Force between a charged conducting hollow sphere and a grounded metal plate The conducting sphere has radius R and potential V.(All of the charges reside on the surface) The distance between the center of the sphere and the plate is d.
I know that the method of images could be used if the sphere were a point charge. However, I feel like assuming that the charge distribution on the sphere to be uniform would be wrong since the conducting sphere must have constant potential everywhere on the surface. Then, how can I find the force?
 A: The problem can be tackled by using an infinite number of image charges. Just remember the places of the image charges in the case of the plate and the sphere.
Let the plane at ground be the $z=0$ plane, and the sphere, radius $R$ and potential $V$, be centred at $z=+d$.
First we place a charge
$$
q = +4\pi\epsilon_0 RV
\quad\text{at}\quad
z = +d
$$
(the center of the sphere) to set the sphere at the potential $V$. This disturbs the potential at the plate though, and so we place now an image charge
$$
q' = -q = -4\pi\epsilon_0 RV
\quad\text{at}\quad
z=-d
$$
to restore ground. This disturbs the potential at the sphere now though, and so we place now an image charge
$$
q'' = -\frac R{2d}q' = 4\pi\epsilon_0 \frac{R^2}{2d}V
\quad\text{at}\quad
z = +d-\frac{R^2}{2d}
$$
Then another image charge to restore ground at the plate:
$$
q'''= -q'' = -4\pi\epsilon_0\frac{R^2}{2d}V
\quad\text{at}\quad
z = -d+\frac{R^2}{2d}
$$
Then another image charge to restore $V$ at the sphere:
$$
q''''= -\frac R{2d}q''' = 4\pi\epsilon_0\frac{R^3}{4d^2}V
\quad\text{at}\quad
z=+d-\frac{R^2}{2d-\frac{R^2}{2d}}
$$
and so on…
We see that the total charge on the sphere is
$$
Q = q + q'' + q'''' + \dots = q\sum_{n=0}^\infty \left(\frac R{2d}\right)^n = \frac{4\pi\epsilon_0 RV}{1-\frac R{2d}}
$$
So the capacitance of the arrangement is
$$
C = \frac QV = \frac{4\pi\epsilon_0 R}{1-\frac R{2d}}
$$
the energy is
$$
W = \frac12CV^2 = \frac{2\pi\epsilon_0 R}{1-\frac R{2d}}V^2
$$
and the force is
$$
\mathbf F = -\hat{\mathbf z}\frac{\partial W}{\partial d} = \frac{\pi\epsilon_0R^2}{(d-\frac R2)^2} \hat{\mathbf z}
$$
