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Poisson's equation in 2 dimensions

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data
    If we lived in a world with 2 spatial dimensions, would the solution to Poisson's equation for a singular charge be the same i.e. would it be 1/r radial dependence?
    I am doing a computer simulation and if I take out the third dimension, the results are manifestly different.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2007 #2

    Dick

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    It would be manifestly different. Look at Gauss's law for flux integral vs source charge. In two dimensions, it's a log.
     
  4. Oct 14, 2007 #3
    Sorry. Could you explain that more.

    In 3 dimensions, Gauss's law says that the flux of E through a surface around a charge q is given by

    [tex] \int _{S} \vec{E} \cdot d\vec{a} = \frac{q}{\epsilon_0} [/tex]

    In two dimensions, it is just over a closed path instead of circle and you replace the da by a d(normal vector to the curve) or something.

    What are you saying is a log in two dimensions?
     
    Last edited: Oct 14, 2007
  5. Oct 14, 2007 #4

    Dick

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    The 1/r dependence in 3d is for the potential U. In your 2d example you can conclude that (take a circle for the curve) E*length of curve is proportional to charge. So E is proportional to 1/r. But E=-dU/dr. So U must be like log(r).
     
    Last edited: Oct 14, 2007
  6. Oct 14, 2007 #5
    But then the potential is infinity at infinity? I think maybe Gauss's law would be different in 2 dimensions.
     
    Last edited: Oct 14, 2007
  7. Oct 14, 2007 #6

    Dick

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    Yes. If you integrate a 1/r force from something to infinity, it diverges. Try one dimension as well. The force is a constant independent of distance. Potential is a linear function of r. Also infinite at infinity. Remind you of anything?
     
  8. Oct 14, 2007 #7
    Not really.

    But try inserting ln( sqrt(x^2 +y^2)) into the two dimensional poisson equation i.e. the gradient only has two components. I get 1/sqrt(x^2 + y^2) = rho/epsilon_0, which is not a solution.

    But I just tried the 3 dimensional solution U = 1/sqrt(x^2 +y^2 +z^2) and I get 0 = rho/epsilon_0 .

    So, what could be wrong? Now I am really confused. I am using a point charge at 0, so the RHS of Poisson's equation should be a delta function?
     
  9. Oct 14, 2007 #8

    Dick

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    You should get 0 in both cases. Your source is only at the origin so the laplacian is correctly giving you the charge density=0, at the origin it's singular. To test that your solution is a delta function you have to integrate it against a test function and use integration by parts.
     
  10. Oct 14, 2007 #9
    Aha, I see now.

    In my model, I was putting the charge in a square with a boundary condition that the sides of the square have a potential of 0. In 3 dimensions it made sense to make the sides of the box around the charge zero as an approximation of infinity, but in 2 dimensions that breaks down apparently. I am not really sure how you would make an accurate numerical model of the potential in two dimensions then since I can't have a boundary condition of infinity.

    Maybe I need to use Neumann boundary conditions or something.

    Thanks.
     
    Last edited: Oct 14, 2007
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