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Griffiths 4.10

  1. Mar 15, 2008 #1
    [SOLVED] Griffiths 4.10

    1. The problem statement, all variables and given/known data
    This question refers to Griffths E and M book.


    2. Relevant equations



    3. The attempt at a solution
    part a was easy.
    For part b, I am trying to calculate V(r) using equation 4.13 and I am facing an absurdly difficult integration which means I must be doing something wrong.
    [tex]V(\vec{r}) = \frac{1}{4 \pi \epsilon_0}\left(\int_0^{2\pi} \int_p^{\pi}\frac{\sigma_b}{|\vec{r}-\vec{r'}|}R^2 \sin \theta' d\theta' d\phi' + \int_0^{2\pi} \int_0^{\pi} \int_0^R \frac{\rho_b}{|\vec{r}-\vec{r'}|}r'^2 \sin \theta' dr' d\theta' d\phi' \right)
    [/tex]
    It is those denominators that are going to kill me!
     
    Last edited: Mar 15, 2008
  2. jcsd
  3. Mar 15, 2008 #2

    kdv

    User Avatar

    Have you tried using eq 3.94?
    The sigma_b and rho_b are not complicated right? So it should not be too hard.
     
  4. Mar 15, 2008 #3
    I don't see how inserting an infinite series into the integral will make it simpler. I don't see any way to make 3.94 not an infinite series in this case.

    Yes sigma_b and rho_b are not complicated.
     
  5. Mar 15, 2008 #4

    kdv

    User Avatar

    The point of using the infinite series is that you can then use the orthonormality of the polynomials to do all the angular integrations trivially! Have you used 3.94 in an actual calculation? That's very powerful. What are sigma_b and rho_b?
     
  6. Mar 16, 2008 #5
    [tex]\rho_b = -2k/r[/tex]
    and
    [tex]\sigma_b = Rk[/tex]

    In equation 3.94 [itex]\theta'[/itex] is the angle between r and r', not the zenith angle.
     
  7. Mar 16, 2008 #6
    Never mind. I got it. kdv, there is a much easier way to do it than to use eqn 3.94. BTW the bound charges I gave are wrong.
     
    Last edited: Mar 16, 2008
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