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?

  • $\begingroup$ Yes, the charge distribution on the sphere would not be uniform because of the presence of the nearby conductor plate. Where's this problem from? It probably doesn't have a nice closed-form solution. On this link electrostatics.org/images/B4.pdf there's a solution to the problem of two conducting spheres. Maybe it can be transformed into a solution to your problem in the same manner as the case for point charges / method of images. $\endgroup$ Mar 25, 2021 at 7:07
  • $\begingroup$ This is the experiment we did in the lab. We were told to find the theoretical force ourselves. Even if there isn't a nice closed, I would appreciate a nice approximation rather than using uniform distribution. $\endgroup$
    – curiouss
    Mar 25, 2021 at 7:44

1 Answer 1


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

$$ \mathbf F = -\hat{\mathbf z}\frac{\partial W}{\partial d} = \frac{\pi\epsilon_0R^2}{(d-\frac R2)^2} \hat{\mathbf z} $$

  • $\begingroup$ Cant we just calculate force of interaction between all charges on q charge to calculate force ? $\endgroup$
    – Orion_Pax
    Mar 5, 2022 at 10:57

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