1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential outside a charged disc

  1. Mar 12, 2007 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Laplace's equation

    3. 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:
    Last edited: Mar 12, 2007
  2. jcsd
  3. Mar 12, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    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: Mar 12, 2007
  4. Mar 12, 2007 #3
    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:
  5. Mar 12, 2007 #4

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
  6. Mar 12, 2007 #5
    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 :(
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?