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Potential near a conducting strip (Conformal maps)

  1. Aug 11, 2017 #1

    1. The problem statement, all variables and given/known data

    We have an infinitely long wire with charge density ##\lambda## located at ##x=0,y=h##. We also have a semi-infinite strip of a conductor ##|x|<a/2, y<0##. We need to find the potential in all space. The hint is to use conformal maps.

    2. Relevant equations

    $$\nabla^2 = 0$$
    (U=constant at the boundary of the conductor)
    $$U = -\frac{\lambda}{2 \pi \epsilon_0} ln(r)$$

    3. The attempt at a solution

    The general idea is to conformal map the strip to a more simple geometry where I could use the method of image charges. I've had two attempts at doing that:

    In first attempt I've simply first used the map ##f(z) = z*i## to rotate the strip and then using ##g(z) = \frac{z+a/2i}{z-a/2i}(-ia/2)## (sending -ia/2 to 0, 0 to ia/2 and ia/2 to inf) effectively strecthing the strip to the first quadrant. Finally I wanted to use ##h(z) = z^2## to map the first quadrant to upper half plane. I must've visualised ##g(z)## the wrong way as this map mapped my wire ##z=ih## inside the upper half plane.(When it was originally outside the conductor)

    The second attempt started in a similar way. I use ##f(z) = zi## to rotate the strip. Then I use ##g(z) = e^{\pi z/a}## to map that to ##{|z| >1 , Re(z)>0}## Using ##h(z) = z^{-2}## I map this to the unit disc and then using ##i(z) = \frac{z+i}{z-i}(-1)## I map the disc to the upper half plane again. The problem this time was the use of the exponential fuction. It made my potential peroidic(which looking at the original problem it shouldn't be), since exponential is periodic in complex.


    Where did my thinking go wrong in these steps? Could I avoid the use of exponential functions and still map to a simple enough geometry to solve analitically somehow?
     
  2. jcsd
  3. Aug 16, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Aug 16, 2017 #3

    haruspex

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    Is it periodic within the original strip, or only if you project the solution beyond the original strip?
     
  5. Aug 19, 2017 #4
    The strip is a piece of a conductor and there is a wire next to it. Isn't potential constant within the conductor? And I need to find the potential outside the strip.
     
  6. Aug 19, 2017 #5

    haruspex

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    You only provided the bare outline of your method beyond all the mappings, so it is not clear to me how you got your potential function. You mentioned method of images, so maybe you meant only the contribution to the potential from the charge distribution within the conductor.
    If you meant the actual potential in the conductor, I don't see how you could have got that. That it is constant would be the boundary condition you were plugging in.
    Maybe if you post more details it will become clear.
     
  7. Aug 21, 2017 #6
    I'm calculating the potential created by both the infinitely long wire and the charges in conductor. Potential outside conductor is a harmonic function so i conformal map that area(outside conductor) to a half plane. Now my problem is an infinite conductor spanning the whole half plane and a wire next to it. I solve that with the method of image charges(another wire inside where the conductor would be).
     
  8. Aug 22, 2017 #7

    haruspex

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    Yes, I got that, but you are still not posting the details of your working beyond the proposed mapping.

    Never mind. Does it help to observe that cosh maps the strip Re(z)>0, 0<Im(z)<π to the upper halfplane, reversibly? That's probably equivalent to your second method.
    I see why it makes it periodic. It maps the next such strip up to the lower half plane, and so on alternating. At the same time, it maps negative x strips the other way. So when you apply the boundary condition that the potential is zero along the real line, to use the method of images, that maps back to saying the potential is zero along all the boundaries of the strips in this grid.
     
    Last edited: Aug 22, 2017
  9. Aug 22, 2017 #8
    Ah alright so let me explain:

    So I have the map k(z) a composition of above maps that maps strip to half plane. Lets say it maps the wire to a+ib. The resulting potential of a wire at a+ib and a condutor in the upper half plane is lambda/(4pi epsilon0)*log((x'-a)^2+(y'-b)^2)/(x'-a)^2+(y'+b)^2), where x'+iy' = k(x+iy). That is how I get the potential.

    Cosh(z) seems like a nice idea, but that is still the sum of two exponents which would be periodic in complex right? So it doesn't solve the problem I have.
     
  10. Aug 22, 2017 #9

    haruspex

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    Yes. But wouldn't similar issues arise with any map that is not 1-1 on the entire plane? E.g. with z2, the inverse map of the zero potential boundary will consist of more than just the original zero potential boundary. And the point charge image used to represent the mapped strip maps back to multiple images.
     
