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Homework Help: Laplace equation for parallel plate condersers

  1. Apr 14, 2007 #1
    I've recently started studying Laplace's equation and it's solution under various simple circumstances in electrostatics. I tried to solve the equation for a parallel plate condenser system, but I couldn't meet the boundary conditions. I had two plates, one placed on xz plane at y=0 (with potential = 0), second parallel to it, at y=d (with potential [itex]V_0[/itex]). I placed them such that they're symmetrical in x and z, i.e., y axis crosses midpoints of plates; therefore the potential should be an even function of x and z. Noting that [itex]V(0,0,0) = 0[/itex] I wrote the solution:

    [tex]A\cosh(kx) \cosh(lz) \sinh(my)[/tex]
    with [tex]k^2 + l^2 + m^2 = 0[/tex] and let [itex]A[/itex] be any complex number.

    I assumed that potential should drop to zero when [itex]x,z \to \pm \infty[/itex], and this's the boundary condition that doesn't meet with my "solution".

    Can anyone help me working out the solution, or forward me to some resource on it?
    Thanks!
     
    Last edited: Apr 14, 2007
  2. jcsd
  3. Apr 15, 2007 #2
    Your formula cannot fit the potential. Of course, k and l are zero (because of symmetry the potential does not depends on x or z).
    But the potential in this problem is a linear function of y. There is no way to fit a sinh into a straight line.
    Then your assumption about the mathematical form of the solution is wrong.
    Let's start with Laplace's equation:

    [TEX]{\partial^2 \varphi\over \partial x^2 } +
    {\partial^2 \varphi\over \partial y^2 } +
    {\partial^2 \varphi\over \partial z^2 } = 0. [/TEX]

    The first and third terms are zero. Then:
    [TEX]{\partial^2 \varphi\over \partial y^2 }= 0. [/TEX]

    Then: [TEX]\varphi = ay+b[/TEX]
     
  4. Apr 15, 2007 #3
    This can't be true because plates are not infinite, and field lines are no longer straight lines when we approach to the edges:

    http://www.regentsprep.org/Regents/physics/phys03/aparplate/plate2.gif

    And how do we say potential itself does not depend on x either z? Apparently they do --even though the assumption that field lines were straight, they should vanish in the outside region between plates, which is defined by x and y.

    I remember this problem is not easily solved (possibly from Feynman lectures) :grumpy:
     
    Last edited by a moderator: Apr 22, 2017
  5. Apr 15, 2007 #4
    Well, if you had said that the plates where finite. I wouldn't have bothered to answer.
    This problem, as Feynman said is not soluble analytically.
     
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