Capacitance matrix is diagonally dominant

Question

I have heard of the following claim “the capacitance matrix of an isolated system of conductors is diagonally dominant” but I couldn’t find a way of proving it myself or any reference for that matter. Does anyone know how to prove it or where to find a rigorous proof?

Answer 1

First of all, the article by John Denker is useful as an explanation of capacity matrices.

Suppose we have two capacitors, one with a capacitance of 10 µF and the other with a capacitance of 20 pF. Arrange them as shown in the figure and join the bottom conductors of the two capacitors to make a three-conductor problem. (The original picture is taken from Wikipedia(jpn)).

Symbolize capacitance as $C_A(=10\mu F), C_B(=20pF)$, and assign conductor numbers as shown in the figure. The capacitance matrix for a three-conductor system is as follows, \begin{equation*} C= \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} = \begin{pmatrix} C_A & -C_A & 0 \\ -C_A & C_A+C_B & -C_B \\ 0 & -C_B & C_B \end{pmatrix} \end{equation*} Since conductors 1 and 3 are not coupled, the corresponding component ($C_{13}$) is set to 0. We chose $C_{33}$ as the representative of the diagonal component and $C_{12}$ as the representative of the off-diagonal component, and calculated the ratio of their magnitudes. \begin{equation*} \text{ratio}=\frac{-C_{12}}{C_{33}}=\frac{10\mu F}{20pF}=5.0\times 10^5 \end{equation*}

Therefore, the statement "the capacitance matrix of an isolated system of conductors is diagonally dominant" is false.