A.3 Capacitance of Two Spheres

  First we study the case of a point charge near a grounded conducting sphere by the  image charge method. Removing the conductor we try to find the position and magnitude of an `image' charge q', as in Fig. A.2, that will make the potential zero on the spherical surface.

  

Figure: A point charge q at a distance d from the center of a grounded conducting sphere of radius a. The boundary condition $\protect\varphi = 0$ on the spherical surface is satisfied by the original charge q and its image q'.

It is clear from the symmetry of the problem, that if such a charge exists it must lie on the line connecting q and the center of the sphere. We begin by forcing the potential to vanish at the points P $_{\text{1}}$ and P $_{\text{2}}$. Then
$$\begin{gather}\frac{q}{d+a} + \frac{q'}{d'+a} = 0,\qquad \frac{q}{d-a} + \frac{q'}{a-d'} = 0. \end{gather}$$
Solving these two equations gives
 $$\begin{gather} q' = -\frac{a}{d}q, \qquad d' = \frac{a^2}{d}. \end{gather}$$

We still have to assure whether this charge arrangement will make the potential zero at a general point P $_{\text{3}}$ on the surface of the sphere. At P $_{\text{3}}$
$$\begin{gather}4\pi\epsilon\varphi = \frac{q}{r_1}+\frac{q'}{r_2}, \end{gather}$$
where
$$\begin{gather}r_1 = \sqrt{(d^2+a^2+2da\cos\theta)}, \qquad r_2 = \sqrt{(d^{'2}+a^2+2d'a\cos\theta)}. \end{gather}$$
Taking q' and d' as in (A.7) does in fact make $\varphi = 0$ at P $_{\text{3}}$, as required. Since this set of charges has a rotation symmetry around the axis P $_{\text{1}}$-P $_{\text{2}}$, the condition is indeed fulfilled for all points on the sphere.

Applying the above result for a successive approximation in a method of images, one can calculate the capacitance of two spheres, which is applied here for two equal spheres. We are replacing both spheres by a set of point charges which will maintain the conductor surfaces as equipotentials.

First we put a charge q at the center of the left sphere, as in Fig. A.3.

  

Figure A.3: Two conducting spheres at distance d and with radius a. One sphere is connected to ground.

This makes the left sphere an equipotential, but not the right sphere. Next we put the `image' charge q' in the right sphere. This makes the right sphere a zero equipotential but destroys the spherical potential on the left. So we put the `image' of the `image' q'' inside the left sphere, to compensate for q'. This makes the left sphere again an equipotential but upsets the right sphere. We continue the process which converges rapidly, until we reach the required precision. The total charge on the left sphere is
$$\begin{gather}q_{\Sigma} = q\left(1+\frac{a^2}{d^2-a^2}+\frac{a^4}{d^4-3d^2a^2+a^4}+ \ldots\right). \end{gather}$$
But only q contributes to its potential. The charges q' and q'' cause the potential, they contribute to the left sphere, to vanish, and the same is true of all following pairs of charges. The potential of the left sphere is therefore
$$\begin{gather}\varphi = \frac{q}{4\pi\epsilon a}. \end{gather}$$
The charges q and q' have a zero contribution to the potential of the right sphere and so do the pairs q'', q''' a.s.o. Since the potential of the right sphere was always compensated to zero the total capacitance of either sphere is
$$\begin{gather}C=\frac{\partial q_{\Sigma}}{\partial\varphi} = 4\pi\epsilon a \left(1+\frac{a^2}{d^2-a^2}+\frac{a^4}{d^4-3d^2a^2+a^4}+\ldots\right). \end{gather}$$
The total capacitance C can be split in a capacitance between the spheres C₁₂ and a capacitance to infinity or self-capacitance of the sphere C₁₀, as shown in Fig. A.4.

 

Table A.1: Charges and location of successive compensation charges.

left sphere		right sphere
charge	distance	charge	distance
	from center		from center
q	0	$q'=-\frac{aq}{d}$	$d'=\frac{a^2}{d}$
$q''=\frac{a^2q}{d^2-a^2}$	$d''=\frac{a^2}{d-\frac{a^2}{d}}$	$q'''=-\frac{a^2}{d^2-2a^2}\frac{aq}{d}$	$d'''=\frac{a^2}{d-\frac{a^2}{d-\frac{a^2}{d}}}$
$q''''=\frac{a^4q}{d^4-3d^2a^2+a^4}$	$d''''=\frac{a^2}{d-\frac{a^2}{d- \frac{a^2}{d-\frac{a^2}{d}}}}$	$q'''''=-\frac{a^4}{d^4-4d^2a^2+3a^4}\frac{aq}{d}$	$d'''''=\frac{a^2}{d-\frac{a^2}{d-\frac{a^2}{d-\frac{a^2}{d- \frac{a^2}{d}}}}}$

 

  

Figure A.4: Two conducting spheres and their capacitances.

$$\begin{gather}C = C_{12} + C_{10}\\ C_{12} = 4\pi\epsilon\frac{a^2}{d}\left(1 +... ...+ \frac{a^2}{d^2 - a^2} - \frac{a^3}{d^3 - 2da^2} + \ldots\right) \end{gather}$$