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Minimal Electric Potential

This example shows how to find the minimal electric potential by solving the equation


on the unit disk Ω={(x,y)|x2+y21}, with V(x,y)=x2 on the boundary Ω. Here, ε is the absolute dielectric permittivity of the material. The toolbox uses the relative permittivity of the material εr=ε/ε0, where ε0is the absolute permittivity of the vacuum. Note that the constant ε0does not affect the result in this example.

For the minimal surface problem, the value ε is given by


Because the dielectric permittivity is a function of the solution V, the minimal surface problem is a nonlinear elliptic problem.

To solve the minimal surface problem, first create an electromagnetic model for electrostatic analysis.

emagmodel = createpde('electromagnetic','electrostatic');

Create the geometry and include it in the model. The circleg function represents this geometry.


Plot the geometry with the edge labels.

axis equal
title 'Geometry with Edge Labels';

Figure contains an axes. The axes with title Geometry with Edge Labels contains 5 objects of type line, text.

Specify the vacuum permittivity value in the SI system of units.

emagmodel.VacuumPermittivity = 8.8541878128E-12;

Specify the relative permittivity of the material.

perm = @(region,state) 1./sqrt(1+state.ux.^2 +^2);

Specify the electrostatic potential at the boundary using the functionV(x,y)=x2.

bc = @(region,~)region.x.^2;

Generate and plot the mesh.

axis equal

Figure contains an axes. The axes contains 2 objects of type line.

Solve the model.

R = solve(emagmodel);
V = R.ElectricPotential;

Plot the electric potential, using the Contour parameter to display equipotential lines.

xlabel 'x'
ylabel 'y'
zlabel 'V(x,y)'
title 'Minimal Electric Potential'

Figure contains an axes. The axes with title Minimal Electric Potential contains 12 objects of type patch, line.