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Electrostatics in 3d vs 2d

  1. Jun 5, 2009 #1
    I have asked this question before in another section of the forum but I still don’t have an answer so I thought I would try here. Ok…here goes..

    In three dimensions, Poissons equation can be used to model an electrostatic problem in which there is a single point charge at the origin. The right hand side of the equation would be represented by the three dimensional Dirac delta function. The solution for this equation gives a the potential function u = 1/(4 Pi r). Taking the gradient of this produces the vector function for the electric field which as expected is an inverse square of the distance.

    Ok…so here is the real part of the question. If we do this in two dimensions the solution now is u = 1/ (2 Pi Log(r) ). What bothers me about this is taking the gradient of this function now produces a field that is no longer an inverse square.

    Is there some physical explanation for the fact that in two dimensions the field decays as 1 over r or is the reason because electrostatics problems really should only be thought of as three dimensional problems?
     
    Last edited: Jun 6, 2009
  2. jcsd
  3. Jun 5, 2009 #2

    Born2bwire

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    A simple explaination why a field decays over distance is the fact that the power is spreading across a larger and larger wavefront, or area. Let's say we have a lossless medium and we send out a spherical wave. The total energy along a wavefront would be the energy density times the area, which is proportional to r2. If the wavefront is to have constant energy, then as the wavefront propagates in space, it's energy density must decrease to make up for the increase in the area. This is why the energy of the wave falls off as 1/r2. This is equivalent to saying the amplitude of the wave (field) falls off as 1/r.

    Now this is in three dimensions but what happens in two dimensions? Now the area of the wavefront is related to r, not r2. So now the associated field must drop off as 1/\sqrt{r}.

    This should give you an idea why you will see a difference in the space loss between different numbers of dimensions.
     
  4. Jun 6, 2009 #3
    Thanks for the explaination. It's getting me closer to the understaning that I want to get with this problem.

    I am sorry to say that I made a mistake in part of the statment. In 2 d the solution involves
    u = (1/2 PI) log r ... ie the potential involves log r and the field drops off like 1/r.

    Another question. In 2d electrostatics why doesen't the potential decay as well?
     
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