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Potential and electric field.

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Charge is uniformly distributed along a circular plate (radius R). surface charge is [tex]\sigma[/tex].
    a) Determine the electric potential on the plates axis distance x from the center of the plate.
    b) Calculate the electric field on the axis distant >x from the center of the plate.

    3. The attempt at a solution

    http://www.aijaa.com/img/b/00132/3756420.jpg [Broken]
    http://www.aijaa.com/img/b/00540/3756421.jpg [Broken]

    I couldn't get the E field any "cleaner" and in a I wasn't sore about the substitution of A (i.e. can I do it like that). fire my methods correct?

    And don't mind my handwriting, MS Paint isn't that responsive.

    Edit: Holy crap! Can I resize the images?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 8, 2009 #2

    tiny-tim

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    Hi Kruum! :smile:

    (have an integral: ∫ and a square-root: √ and a pi: π and an epsilon: ε and a sigma: σ :wink:)

    In a), yes that's the way to get dA (but haven't you lost the 2 in 2πr dr?);

    but your integral is wrong …

    you should be able to integrate that just by looking at it, but if you can't, try susbtituting u = √(x2 + r2) :wink:
     
  4. Mar 8, 2009 #3
    Thanks for having such fast fingers, tiny-tim! :biggrin:

    You can call me stupid, but isn't the area of a circle [tex] \pi r^2[/tex], hence my [tex] \pi rdr[/tex]. So wouldn't [tex]2 \pi rdr[/tex] be the volume of a sylinder?

    I can't believe it! I missed the minus... And the 2 should be 8.
     
  5. Mar 8, 2009 #4

    tiny-tim

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    Yes it is :smile:

    but you want the area of a ring, of thickness dr :wink:
    uhh? what minus? :confused:
     
  6. Mar 8, 2009 #5
    Of course! Thanks...

    Nevermind... I've been recapping too much, I can't think clearly anymore! :redface: I rememberd that s=√x2-r2.

    But that electric field equation cannot be simplified?

    Need... to... take... a... break...! :zzz:
     
  7. Mar 8, 2009 #6

    tiny-tim

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    Looks like s=√x2+r2 to me …

    anyway, try that substitution I mentioned :smile:
     
  8. Mar 8, 2009 #7
    Isn't it [tex] \frac{ \sigma }{2 \epsilon _0} \int_0^R \frac{r}{ \sqrt{x^2+r^2}}dr= \frac{ \sigma }{4 \epsilon _0} ln( \frac{ \sqrt{x^2+r^2}}{x})[/tex]
     
  9. Mar 8, 2009 #8

    tiny-tim

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    Nope!
     
  10. Mar 8, 2009 #9
    Okay, now I'm completely lost! Can you point out where the error is? I've been using the fact that [tex] \int \frac{f'}{f}=ln(f)+C[/tex]. There is an ln in the answer, right?
     
  11. Mar 8, 2009 #10
    I used the substitution method and got [tex] \frac{ \sigma }{2 \epsilon _0} \int_0^R \frac{r}{ \sqrt{x^2+r^2}}dr= \frac{ \sigma }{2 \epsilon _0} ln( \frac{ \sqrt{x^2+R^2}}{x})[/tex].
     
  12. Mar 8, 2009 #11

    tiny-tim

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    no, no ln …

    hmm … ∫2r dr/(x2 + r2) = ln(x2 + r2) …

    (and anyway ln(√(x2 + r2)/x) = (1/2)ln(x2 + r2) - lnx)

    what's your f? :confused:
     
  13. Mar 8, 2009 #12
    Ignore that. So there's no ln in the answer? I'm just too tired to think anymore...
     
  14. Mar 8, 2009 #13

    tiny-tim

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    get some sleep :zzz: …
     
  15. Mar 8, 2009 #14
    Looks like some food and a look at my maths book helped clear my mind... Now I get from the integral [tex] \frac{ \sigma}{2 \epsilon _0}( \sqrt{x^2+R^2}-x)[/tex]. So the E field would be [tex] \frac{ \sigma}{2 \epsilon _0x}( \sqrt{x^2+R^2}-x)[/tex]
     
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