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Homework Help: Finding the potential for a charged disk on a test charge

  1. Apr 6, 2006 #1
    this is my problem and it has a twist:

    find the electric field of a uniform disk of charge (assuming it's lying on the x, y plane) on a test charge located at the center of the disk on the z-axis but wait... here is the twist: displace the test charge a small displacement to the left. :eek:

    Well, it's almost like your regular symmetric disk of charged, but this one is not symmetric.

    I have no clue where to start. My teacher gives this as a critical and challenging problem for extra credit.

    Please, give me some help to get me started.

  2. jcsd
  3. Apr 6, 2006 #2


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    It is is in principle exactly solvable. I would at least write down the integral expression. Then, since you know the displacement is very small, you can make an approximation to first order in the displacement to simplify the integral.

    EDIT: Better yet. Use that dV=(dV/dr)dr
    Last edited: Apr 6, 2006
  4. Apr 6, 2006 #3
    principle of exactly solvable?

    can you explain me what this is or where I can find information on it? The internet doesn't help so far...

  5. Apr 7, 2006 #4


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    Start by writing down the expression (integral) for the potential in cylindrical coordinates. (This integral is solvable exactly)

    I was proposing a first order expansion in the displacement, but I noticed that's actually messier than the exact way.
    Last edited: Apr 7, 2006
  6. Apr 7, 2006 #5
    My work so far...

    http://i15.photobucket.com/albums/a365/jimykid86/extracredit.jpg [Broken]

    how do i find c?

    Last edited by a moderator: May 2, 2017
  7. Apr 8, 2006 #6
    My work so far 2...

    http://i15.photobucket.com/albums/a365/jimykid86/extracredit2001.jpg [Broken]

    what do I do now?

    Last edited by a moderator: May 2, 2017
  8. Apr 8, 2006 #7


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    You got the expression for the distance (almost) right, I`ll name it [itex]\rho[/itex]:

    [tex]\rho=\sqrt{(r\cos \phi-a)^2+r^2\sin^2 \phi+z^2}=\sqrt{r^2-2ra\cos \phi+z^2+a^2}[/tex]

    [tex]=\sqrt{(r-a\cos \phi)^2+a^2\sin^2\phi+z^2}[/tex]
    So the integral becomes:

    [tex]V=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\int_0^R\frac{\sigma rdrd\phi}{\sqrt{(r-a\cos \phi)^2+a^2\sin^2\phi+z^2}}[/tex]

    You can integrate this wrt r, but not wrt phi. Maybe a power series expansion will work.

    It's quite messy no matter how you look at it, but I guess you gotta do something for bonus points right? :biggrin:
    Last edited: Apr 8, 2006
  9. Apr 8, 2006 #8
    thank you Galileo

    haha... yeah... it will be a mess, but I'll try hard. thank you for your help

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