  11. Aug 23, 2017 #10
    Are you talking about ##sqrt(z)## having 2 solutions? Perhaps that could cause some problems but I don't use such map in my solution(I don't need an inverse to my map)
     
  12. Aug 23, 2017 #11

    haruspex

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    You perform a mapping and find a solution in the mapped plane. To apply that solution to the original problem you must invert the mapping.
    I don't see how you can use the result from the mapped plane in any other way. I know you are using exp, not z2, but both have this problematic feature. Indeed, this ambiguous inverse seems to be where the periodicity comes from.

    I considered a simpler problem: first quadrant is a conductor, point charge q at -1+i. With the mapping z2 this becomes conducting upper half plane, q at -2i, so image -q at 2i gives zero potential on the real line. But inverting the map gives an image charge -q at 1+i. That gives a zero potential on the +Im half axis, but not on +Re. Also, note that if the original problem had been a conductor covering first and third quadrants with +q/2 charges at -1+i and 1-i then the z2 map would have produced the identical solution, but now the inverse map produces a sensible result: -q/2 charges at 1+i and -1-i.

    I did wonder whether the trick is to take a linear combination of all the alternative inverse maps, a bit like summing linear combinations of solutions to homogeneous differential equations, but I could not make that work for this simpler problem either.

    Do you have any examples of this map & image method working where the map is not 1-1 on the plane?
     
  13. Aug 24, 2017 #12
    Don't have any examples :(.

    I'm still not sure why we'd need to invert the map. Sure we would if we wanted to know where those image charges would have to be placed in our original problem but do we need to know that? Once we know the potential V(k(z)),we can just use that right?

    Interesting point with your simpler problem. I don't have the answer to that one.
     
  14. Aug 24, 2017 #13

    haruspex

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    I would say that is still effectively an inversion, but no matter. Suppose you apply that to my simple quadrant problem, asking what the potential is at -1-i. The map takes that to 2i, which is inside the conductor.
     
  15. Aug 25, 2017 #14

    TSny

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    I'm not experienced with conformal mapping techniques. But it looks like a mapping of the "Schwarz-Christoffel" type might be appropriate. I tried it and it seems to work (i.e., it gives me equipotential lines for the strip and line charge that look right).

    https://en.wikipedia.org/wiki/Schwarz–Christoffel_mapping
    Note that in the example at this link, the upper half plane is mapped into the interior region of the strip. But you can modify the transformation so that it maps the upper half plane into the exterior region of the strip.
     
  16. Aug 25, 2017 #15
    Hmmm I see. That seems weird. The ##z^2## map still preserves harmonic functions right? If yes then maybe thats not even wrong, we just must look at the equation strictly in a mathematic way and forget about how a charge could be inside the conductor.
     
  17. Aug 25, 2017 #16

    haruspex

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    For the strip, this gives the cosh mapping we already tried. At the link, it looks like the strip would be the non-conducting region with the rest of the plane conducting. That case is ok because in forming the image of the non-conducting region (i.e. mapping to the half plane) no overlap occurs. The solution is exactly the same as if the original layout were a planeful of non-conducting strips (going to left and right from the i axis) separated only by conducting wires.
    Likewise, in the simpler problem I looked at, a conducting quadrant with an outside charge, if we swap that around to be a non-conducting quadrant with an internal charge, the z2 mapping works.
    In both problems, the difficulty arises when the non-conducting region is concave. The same maps don't work because they overlay regions that clearly would not have the same potential distribution.
    Is there some trick in modifying the map that we're missing?
     
  18. Aug 25, 2017 #17

    TSny

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    To go from concave to convex, change the "bending angles" ##\beta## and ##\gamma## from ##\pi/2## to ##3\pi/2##.
    upload_2017-8-25_17-37-58.png
     
    Last edited: Aug 25, 2017
  19. Aug 25, 2017 #18

    haruspex

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    Very neat!
    I get a mix of functions for the inverse map. Choosing -1 and 1 as the images of the two corners and taking those corners to be at 0 and i: ##z=f^{-1}(w)=K(\frac 12w\sqrt{w^2-1}-arcosh(w))##, where ##K=\frac 1{\pi-\frac i{\sqrt 2}}##.
    I hope that's wrong, because it makes it hard to figure out where the point charge maps to.
     
  20. Aug 25, 2017 #19

    TSny

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    You're getting similar to what I got. I chose ##K = i## and got ##z = \frac{1}{2}\left( w\sqrt{1-w^2} + \sin^{-1}w \right)##. The conducting strip then has the right orientation and position. The strip has a width ##\pi/2## which can easily be changed with an overall real scale factor.

    Any point on the positive imaginary ##w## axis gets mapped to the positive imaginary ##z## axis. So, the point charge gets mapped to where we want.
     
  21. Aug 25, 2017 #20

    TSny

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    Here's an example of how a rectangular patch in the upper half of the w-plane gets mapped to the z-plane.
    upload_2017-8-25_19-54-10.png
     
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