Voltage matrix for two nearby conductors

Each conductor above a perfectly conducting earth produces a potential at the other conductor. With the method of images, these effects are written in matrix form: $[V] = [P][Q]$. The off-diagonal terms show the proximity effect of the nearby conductor.

Assumptions

• Long, straight, parallel conductors of radii $r_1$ and $r_2$, with charges $+q_1$ and $+q_2$ C/m.
• The earth is a perfect conductor at zero potential. It is replaced by images $-q_1$ and $-q_2$ at depths $H_1$ and $H_2$ below ground.
• $A_{12}$ = distance between conductors 1 and 2; $I_{12}$ = distance from conductor 1 to the image of conductor 2 (equal to that from 2 to the image of 1).

        +q1 o . . A12 . . o +q2
            |  .         .|
         H1 |    .  I12   | H2
  //////////////////////////////// ground
         H1 |        .    | H2
        -q1 o             o -q2
           (images)

Step 1: Potential due to a line charge

The field of a line charge $q$ at distance $x$ is $E = \dfrac{q}{2\pi\varepsilon_0 x}$. The potential difference between distances $x_1$ and $x_2$ is

$$V_{x_1} - V_{x_2} = \frac{q}{2\pi\varepsilon_0}\ln\frac{x_2}{x_1}$$

For a charge $+q$ and its image $-q$, the potential at a point is

$$V = \frac{q}{2\pi\varepsilon_0}\ln\frac{\text{distance from image}}{\text{distance from charge}}$$

This is zero on the ground plane, as required.

Step 2: Potential of conductor 1

• Due to $+q_1$ and its image (distances $r_1$ and $2H_1$): $\dfrac{q_1}{2\pi\varepsilon_0}\ln\dfrac{2H_1}{r_1}$
• Due to $+q_2$ and its image (distances $A_{12}$ and $I_{12}$): $\dfrac{q_2}{2\pi\varepsilon_0}\ln\dfrac{I_{12}}{A_{12}}$

$$V_1 = \frac{q_1}{2\pi\varepsilon_0}\ln\frac{2H_1}{r_1} + \frac{q_2}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}}$$

Step 3: Potential of conductor 2

$$V_2 = \frac{q_1}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}} + \frac{q_2}{2\pi\varepsilon_0}\ln\frac{2H_2}{r_2}$$

Step 4: Voltage matrix

$$\begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \frac{1}{2\pi\varepsilon_0} \begin{bmatrix} \ln\dfrac{2H_1}{r_1} & \ln\dfrac{I_{12}}{A_{12}} \\[8pt] \ln\dfrac{I_{12}}{A_{12}} & \ln\dfrac{2H_2}{r_2} \end{bmatrix} \begin{bmatrix} q_1 \\ q_2 \end{bmatrix}$$

or $[V] = [P][Q]$, with Maxwell's potential coefficients

$$P_{11} = \frac{1}{2\pi\varepsilon_0}\ln\frac{2H_1}{r_1},\quad P_{22} = \frac{1}{2\pi\varepsilon_0}\ln\frac{2H_2}{r_2},\quad P_{12} = P_{21} = \frac{1}{2\pi\varepsilon_0}\ln\frac{I_{12}}{A_{12}}$$

Here $1/(2\pi\varepsilon_0) = 18 \times 10^9$ m/F.

Interpretation (proximity effect)

• $P_{11}$ and $P_{22}$ are the self terms: the potential a conductor would have alone above ground.
• $P_{12}$ is the mutual term. Because $I_{12} > A_{12}$, it is positive. The nearby positively charged conductor raises the potential of conductor 1, and the effect grows as the conductors come closer (smaller $A_{12}$).
• Inverting gives the capacitance (charge coefficient) matrix: $[Q] = [P]^{-1}[V] = [C][V]$. This is used to find conductor charges and surface gradients. For $n$ conductors, the same rule gives an $n \times n$ matrix.