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Boundary conditions of 2 conductors

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data


    Ignore the text in German. You just need to see the picture. 2 conductors both with potential 0 are given. [itex]\alpha[/itex] is the angle between the conductors. (r, [itex]\varphi[/itex]) are polar coordinates pointing to a point in the plane.

    2. Relevant equations

    What we need to do is solve the Laplace equation:

    Δ[itex]\phi[/itex] = 0

    for the boundary conditions which are implied by the picture.

    3. The attempt at a solution

    My attempt at a solution was to separate the variables in polar coordinates:

    [itex]\phi(r, \varphi) = R(r)\Psi(\varphi) [/itex]

    I think I did this correctly. It gives me the 2 ODEs:

    [itex]R(r) = c_1 e^{k \cdot ln(r)} + c_2 e^{-k \cdot ln(r)}[/itex]

    [itex]\Psi(\varphi) = c_3 sin(k \varphi) + c_4 cos(k \varphi)[/itex]

    Now my problem is that I can only see 2 boundary conditions in the above picture:

    [itex]\phi(r, 0) = 0[/itex] and [itex]\phi(r, \alpha) = 0[/itex]

    But this only gives me the constants [itex]c_4[/itex] and k. How do I get the other constants? Are there any more boundary conditions which I'm blind to see?

    Thanks in advance.
  2. jcsd
  3. Nov 17, 2013 #2


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    Should the potential blow up as ##r\rightarrow \infty##? What does this tell you about ##c_1##?

    Sweeping out ##\varphi## takes you from one conducting plate to the other and both plates are grounded. What does that tell you about ##c_4##?
  4. Nov 17, 2013 #3
    I already did your second question. It tells me [itex]c_4[/itex] = 0.

    The first one I didn't see however. The potential should disappear of course in infinity. This tells me [itex]c_1 = 0[/itex].

    This would still leave me with 2 unknown constants however. [itex]c_2[/itex] and [itex]c_3[/itex]

    What about r → 0? Can I say that the potential has to disappear there as well? Probably not otherwise the entire R(r) term would be 0.

    Hmmm ...
  5. Nov 17, 2013 #4
    The next question would be to approximate the potential for small r. So from that question alone I'd say that it should depend on r? lol
  6. Nov 17, 2013 #5


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  7. Nov 17, 2013 #6
    I found out that another constant can be determined by the equation Δ [itex](R(r) \Psi(\varphi)) = 0[/itex] itself. It gives [itex]c_3 = -1[/itex].

    One constant remains though. So now I have:

    [itex]\Phi(r, \varphi) = -C e^{-\frac{n \pi}{\alpha} \cdot ln(r)} sin(\frac{n \pi}{\alpha} \varphi)[/itex]

    Oh well ... Close enough I guess. :redface:

    Could you help me with the approximation for little r? I guess it's supposed to be a Taylor Series approximation but if I want to develop around [itex]x_0[/itex] = 0 for the very first summand [itex]f(x_0)[/itex] doesn't compute because of ln(0) ...
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