We assume that there is no current and a time independent charge distribution. Then the electric field \(\vec E\) satisfies

\begin{equation} \label{eq:Maxwell} \mathrm{div}\vec E=\rho/\epsilon,\quad \mathrm{curl}\vec E=0 \end{equation}where \(\rho\) is the charge density and \(\epsilon\) is called the permittivity of free space. From the second equation in \eqref{eq:Maxwell}, we can introduce the electrostatic potential such that \(\vec E=-\nabla \phi\). Then we have Poisson's equation \(-\Delta \phi=f\), \(f=-\rho/\epsilon\). We now obtain the equipotential line which is the level curve of \(\phi\), when there are no charges except conductors \(\{C_i\}_{1,\cdots,K}\). Let us assume \(K\) conductors \(C_1,\cdots,C_K\) within an enclosure \(C_0\). Each one is held at an electrostatic potential \(\varphi_i\). We assume that the enclosure \(C_0\) is held at potential 0. In order to know \(\varphi(x)\) at any point \(x\) of the domain \(\Omega\), we must solve

$$-\Delta \varphi =0\quad \textrm{in }\Omega,$$where \(\Omega\) is the interior of \(C_0\) minus the conductors \(C_i\), and \(\Gamma\) is the boundary of \(\Omega\), that is \(\sum_{i=0}^N C_i\). Here \(g\) is any function of \(x\) equal to \(\varphi_i\) on \(C_i\) and to 0 on \(C_0\). The boundary equation is a reduced form for:

$$\varphi =\varphi _{i}\;\text{on }C_{i},\;i=1...N,\varphi =0\;\text{on }C_{0}.$$**Example**

First we give the geometrical informations. Here they are in mathematical and in FreeFem++ form. The FreeFem++ definition of $C_1$ and $C_2$ can be modified to view the effect of different geometries.

\(C_0=\{(x,y);\; x^2+y^2=5^2\}\) is a circle with center at (0,0) and radius 5,

border C0(t=0,2*pi) { x = 5 * cos(t); y = 5 * sin(t); }

\(C_1=\{(x,y):\; \frac{1}{0.3^2}(x-2)^2+\frac{1}{3^2}y^2=1\}\) first elliptical hole,

\(C_2=\{(x,y):\; \frac{1}{0.3^2}(x+2)^2+\frac{1}{3^2}y^2=1\}\) second elliptical hole.

Let \(\Omega\) be the disk enclosed by \(C_0\) with the elliptical holes enclosed by \(C_1\) and \(C_2\). Note that \(C_0\) is described counterclockwise, whereas the elliptical holes are described clockwise, because the boundary must be oriented so that the computational domain is to its left.

mesh Th = buildmesh(C0(60)+C1(-50)+C2(-50));

plot(Th,ps="electroMesh.eps"); // see figure [electroMesh]

P1 FE-space, unknown and test function :

fespace Vh(Th,P1);

Vh uh,vh;

Definition of the problem :

problem Electro(uh,vh) = int2d(Th)( dx(uh)*dx(vh) + dy(uh)*dy(vh) ) // bilinear

+ on(C0,uh=0) // boundary condition on $C_0$

+ on(C1,uh=1) // +1 volt on $C_1$

+ on(C2,uh=-1); // -1 volt on $C_2$

Solve the problem :

Electro;

plot(uh,ps="electro.eps",wait=true); // see figure [electro]

Figure [electroMesh]: Disk with two elliptical holes |
Figure [electro]: Equipotential lines, where \(C_1\) is located in the right hand side |