# Potential by Green's function

1. Jan 23, 2009

### gop

1. The problem statement, all variables and given/known data

Consider a potential problem in the half space z>=0 with Dirichlet boundary conditions on the plane z=0.
If the potential on the plane z=0 is specified to be V inside a circle of radius a centered at the origin, and Phi=0 outside that circle, show that along the axis of the circle (rho=0) the potential is given by

$$\Phi=V(1-\frac{z}{\sqrt{a^2+z^2}})$$

2. Relevant equations

3. The attempt at a solution

I used the Green's function (by method of images)

$$G(x,x')=\frac{1}{|x-x'|}-\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}$$

Then I converted the formula to cylindrical coordinates and computed the normal derivative -dG/dz. Then I set rho=0 and finally I got

$$\Phi = \frac{-z}{2\pi} \int_0^{2\pi} \int_0^a \frac{V}{(p'^2+z^2)^{3/2}} = \frac{-Va}{z \sqrt{a^2+z^2}}$$

I'm somewhat certain that the computations are correct I checked them two times in a CAS so I don't really know what I did wrong.

2. Feb 17, 2010

### Jeansua

This thread has been around for a while, but hopefully this reply will help future users. Your only mistake was forgetting that in polar coordinates, the area element is \rho d\rho d\phi as opposed to just d\rho d\phi. With the extra \rho up top, you get the right answer (and the integral becomes easier, to boot!).