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Laplace's Equation and the potential above the xy-plane

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Essentially it gives the potential above the xy-plane as gif.gif and I am tasked with verifying it satisfies laplace's equation, determining the electric field, and describing the charge distribution on the plane.

    2. Relevant equations
    gif.gif then gif.gif
    gif.gif
    gif.gif

    3. The attempt at a solution
    As far as I can tell it does not satisfy laplace's equation because when I take the gradient of the potential twice I get gif.gif which is non-zero. So taking the second equation up there I found gif.gif and using the third I found gif.gif .

    In the case of the electric field I would assume the field in the x direction wouldn't matter (or at least shouldn't be there) because it is parallel to the field, but I'm not really sure if I'm doing this correctly.
     
  2. jcsd
  3. Sep 14, 2015 #2

    SteamKing

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    I think you are missing something here.

    The Laplacian operator does not return a vector; it returns a scalar. The function φ here is a scalar function.

    https://en.wikipedia.org/wiki/Laplace's_equation
     
  4. Sep 14, 2015 #3
    But when you take the gradient isn't it this?
    al%20y%7D%5Cphi%5Chat%7By%7D+%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cphi%5Chat%7Bz%7D.gif
    So taking the couble gradient would be:
    7D%20+%20%5Cfrac%7B%5Cpartial%5E%7B2%7D%7D%7B%5Cpartial%20z%5E%7B2%7D%7D%5Cphi%5Chat%7Bz%7D.gif
    Or is the laplacian not strictly the double gradient? If there's no directionality then it does equal zero and there for the charge distribution on the plane is 0 and the field is 0 right?
     
  5. Sep 14, 2015 #4

    SteamKing

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    The Laplacian is defined as the divergence of the gradient, or Δφ = div grad φ = ∇ ⋅ ∇φ = ∇2φ
     
  6. Sep 14, 2015 #5
     
  7. Sep 14, 2015 #6
    Sorry about that last reply. Okay I get it now, so 7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Cphi%2C%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cphi%3E.gif which is then equal to 5E%7B2%7D%7D%5Cphi%20+%5Cfrac%7B%5Cpartial%5E%7B2%7D%7D%7B%5Cpartial%20z%5E%7B2%7D%7D%5Cphi.gif . So would that mean the the charge distribution is 0?
     
  8. Sep 14, 2015 #7

    SteamKing

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    It would seem so.

    I know a little about the math behind potential theory, but I'm not au fait with electrostatics. If anyone else wants to jump in here ...
     
  9. Sep 14, 2015 #8
    Would that mean the charge distribution is zero?

    Yes.

    A good way to check this is to apply Gauss' law to the electric field you calculated. It is extremely simple to show that the divergence of E goes to zero in this case, so that automatically implies that net charge is zero.

    You stated in your initial post that the field above the xy-plane is given. Gauss' law allows us to create a Gaussian surface over the area of this region, so showing that ∇⋅E goes to zero shows us that there is no charge enclosed in the surface. This simply implies that the electric field (and thus the potential) is due to some charge distribution outside of the specified region or is simply an external field that is applied.

    Gauss' law can be tricky like this sometimes, but it is helpful to think of the integral form, noting that ∫E⋅dA = QEnclosed0. This integral is around a surface (the Gaussian surface) so when you use the law to calculate the total charge, you are finding the total charge enclosed in the region. So what you have shown (correctly) is that the total charge distribution in this region is zero.

    Note: If any of this isn't explained well, others feel free to jump in and correct me or elaborate.[/QUOTE]
     
  10. Sep 14, 2015 #9
    [/QUOTE]
    Thank y'all so much! Just needed to make sure my reasoning was sound :)
     
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