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Homework Help: Potential of a Charged Disk

  1. Feb 3, 2009 #1
    I've attached a .pdf with the problem explanation, questions, and answers. I can't figure out for the life of me how to get these answers though. I've setup a countless number of integrals and can't seem to get it right. Any help you guys can give me would be greatly appreciated.

    Not sure if you guys can see the .pdf I posted since its saying its still pending approval. So you can view it here: http://evergreenwebdesigns.com/playground/HW/Problem.pdf [Broken]

    Attached Files:

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 3, 2009 #2
    Show one of your attempts at the problem. This provides a starting point for help.
  4. Feb 3, 2009 #3
    Chrisk, I've attached my attempt at this problem. Hope you don't mind the .pdfs it's easier for me to type the equations in word rather than on the physics forums. Thanks for your any help you can give.

    Attached Files:

  5. Feb 4, 2009 #4
    Thank you for posting the attempt at the solution. Check your expression for dq. You are on the right path of taking a ratio.
  6. Feb 4, 2009 #5
    I checked the pdf but cant see any attempts for solution.

    It does not contains a complicated physics , a little bit mathematics.

    I can give a starting point;

    *Take a ifinitesmall charge on surface of the disk [tex] dq= \lambda ds [/tex]

    [tex]\lambda[/tex]: surface charge density
    [tex]ds[/tex]: surface element (in cylindrical coordinates)

    **Find the potential of that infinitesmall charge at point z.
    ***Than integrate it for all surface.
  7. Feb 4, 2009 #6
    You need to assign a surface charge density, then divide the disk into infinitesimal elements. Because of azimuthal symmetry, you can divide the disk into infinitesimal rings. The potential contribution from each ring should be evaluated, then integrated over the radius of the disk.

    You could also use solve the problem at an arbitrary point using legendre polynomials, then set the polar angle equal to zero.
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