# Cylindrical potential problem using Bessel functions

1. Oct 10, 2009

### Pengwuino

Jackson 3.12: An infinite, thin, plane sheet of conducting material has a circular hole of radius a cut in it. A thin, flat disc of the same material and slightly smaller radius lies in the plane, filling the hole, but separted from the sheet by a very narrow insulating ring. The disc is maintained at a fixed potential V, while the infinite hseet is kept at zero potential.

a) Using appropriate cylindrical coordinates, find an intergral expression involving Bessel functions for the potential at any point above the plane.

Attempt!

So considering there is no$$\phi$$ dependence, I know the potential $$\Phi = Z(z)R(\rho)\Phi(\phi)$$ reduces down to 2 equations: $$Z(z) = Asinh(kz) + Bcosh(kz)$$ and $$R(\rho) = CJ(k\rho) + DN(k\rho)$$.

By the fact that V is finite at $$\rho = 0$$, D = 0.

So the solution looks like

$$\sum\limits_{m = 0}^\infty {\sum\limits_{n = 1}^\infty {(A_{mn} \sinh (k_{mn} z) + B_{mn} (\cosh (k_{mn} z))J_m (k_{mn} \rho )} }$$ where $$k_{mn} = \frac{x_{mn}}{a}$$ where $$x_{mn}$$ is the n-th root for the m-th Bessel function.

At z=0, I have

$$V_0 = \sum\limits_{m=0}^\infty{\sum\limits_{n=1}^\infty{B_{mn}J_m(k_{mn}\rho)}$$. At this point, i integrate both sides by $$\int_0^a {\rho J_m (k_{mn'} \rho )d\rho }$$ and get:

$$\begin{array}{l} \int_0^a {V_0 J_m (k_{mn'} \rho )\rho d\rho = } \int\limits_0^a {\sum\limits_{m = 0}^\infty {\sum\limits_{n = 1}^\infty {B_{mn} J_m (k_{mn} \rho )J_m (k_{mn'} \rho )\rho d\rho } } } \\ \int_0^a {V_0 J_m (k_{mn'} \rho )\rho d\rho = } \frac{1}{2}\sum\limits_{m = 0}^\infty {\sum\limits_{n = 1}^\infty {B_{mn} |J_{m + 1} (k_{mn} )|^2 \delta _{nn'} } } \\ \end{array}$$

Now I'm quite new to Bessel functions so at this point, I'm not sure what to do. It seems like I still have one series on the right and a number on the left. I ran the left side through mathematica and it said it was simply $$V_0$$ but I don't know enough about Bessel functions to tell if that's right or not. Any tips or problems to be pointed out would be greatly appreciated :)

Last edited: Oct 10, 2009
2. Oct 10, 2009

### drizzle

Last edited by a moderator: Apr 24, 2017
3. Oct 10, 2009

### Pengwuino

Yah that doesn't really help at all. He just immediately goes to the integral form and drops a summation which considering how the general solution is formed for these types of problems, doesn't make too much sense.