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Potential Difference of Inverted Hemispherical Shell

  1. Sep 10, 2015 #1
    The problems states: An inverted hemispherical bowl of radius R carries a uniform surface charge density σ. Find the potential difference between the north pole and the center.

    I was able to do the problem and got the correct answer the book gives, which is (Rσ/2ε0)(√2 - 1). My professor, however asks that we do an assessment, or check, of our answer for every problem. This is where I am having troubles. I figured I could use Poisson's Equation to show that the Laplacian of the answer is -σ/ε0, but I can't seem to get this. Any other ideas?
     
  2. jcsd
  3. Sep 15, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Sep 18, 2015 #3

    rude man

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    Well, sice no one else has responded in a week ..for what it's worth, probably not much ...

    Poisson's equation is ∇2V = -ρ/ε with ρ = volume, not surface, density, so your units would be incorrect.

    Also, the entire volume inside the hemisphere has ρ = 0. But I don't see that solving for ∇2V = 0 would be pleasant, since V varies not only along the z axis (assuming the hemisphere axis is the z axis and the bottom rests symmetrically on the x-y plane), but also with x and y (or similar problem if you went cylindrical or spherical.) The only alternative that comes to mind is solving for the E field directly and integrating from the pole to the bottom center.
     
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