Is there an integral representation of the capacitance matrix?

Question

Suppose we have conductors $\{S_1, S_2,\ldots,S_N\}$ with boundaries $\{\partial S_1, \partial S_2,\ldots,\partial S_N\}$. These conductors are suspended in vacuum. Is there an integral over these bodies and/or their boundaries that provides the capacitance matrix of the system?

Answer 1

I'll give a try, even if this is not my main field of interest, since I worked a bit on boundary element methods. I guess that you can only find a solution using a surface integral if the permittivity $\varepsilon$ is uniform in the region of space where you're inserting the conductors of the capacitor.

Governing equations. Let's denote with $\Omega_0$ the region of space outside the conductors, $\Omega_i$ the volume of the $i^{th}$ conductor, whose surface is labelled $\partial \Omega_i$.

• Assuming the laws of electrostatics hold \begin{aligned} \nabla \cdot \mathbf{d} & = \rho \\ \nabla \times \mathbf{e} + \underbrace{\partial_t\mathbf{b}}_{\partial_t \equiv 0} & = 0 \qquad \rightarrow \qquad \mathbf{e} = - \nabla v \ , \end{aligned}

• assuming that the permittivity $\varepsilon$ is uniform in $\Omega_0$ and the constitutive law reads $\mathbf{d} = \varepsilon \mathbf{e}$

• assuming there is no free charge in the space $\Omega_0$, $\rho = 0$ in $\Omega_0$, so that we get a Laplace equation for the potential

  $- \nabla^2 v = \rho = 0$

• remembering that there is no free charge inside a conductor, except for a very thin layer at the surface, and the electric field is $\mathbf{e} = 0$ inside, we can:

  • recall that the surface of a conductor is equipotential, i.e. the value of the electric potential on each surface is uniform,

    $v(\mathbf{x})|_{\mathbf{x} \in \partial \Omega_i} = v_i$

  • use the jump conditions at the interface to get another boundary condition relating the normal derivative of the

    $\varepsilon \mathbf{\hat{n}} \cdot \mathbf{e} = \varepsilon \mathbf{\hat{n}} \cdot \nabla v = \varepsilon \partial_n v = \sigma$.

Now, putting everything together, we get a Laplace equation for $v$ and its boundary conditions

\begin{cases} -\nabla^2 v = 0 \quad \text{in} \Omega_0 \\ v(\mathbf{x})|_{\Gamma_i} = v_i \\ \varepsilon \partial_n v(\mathbf{x})|_{\partial \Omega_0} = \varepsilon \mathbf{\hat{n}} \cdot \nabla v(\mathbf{r}) = \sigma_i(\mathbf{x}) \end{cases}

Solution using Green's function. It's now possible to find a solution by collocation, after having exploited the properties of the Green's function for Laplace equation, $-\nabla^2 G(\mathbf{r}, \mathbf{r}_0) = \delta(\mathbf{r} - \mathbf{r}_0)$.

Introducing the factor $E(\mathbf{r}_0)$ that takes care if the point $\mathbf{r}_0$ is inside the domain, on the boundary, or outside the domain, and performing integration by part, we can write

\begin{aligned} E(\mathbf{r}_0)v(\mathbf{r}_0) &= \int_{\Omega_0} \delta(\mathbf{r}-\mathbf{r}_0) v(\mathbf{r}) dV_{\mathbf{r}} = \\ & = - \int_{\Omega_0} \nabla^2 G(\mathbf{r}, \mathbf{r}_0) v(\mathbf{r}) dV_{\mathbf{r}} = \\ & = - \int_{\partial \Omega_0} v(\mathbf{r}) \mathbf{\hat{n}} \cdot \nabla G(\mathbf{r}, \mathbf{r}_0) dS_{\mathbf{r}} + \int_{\partial \Omega_0} G(\mathbf{r}, \mathbf{r}_0) \underbrace{\mathbf{\hat{n}} \cdot \nabla v(\mathbf{r})}_{=\frac{\sigma}{\varepsilon}} dS_{\mathbf{r}} - \int_{\Omega_0} \underbrace{\nabla^2 v(\mathbf{r})}_{=0} G(\mathbf{r}, \mathbf{r}_0) dV_{\mathbf{r}} \end{aligned}.

