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Laplacian over a radial function for charge density

  1. Apr 26, 2012 #1
    ScreenShot2012-04-26at112622AM.png


    As you probably can see from the above shot, I'm determining charge density via the Laplacian over the potential (phi). I understand the mathematical steps, just confused on the factor of 4pi that pops up in the denominator. I think I understand why you would do that and here's my reasoning...

    You're taking a function that is radially outward and operating on it. Since we are assuming this is a spherically symmetric area we can neglect the aspect of the spherical coordinate system that would have symmetry and only focus on the radial change. Since again, you assume spherical symmetry the 4pi covers it. Would this be (semi) sound reasoning? Also, where did the permittivity constant go (epsilon naught)?

    Thanks.
     
  2. jcsd
  3. Apr 26, 2012 #2
    This is the (time-averaged) potential of a hydrogen atom. The 4pi comes from the potential itself, and then pops out of all the differentials because it is constant. Same as where the epsilon_0 comes from. By the way, the answer is actually wrong -- there should be a delta function in there too.
     
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