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Homework Help: Solve Poisson's Equation

  1. Jan 20, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\nabla[/tex][tex]^{2}[/tex]V=([tex]\rho[/tex])/([tex]\epsilon[/tex][tex]_{0}[/tex])


    2. Relevant equations

    -

    3. The attempt at a solution

    I have a function of x which i can supstitute into Charge density and boundary conditions, however my problem with this is very simple. How to i manipulate the equation to get a funvtion for V. Obviously i need to get rid of the div of the div of v and i assume this has to be done through some form of intergration?

    I assume this isn't as simple as intergrating both sides twice?

    I would appreciate any guidance that can be given on this.

    Regards.

    Leo
     
  2. jcsd
  3. Jan 20, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    In general, differential equations in physics are hard to solve. From what you posted, I understand that the function [itex]\rho = \rho(x)[/itex] is given, and you are working in one dimension only.

    In that case,
    [tex]\nabla^2 V = \frac{d^2 V}{dx^2}[/tex]
    so you can solve your equation (I'm wiping the constant into rho, for notational convenience)
    [tex]\frac{d^2 V}{dx^2} = \rho(x)[/tex]
    by integrating twice.

    For example, when [itex]\rho(x) = \epsilon_0 x^2[/itex] you would simply get
    [tex]V''(x) = x^2[/tex]
    so
    [tex]V'(x) = \frac13 x^3 + c[/tex]
    and
    [tex]V(x) = \frac{1}{12} x^4 + c x + k[/tex]

    The only snag might be that the integration is very hard to do: just pick a nasty function like [tex]\rho / \epsilon_0 = \sqrt{1 + x e^x}[/tex] and your scr*wed :)

    Note that once you go to two dimensions, things already get far less trivial, you would have to solve
    [tex]\frac{\partial^2V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = \rho(x, y)[/tex]
    which takes a little more than two integrations (usually, you already need things like symmetry and polar coordinates here to make anything of it).
     
  4. Jan 20, 2010 #3
    That makes makes perfect sense! Thank you for your help!
     
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