Given the electrical potential on a conductor surface S_i (the equipotential surface or the point P chosen on surface i) contained in a system of conductors j = 1, 2, ..., n:

${\displaystyle \phi _{i}=\sum _{j=1}^{n}{\frac {1}{4\pi \epsilon _{0}}}\int _{S_{j}}{\frac {\sigma _{j}da_{j}}{R_{ji}}}{\mbox{ (i=1, 2..., n)}},}$ 

where R_ji = |r_i - r_j|, i.e. the distance from the area-element da_j to a particular point r_i on conductor i. σ_j is not, in general, uniformly distributed across the surface. Let us introduce the factor f_j that describes how the actual charge density differs from the average and itself on a position on the surface of the j-th conductor:

${\displaystyle {\frac {\sigma _{j}}{\langle \sigma _{j}\rangle }}=f_{j},}$ 

or

${\displaystyle \sigma _{j}=\langle \sigma _{j}\rangle f_{j}={\frac {Q_{j}}{S_{j}}}f_{j}.}$ 

Then,

${\displaystyle \phi _{i}=\sum _{j=1}^{n}{\frac {Q_{j}}{4\pi \epsilon _{0}S_{j}}}\int _{S_{j}}{\frac {f_{j}da_{j}}{R_{ji}}}.}$ 

It can be shown that ${\displaystyle \int _{S_{j}}{\frac {f_{j}da_{j}}{R_{ji}}}}$  is independent of the distribution ${\displaystyle \sigma _{j}}$ . Hence, with

${\displaystyle p_{ij}={\frac {1}{4\pi \epsilon _{0}S_{j}}}\int _{S_{j}}{\frac {f_{j}da_{j}}{R_{ji}}},}$ 

we have

${\displaystyle \phi _{i}=\sum _{j=1}^{n}p_{ij}Q_{j}{\mbox{ (i = 1, 2, ..., n)}}.}$ 

In this example, we employ the method of coefficients of potential to determine the capacitance on a two-conductor system.

For a two-conductor system, the system of linear equations is

${\displaystyle {\begin{matrix}\phi _{1}=p_{11}Q_{1}+p_{12}Q_{2}\\\phi _{2}=p_{21}Q_{1}+p_{22}Q_{2}\end{matrix}}.}$ 

On a capacitor, the charge on the two conductors is equal and opposite: Q = Q₁ = -Q₂. Therefore,

${\displaystyle {\begin{matrix}\phi _{1}=(p_{11}-p_{12})Q\\\phi _{2}=(p_{21}-p_{22})Q\end{matrix}},}$ 

and

${\displaystyle \Delta \phi =\phi _{1}-\phi _{2}=(p_{11}+p_{22}-p_{12}-p_{21})Q.}$ 

Hence,

${\displaystyle C={\frac {1}{p_{11}+p_{22}-2p_{12}}}.}$ 

Note that the array of linear equations

${\displaystyle \phi _{i}=\sum _{j=1}^{n}p_{ij}Q_{j}{\mbox{ (i = 1,2,...n)}}}$ 

can be inverted to

${\displaystyle Q_{i}=\sum _{j=1}^{n}c_{ij}\phi _{j}{\mbox{ (i = 1,2,...n)}}}$ 

where the c_ij with i = j are called the coefficients of capacity and the c_ij with i ≠ j are called the coefficients of electrostatic induction.^[1]

For a system of two spherical conductors held at the same potential,^[2]

${\displaystyle Q_{a}=(c_{11}+c_{12})V,\qquad Q_{b}=(c_{12}+c_{22})V}$ 

${\displaystyle Q=Q_{a}+Q_{b}=(c_{11}+2c_{12}+c_{bb})V}$ 

If the two conductors carry equal and opposite charges,

${\displaystyle \phi _{1}={\frac {Q(c_{12}+c_{22})}{(c_{11}c_{22}-c_{12}^{2})}},\qquad \quad \phi _{2}={\frac {-Q(c_{12}+c_{11})}{(c_{11}c_{22}-c_{12}^{2})}}}$ 

${\displaystyle \quad C={\frac {Q}{\phi _{1}-\phi _{2}}}={\frac {c_{11}c_{22}-c_{12}^{2}}{c_{11}+c_{22}+2c_{12}}}}$ 

The system of conductors can be shown to have similar symmetry c_ij = c_ji.

• James Clerk Maxwell (1873) A Treatise on Electricity and Magnetism, § 86, page 89.