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## Homework Statement

Find the potential outside a thin, circular disc with charge q.

## Homework Equations

Laplace's equation

DivgradU=0

## The Attempt at a Solution

Since the problem implies symmetry around phi its obvious the solution doesn't depend on phi.

I separate the differential equation into U=A(r)B(z)

Then its easy to find the solutions

[tex]B(z)=c(\lambda)e^{-\lambda z}[/tex]

[tex]A(r)=J_0(\lambda r)[/tex]

where J_0 is the 0'th bessel function.

So I have

[tex]U(r,z)=c(\lambda)e^{-\lambda z}J_0(\lambda r)[/tex]

The general solution then becomes

[tex]\int_{-\infty}^{\infty}C(\lambda)e^{-\lambda z}J_0(\lambda r) d\lambda[/tex]

The problem is that I don't have the slighest clue on how to find C. I know that far away from the disc the potential should tend toward the potential outside a point charge.

So if we work along the r plane and put z=0 we have the equation, when r is large

[tex]\int_{-\infty}^{\infty}C(\lambda)J_0(\lambda r) d\lambda= \frac{q}{4\pi\epsilon r} [/tex]

If I instead work on the z axis I know that the potential is given by.

[tex]\int_{-\infty}^{\infty}C(\lambda)e^{-\lambda z} d\lambda= \frac{q}{2\pi\epsilon a^2}(\sqrt{z^2+a^2}-|z|) [/tex]

How do I transform this to solve c?

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