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How to find an electric potential in anisotropic, inhomogeneous medium

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Agent007
#1
Nov25-12, 11:52 PM
P: 1
Hello!

(I am sorry for probable mistakes. English is not my native language. I have never written anything about mathematics and physics in English.)

I have an electrostatic problem. I need to find an electric potential [itex]\psi[/itex] ([itex]\vec{E}=-\nabla\psi[/itex]) in anisotropic, inhomogeneous medium.

Let's introduce a cylindrical coordinate system ([itex]\rho[/itex], [itex]\varphi[/itex], z).

The only source of the field is the linear charge on the endless thread:
[itex]\rho=\lambda\delta(\rho).[/itex]

Here [itex]\rho[/itex] is the volume charge density, [itex]\lambda[/itex] is a constant that describes the linear charge density.

1. If [itex]\rho<a[/itex], medium is homogeneus and anisotropic. Permittivity [itex]\widehat{\varepsilon}[/itex] is the given Hermitian matrix (3 x 3). All its entries are non-nil, some of them depend on the polar angle [itex]\varphi[/itex] so [itex]\widehat{\varepsilon}=\widehat{\varepsilon}( \varphi )[/itex].

From Gauss's flux theorem we obtain ([itex]\rho<a[/itex]):
div([itex]\widehat{\varepsilon}(\varphi)\nabla\psi[/itex])=-4[itex]\pi\rho[/itex].

This is the hyperbolic partial differential equation due to properties of [itex]\widehat{\varepsilon}[/itex].

2. If [itex]\rho\geq a[/itex], medium is homogeneus and isotropic. Permittivity [itex]\varepsilon=1[/itex], its a scalar.

From Gauss's flux theorem we obtain ([itex]\rho\geq a[/itex]):
div([itex]\nabla\psi[/itex])=-4[itex]\pi\rho[/itex]=0.

This is the elliptic partial differential equation.
________

So I have to solve these equations. Unfortunately, it's impossible to separate variables in the area [itex]\rho<a[/itex]. The only thing that may help is that nothing depends on z.

I have the boundary conditions:
[itex]\psi(a-0)=\psi(a+0)[/itex],
[itex]\widehat{\varepsilon}\frac{\partial\psi}{\partial \rho}(a-0)=\frac{\partial\psi}{\partial\rho}(a+0)[/itex],
[itex]\psi(\rho\rightarrow\infty)\rightarrow 0[/itex].

If somebody has any ideas, it will be great!
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