Approximation/discretization. Now, you can use an approximation to find the charge distribution on each surface. As an example, you can divide the surfaces of the conductors in $N$ elements, $E_e$, $e = 1:N$. Assuming you know the potential on the surface $v_i$ (you can prescribe them in an experiment), you have $N$ unknowns, namely $\sigma_e$, for $e=1:N$. Thus, you use collocation method and choose $N$ different points $\mathbf{r}_0$ (usually collocated with the centers of your elements) to get a system of $N$ equations.

Briefly, we can recast the equation above for an element $e \in \Gamma_i$

\begin{equation} E(\mathbf{r}_e) v_i = - \sum_j v_j \int_{\Gamma_j} \mathbf{\hat{n}} \cdot \nabla G(\mathbf{r}, \mathbf{r}_e) dS_{\mathbf{r}} + \sum_{f = 1:N} \int_{E_f} G(\mathbf{r}, \mathbf{r}_e) \dfrac{1}{\varepsilon} \sigma_f dS_{\mathbf{r}} \qquad \text{for $e = 1:N$} \end{equation}

Assuming uniform charge density on each element, we can recast the equation above in matrix form as: \begin{equation} [ \underline{\underline{E}}_{ei} + \underline{\underline{D}}_{ei} ] \underline{V}_i = \underline{\underline{S}}_{ef} \underline{\sigma}_f \qquad , \qquad (\underline{\underline{E}}+\underline{\underline{D}}) \underline{V} = \underline{\underline{S}} \underline{\sigma}. \end{equation}

Now, it's possible to find the vector of the source density over each element $\sigma$, solving this linear system

\begin{equation} \underline{\sigma} = \underline{\underline{S}}^{-1} (\underline{\underline{E}}+\underline{\underline{D}}) \underline{V} \ . \end{equation}

Once you know the charge density, you can find the total charge on every conductor by integration,

\begin{equation} Q_i = \int_{\Gamma_i} \sigma(\mathbf{r}) \ , \end{equation}

that can be written with matrix formalism as

\begin{equation} \underline{Q} = \underline{\underline{P}} \underline{\sigma} \end{equation}

with the matrix $\underline{\underline{P}}$ performing the numerical integration.

Putting everything together, it's possible to relate the total charges on the conductors with their potentials as

\begin{equation} \underline{Q} = \underline{\underline{P}} \underline{\sigma} = \underline{\underline{P}} \underline{\underline{S}}^{-1} (\underline{\underline{E}}+\underline{\underline{D}}) \underline{V} \ , \end{equation}

i.e.

\begin{equation} \underline{Q} = \underline{\underline{C}} \underline{V} \end{equation}

with the capacitance matrix

\begin{equation} \underline{\underline{C}} = \underline{\underline{P}} \underline{\underline{S}}^{-1} (\underline{\underline{E}}+\underline{\underline{D}}) \ . \end{equation}

Remark. If you want non-zero values of the potential at the infinity (remember that potential is defined up to an additive constant), you need to treat the ``doublet'' integral contribution, \begin{equation} - v_{\infty} \underbrace{\int_{\partial \Omega_{\infty}} \mathbf{\hat{n}} \cdot \nabla G(\mathbf{r}, \mathbf{r}_0) dS_{\mathbf{r}}}_{-1} = v_{\infty} \ . \end{equation} from the surface at the infinity as well, since the integral doesn't vanish but equals a round angle divided by $2 \pi$, i.e. equals $1$.

Link. You can find a preliminary version of a script evaluating the capacitance matrix for 2D problem in this Colab notebook: https://colab.research.google.com/drive/1jC8tPdOELI3cmCPXWLWKMsMTBgoqY1BF?usp=sharing