Potential outside a charged disc

  • Thread starter Azael
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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 doesnt 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 dont 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?:confused:
 
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Answers and Replies

  • #2
dextercioby
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Azael;1270416 So if we work along the r plane and put z=0 we have the equation said:
\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}k(\sqrt{z^2+a^2}-|z|) [/tex]

How do I transform this to solve c?:confused:
First, it should be something like

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

Then, is "k" a Bessel function ? I'd say it looks that this second equation should be asking for a reverse Laplace transformation...

EDIT:Apparently the tex code is causing trouble. Just click the code to see the minor correction (a minus sign in the exp)
 
Last edited:
  • #3
253
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First, it should be something like

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

Then, is "k" a Bessel function ? I'd say it looks that this second equation should be asking for a reverse Laplace transformation...

EDIT:Apparently the tex code is causing trouble. Just click the code to see the minor correction (a minus sign in the exp)
Ops, the k was just a typo and it shouldnt be there at all:blushing: Yoru right its a minus in the exp aswell. Edited my first post so its correct.

Is there any other way beside laplace transform to find C? Laplace transforms isnt even part of this class and hasnt been part of any previous class so its odd if we would be given a problem that require them.

In arfken there is a hint that the problem can be solved by fourier transform but I havent been able to come up with a way.:confused:
 
  • #4
Meir Achuz
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I think the problem is easier in spherical coords.
Your result for the z-axis potential can be exanded to all angles using Legendre polynomials.
 
  • #5
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Il have to give spherical coordinates a try tomorrow! Thanks for that tip didnt think about that.

But I hope I can figure it out in cylindrical coordinates aswell. I hate beeing stuck :(
 

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