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Neumann's problem, electrostatics, proof

  1. Mar 2, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    In the Neumann's problem ##\begin{cases} \triangle \Phi =-4\pi \rho \\ \frac{\partial \Phi }{\partial n}=g \end{cases}##, what condition must g satisfy?


    2. Relevant equations
    I'm given the answer, it's ##\int _ \Omega \rho dV=-\int _{\partial \Omega } g dS##.


    3. The attempt at a solution
    I'm trying to demonstrate that the given answer is indeed true.
    So let Omega be the region of interest (in ##\mathbb{R} ^3##, mathematically) and ##\partial \Omega## be its contour.
    Now ##\int _\Omega \rho dV=Q##, the total charge enclosed in Omega. So I'm left to show that ##\int _{\partial \Omega } g dS## is worth minus the total charge enclosed in Omega. To me it just looks like the Gauss theorem. If ##g=\frac{\partial \Phi }{\partial n }=|\vec E _{\text{normal to the surface}}|##, then I must show that ##\int _{\partial \Omega } g dS=-Q##.
    So I replace g by its value, it gives me ##\int _{\partial \Omega } |\vec E _{\text{normal}}| dS=-Q##.
    If I use Gauss theorem, I get that ##\int _\Omega \rho dV= \int _{\partial \Omega } \vec E \cdot \hat n dS##. (where I used one of Maxwell's equation, ##\nabla \cdot \vec E= \rho##, I omitted a constant)
    So in order to get what I should, ##|\vec E _{\text{normal }}|## must equal ##\vec E \cdot \hat n##. If the normal points inward Omega then I get what I should. However if the normal points outward Omega (as it generally does), then I get that ##\int _{\partial \Omega } gdS=Q## which has a wrong sign.

    Do you see any mistake/error(s) in what I've done?
     
  2. jcsd
  3. Mar 2, 2013 #2

    TSny

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    I believe your missing negative sign comes from neglecting the negative sign in ##\vec{E} = -\vec{\nabla}\Phi##. Thus, ##\partial{\Phi}/\partial{n} = -E_n##.
     
  4. Mar 2, 2013 #3

    fluidistic

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    I see, thank you very much TSny!
     
  5. Mar 2, 2013 #4

    fluidistic

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    I realize that the problem was either "what does g must satisfy for the solution to exist?" or "what does g must satisfy for the solution to be unique?".
    And the answer given is ##\int _ \Omega \rho dV=-\int _{\partial \Omega } g dS## that I showed to be true, but I don't see why this imply that it guaranties that Phi (the potential) either exist (and satisfies Neumann's problem) or is unique (and satisfies Neumann's problem).
    Edit: I'm guessing it has to do with if g does not satisfy the relation given as answer, then Phi does not satisfies Poisson's equation.
    In this case the original question was probably "what does g must satisfy for the solution to exist?", though I'm not 100% sure.
     
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