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Potential on an Infinite Strip

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A potential satisfies ##\nabla^{2}\Phi=0## in the 2d slab ##-\infty<x<\infty##, ##-b<y<b##, with boundary conditions ##\Phi(x,b)=V_{s}(x)## on the top and ##\Phi(x,-b)=-V_{s}(x)## on the bottom, where ##V_{s}(x)=-V_{0}## for ##-a<x<0##, and ##V_{s}(x)=V_{0}## for ##0<x<a##, and repeats periodically outside this window.

    2. Relevant equations


    3. The attempt at a solution
    So this is clearly a separation of variables problem, and I'm splitting into two parts - one which treats the bottom boundary condition, and one which treats the top boundary condition (while the opposite condition is set to zero). However, I feel really unsure of what I'm doing. So far I have the potential in relation to the bottom BC as ##\Phi=\sum_{n}A_{n}\sin\frac{n\pi x}{a}\sinh\left(\frac{n\pi y}{a}\right)##. I make no claim that this is entirely right. I suppose I'm not entirely sure how to set up the general function in the first place, or how to evaluate the coefficients while taking into account the periodic nature of the BC. Any nudges to get me started on the right foot?
     
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  3. Sep 18, 2015 #2

    Orodruin

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    Your proposed potential will work fine for the full problem because of its symmetries, but not for the split problem where you put one of the boundary conditions to zero. Can you see why?

    Once you have argued for that, you must fix the constants ##A_n##, what information do you have available for that?
     
  4. Sep 18, 2015 #3
    I think I see why it'd work for the full problem. The sine term is going to give me something that falls on both the upper and the lower boundary, i.e. something like a square wave, right?

    As far as fixing the constants is concerned, I know the condition that has to be met (i.e., that ##\Phi(x,-b)=-V_{s}## and ##\Phi(x,b)=V_{s}##. So it'd be a matter of doing an integral of the form ##\int V_{s}\sin\frac{n\pi x}{a}##, right? But the region of integration would have to be split into intervals depending on n to accommodate the changing sign of the constant as it varies from -a to 0 and from 0 to a, etc.
     
  5. Sep 19, 2015 #4

    Orodruin

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    You only need to do the integration on one segment of the boundary since you have done a sine expansion.
     
  6. Sep 19, 2015 #5
    Ah, I see, the sine already incorporates the periodicity. Let's see, calculating the coefficients then, we should have
    ##\int_{-a}^{a}V_{0}\sin\frac{n\pi x}{a}\,\mathrm{d}x=\int_{-a}^{a}\sum_{n}A_{n}\sin\frac{n\pi x}{a}\sinh\frac{n\pi b}{a}\sin\frac{n'\pi x}{a}\,\mathrm{d}x##,
    which, if I did my integrals correctly, should give ##A_{n}=\frac{4aV_{0}}{n\pi}\frac{1}{\sinh(n\pi b/a)}##. And with the coefficients found, that should complete the potential expansion.
     
  7. Sep 19, 2015 #6

    Orodruin

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    Yes, with the additional comment that it is enough to consider ##0 < x < a## as the sines are anti-symmetric just as the boundary condition. The addition of the range ##-a < x < 0## only doubles both integrals.
     
  8. Sep 19, 2015 #7
    Ahhh, I see that now. Thank you so much for your help!
     
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