r}\hat r-\frac{1}{r}\frac{\partial \Phi}{\partial \theta}\hat \theta$$
Because of the rotational symmetry of the induced charges, on which we want to find the force, the force will be in the $z$ direction: (I omitted the lengthy calculations)
$$\mathbf{F}_{\text{total}}=\hat z \frac{Q^2}{4\pi \epsilon_0} \left[\frac{z}{(R_r^2+z^2)^2}-\frac{1}{R_r^2\sqrt{z^2+R_r^2}}\cases{{\times f_{\text{out}}(r,\theta)}\\{\times f_{\text{in}}(r,\theta)}} \right]$$
Where $f_{\text{out}}$ is used when $r>R_r$ and $f_{\text{in}}$ is used when $r<R_r$. These are dimensionless functions that appear when differentiating the above potential:
$$f_{\text{out}}=\sum_{l=0}^\infty\mathrm{P}_{2l}(0)\left(\frac{R_r}{r}\right)^{2l+2}\left( (2l+1)\mathrm{P}_{2l}(\cos \theta)\cos\theta +\mathrm {P}^1_{2l}(\cos \theta)\sin\theta \right)$$
$$f_{\text{in}}=\sum_{l=0}^\infty-\mathrm{P}_{2l}(0)\left(\frac{r}{R_r}\right)^{2l-1}\left(2l\mathrm{P}_{2l}(\cos\theta)\cos \theta-\mathrm{P}^1_{2l}(\cos \theta) \sin \theta \right)$$
and $r$ and $\theta$ are functions of $z$:
$$r=\left| z-\frac{R_s^2}{\sqrt{z^2+R_r^2}} \right|$$
$$\cos \theta=\frac{z-\frac{zR_s^2}{z^2+R_r^2}}{r}$$
$$\sin \theta=\frac{\frac{R_rR_s^2}{z^2+R_r^2}}{r}$$
$\mathrm {P}^1_{2l}(\cos \theta)$ s are associated Legendre functions, and appear when differentiating $\mathrm {P}_{2l}(\cos \theta)$ w.r.t. $\theta$.
The reason of summing over even terms (indexes are $2l$ instead of $l$ ) is that $\mathrm {P}_{l}(0)$ is nonzero for even terms, as is equal to:
$$\mathrm {P}_{l}(0)= \frac 1 {2^l} \sum_{k=0}^l {l\choose k}^2 (-1)^{l-k}$$
It seems impossible to say anything intuitively, but the above equations can be used to simulate the system (with a rather good accuracy, Since the higher order terms in the above series go as $r^{2l-1}$ instead of $r^l$). 
A: There is only one equilibrium point, the centre. By finding minima of the potential it can be shown that the equilibrium points are exactly those points for which, if one produces a 'central vector' from the centre of the sphere to the ring, the angle $\theta$ that central vector makes with the axis of the sphere has the property $\cos\theta=0.$ This is just the condition that the centre of the sphere is in the same plane as the ring.
