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E-field of a Semi-spherical charge distribution with a hole

  1. Apr 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Using direct integration, compute the electric field at the point (0,0,Z) of a semi-espherical (z > 0) charge distribution of radius R < Z and density ##\rho=constant## having a spherical hole of radius ##r_h<R## centered at (0,0,##r_h##).

    2. Relevant equations

    Coulomb's law in differential form:
    ## d\vec{E} = \frac{Kdq}{|\vec{r} - \vec{r}_{dq}|^2}\frac{\vec{r} - \vec{r}_{dq}}{|\vec{r} - \vec{r}_{dq}|}##

    ##\vec{r} = Z \hat{k} ## (the field point)

    ##\vec{r}_{dq} = r\sin(\theta)\cos(\phi)\hat{\imath} + r\sin(\theta)\sin(\phi)\hat{\jmath} + r\cos(\theta)\hat{k}## (the source point)

    ## r = 2r_h\cos(\theta) ## (the spherical hole in spherical coordinates )

    3. The attempt at a solution

    By symmetry, the only relevant integral is:

    ## \vec{E}(z) = 2\pi K \rho \int_{\theta=0}^{\theta=\pi/2} d\theta \int_{r=2 r_h \cos{\theta}}^{r=R} dr \left[ \frac{ Z-r\cos(\theta) }{ (r^2 - 2 r Z \cos(\theta) + Z^2)^{3/2} } \right]r^2 \sin(\theta) ##

    The question is how to perform the integral. Using brute force via the wolfram alpha site
    finished with Standard computation time exceeded...

    https://www.wolframalpha.com/input/...,+{r,+2*rh*Cos[theta],+R}],+{theta,+0,+Pi/2}]

    Thanks in advance for your help,

    Sergio
    PS. By the way, the formulation seems to be right as one can get the right ##\vec{E}## field of the full semi-sphere
     
  2. jcsd
  3. Apr 3, 2016 #2

    BvU

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    I wonder about the lower limit of ##dr##. Why not simply ##r_h## ?
     
  4. Apr 3, 2016 #3
    It is not a fix lower limit. It is a spherical hole.
     
  5. Apr 4, 2016 #4

    BvU

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    But surely all points on the surface of a spherical hole that is centered at ##(0,0,r_h) ## are at a distance ##r_h## away from the origin ?
     
  6. Apr 4, 2016 #5
    Yes, but the charge source vector is not measure respect the hole. The picture attached "might" help to make clear the narrative of the problem.

    I found the solution via power expansion. Nevertheless, I am still curios on why direct integration does not seem to work here (there is a divergence in the result due to a logarithm term, that in spite the integrand behaves well in the region of integration).
    This might be the reason why textbooks does not have this neither as a solved problem nor as proposed one. It is a very rich problem to ilustrate vectors, integration that does not seems to work (or are very difficult) and solution via power series.

    Perhaps changing the origin to be at the center of the hole could make the integration less hard, but the asymmetry it generates does not seem to support that.





    Sergio charge_distribution.png
     
  7. Apr 4, 2016 #6

    BvU

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    O, I see o:)o:)o:), pretty dumb of me, sorry. I take it the picture came with the exercise ?
    By now I understand relevant equations 1,2,3 And I think I understand ##
    r = 2r_h\cos(\theta)## as the lower bound of ##r##.
    I know it says 'by direct integration', but does that exclude an integration over a charged half-sphere (which you apparently have already) minus an integration over the ##r_h## sphere with opposite charge density (which indeed is easier with the origin at ##(0,0,r_h)## ?
     
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