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Find function with given boundary conditions

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data
    I have an infinite plate of which two electrodes are attached at a distance ##2a## and the electric potential between them is ##U##.
    Now I have to find a function ##\phi (x,y)## that satisfies Laplace's equation ##\nabla ^2 \phi =0## and is equal to ##0## at all possible infinities but at the same time equal to ##\pm \frac U 2## at (or at least very close to) the electrodes.


    2. Relevant equations



    3. The attempt at a solution
    This problem was once already on this forums but he couldn't find and ending. The idea then was to find the function that satisfies all given conditions, but since we never studied Green's functions I assume I don't have to do that, therefore I am now (and also since physicists do that) trying to simply guess the right function.

    Green's Function in 2D is something like ##G(x,y)=\frac{1}{4\pi}ln(x^2+y^2)##, but in my case having two electrodes at points ##(\pm a,0)## and knowing that my function has to be ##0## in infinity, it HAS to look similar to this:

    ##G(x,y)=G_1(x,y)-G_2(x,y)=\frac{1}{4\pi}ln((x-a)^2+y^2)-\frac{1}{4\pi}ln((x+a)^2+y^2)=\frac{1}{4\pi}ln(\frac{(x-a)^2+y^2}{(x+a)^2+y^2})##

    Capture.PNG

    This is almost great, because ##G(\pm \infty, \pm \infty)=0## but sadly ##G(\pm a, 0)## will NEVER be ##\frac U 2## because in point ##(\pm a, 0)## my function has a pole and therefore goes towards ##\pm \infty##.

    So my idea to approximate my electrodes as point charges may not be the best one, therefore I defined a radius ##R## of the electrodes to move away from the pole. This helps a bit, but maybe not enough. My new function therefore looks something like:

    ##\left\{\begin{matrix}
    \frac U 2, (x,y)\in K((a,0),R)\\
    -\frac U 2,(x,y)\in K((-a,0),R)\\
    \frac{1}{4\pi}ln(\frac{(x-a)^2+y^2}{(x+a)^2+y^2}),\text{For all other (x,y)}
    \end{matrix}\right.##

    with boundary conditions ##G(\pm \infty, \pm \infty)=0## but also ##G((x,y)\in K((a,0),R))=\frac U 2## and ##G((x,y)\in K((-a,0),R)=-\frac U 2##. Well, this is to me a bit hard to solve (actually, I have no idea how to do it), so I tried to say that ##R<<a## or even better ##R<<1## which can approximate my boundary conditions to ##G(a-R,0)=\frac U 2## and ##G(-a+R,0)=-\frac U 2##.

    However, this limits me to only certain radius of electrodes. Therefore I seriously doubt my method is ok. Because from conditon ##G(a-R,0)=\frac{1}{2\pi}ln(\frac{R}{R-2a})=\frac U 2## I get an equation for radius ##R##:
    ##R(a,U)=2a\frac{1}{1-exp(-\pi U)}##

    Speaking as a really stupid person: This confuses me. Why would the radius of my electrodes have a minimum value? My electrodes are as big as they are. Why is my radius a function of ##a## and ##U##? I would really like to get a second opinion about that.

    Also, I am sure this is wrong because my radius rapidly approaches to a point where ##R<<a## is no longer true!
    Capture1.PNG
    (Red dashed lines are ##2a##, the ##x## axis is voltage ##U## and the blue line is radius ##R##. Picture is therefore for three different values of ##a##)
    Capture2.PNG
    Also 3D

    So my problems are:
    1. Is this wrong? If yes, why?
    2. How do I proceed?

    In case nobody understands my crazy English here...: My problem is to guess a function that satisfies Laplace's equation, is ##0## at all possible infinities and has a value of ##\pm \frac U 2## at electrodes.
    I assume is has to look something like ##\frac{1}{4\pi}ln(\frac{(x-a)^2+y^2}{(x+a)^2+y^2})## but this function has to be somehow manipulated in order to be ##\pm \frac U 2## in point ##(\pm a,0)##. My question: How?
     
  2. jcsd
  3. Aug 31, 2014 #2
    Nevermind, the solution is really simple. :)

    You just have to put another constant in front of the logarithm, than everything works perfectly.
    The constant is than a function of ##a##, ##U## and ##R##. Which makes sense!

    Thanks, for all the help! :D
     
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