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Potential between two parallel planes

  1. Jul 18, 2011 #1
    Hello ! I have an exam this Wednesday... your help will be appreciated...

    There are two types of questions I can't figure out how to answer...
    the first is this one:

    Find the potential between two parallel planes. The first is in x=0, the second in x=L.
    [itex]\Phi(x=L)=\Phi_0 |sin(ky)|[/itex]
    [itex]\Phi (x=0)=0[/itex]

    What I thought I should do is look for a solution of the sort:


    Then, I should start checking which of the terms should vanish.

    But when I looked at the published solution, it was completely different:

    [itex]|sin ky|=\sum_{n=0}^{\infty}{A_ncos(2kny)}

    [itex]\rightarrow |sin(ky)|=\frac{2}{\pi}+\sum{\frac{4}{\pi(1-4n^2)}}cos(2kny)[/itex]

    So now we seek a solution for Laplas eq. this way:

    [itex]\frac{d^2}{dx^2}C_n =(2kn)^2C_n[/itex]

    I don't understand this solution at all.
    Why should one expand this function, |sin(ky)| ?
    and what is the last equation:
    [itex]\frac{d^2}{dx^2}C_n =(2kn)^2C_n[/itex]

    Thank you so much!
  2. jcsd
  3. Jul 18, 2011 #2


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    That isn't the most general solution to Laplace's equation (which with your factorization reverts to: [itex] X^{''}/X + Y^{''}/Y + Z^{''}/Z = 0[/itex]:
    [itex]X'' = aX[/itex], [itex]Y''=bY[/itex], [itex]Z''=cZ[/itex], [itex] a+b+c=0[/itex].
    Your set of solutions doesn't cover the case where say [itex]a=0, b+c=0[/itex]. You can try to enumerate every combination of cases but it is better to start with your boundary conditions for some guidance. You see that [itex]Y(y)[/tex] is going to be non-trivial and start there.

    It looks like the published solution simply expresses the expansion of Y solutions using coefficients which are functions of the other independent variable(s) and re-applies the Laplace equation.
    Firstly since the boundary conditions and original equation are symmetric under z translation the solution is independent of z.

    The text is expanding solutions of [itex]Y^{''}=bY[/itex] in terms of the cosine expansion of the boundary condition (using the fact that it is an even function of y and independent of z). (One might also have tried a hyperbolic cosine expansion if the boundary condition where even and asymptotically exponential. Cosh is the even component of exp.)

    By the linearity of Lap.Eqn. Each term in the expansion [itex] \Phi(x,y) = C_0(x) + C_1(x)\cos(2ky) + \cdots[/itex], must independently satisfy Laplace's eqn.

    You get that last equation for all terms even the first term where n=0 which is the exceptional case I mentioned above.

    Matching your attempt:
    [itex]\Phi(x,y) = \sum_{some\,a,b} X_a(x)Y_b(y),\quad a+b=0[/tex]
    [itex]=\sum_n C_n(x)\cos(2kny)[/itex]
    with [itex] b = -(2kn)^2[/itex] and so [itex]a = (2kn)^2[/itex].
  4. Jul 19, 2011 #3
    Thank you for the detailed answer!
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