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A problem with an integral

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data
    While trying to find the electric field at a distance z from the center of a spherical surface that carries a uniform surface charge ##\sigma## I got stuck with the following integral (which I'm quite sure is correct):
    [tex]\int_0^\pi \frac{(z-Rcos\theta)sin\theta d\theta}{(R^{2}+z^{2}-2Rzcos\theta)^{3/2}}[/tex]

    2. Relevant equations



    3. The attempt at a solution
    The only idea I had is to let ##u=cos\theta ; du=-sin\theta d\theta##,then the integral becomes:
    [tex]\int_0^\pi \frac{(Ru-z)du}{(R^{2}+z^{2}-2Rzu)^{3/2}}[/tex]
    and now I'm stuck. This looks very similar to a partial fractions problem except there is a square root in the denominator and if I understand correctly I'm not allowed to use partial fractions in this case.

    Any help will be appreciated!
     
  2. jcsd
  3. Aug 11, 2014 #2
    You want to integrate with respect to which variable?
     
  4. Aug 11, 2014 #3
    preferably ##\theta## but if my substitution of ##u## is of any good than ##u## will do also.
     
  5. Aug 11, 2014 #4

    Simon Bridge

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    That looks sort-of familiar ... if you're sure, I'll start from there.

    Remember that z does not depend on u in your last integral, so it has form $$\int \frac{ax-b}{(c-dx)^{3/2}}\;\text dx$$
    ... your main problem is to get rid of the square-root in the denominator.
    How would you normally do that?
     
    Last edited: Aug 11, 2014
  6. Aug 11, 2014 #5
    substitute [tex]x=\frac{c}{d}sinu[/tex]
    EDIT: no, that's not correct, no 2nd power... perhaps multiply and divide by the square root. Am I allowed to use partial fractions with a square root in the numerator?
     
    Last edited: Aug 11, 2014
  7. Aug 11, 2014 #6

    Simon Bridge

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    You do want to get rid of the root before attempting partial fractions.
    How about ##u^2=c-dx## or ##dx=c\sin^2\phi##

    Aside: I probably should not have used "d" as a constant :(
     
  8. Aug 11, 2014 #7
    Alright, substituting ##v^2=c-gx## (replacing ##d## with ##g##) I am finally able to integrate and simplify to get the desired result:
    [tex]\int\frac{\frac{ac-av^{2}}{g}-b}{v^{3}}
    \frac{-2v}{g}
    dv = \int\frac{2a}{g^{2}}dv
    - \int\frac{2ac}{g^{2}v^{2}}dv
    + \int\frac{2b}{gv^{2}}dv
    =\frac{2av}{g^{2}}
    + \frac{2ac}{g^{2}v}
    - \frac{2b}{gv}
    = \frac{1}{z^{2}}\left(\frac{R+z}{|R+z|}-\frac{R-z}{|R-z|}\right)[/tex]

    Thank you very much Simon!
     
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