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Homework Help: Electrostatics in 2 and 1 Dimensions

  1. Sep 19, 2010 #1
    Hi, I'm having a bit of a hard time stumbling over the concepts of the following problem:

    1. The problem statement, all variables and given/known data
    In Electrostatics:
    How do you modify the div and curl of the electric field from 3D to 2D?
    What are the 2D and 1D versions of Coulomb's Law?

    2. Relevant equations
    In 3D (sorry, no latex here): del dot E = 4*pi*rho
    Del Cross E = 0
    Coulomb's law in 3D: F= ((1/4*pi*eps)qQ/r^2) r hat
    wow, that looks ugly, sorry about that.

    3. The attempt at a solution
    I would think the divergence wouldn't change from 3D to 2D since it lives in two planes in both case. BUT what about the 4*pi*rho? it looks suspiciously like it's geometry dependent... and if so, could change in 2D...

    As for the Curl: My best guess would be that it would be zero, since there is no perpendicular plane for it to live. Or would it be a scalar? I don't even know how to set up the determinant for this... or.... can you even use a det to find the curl (i.e. is that method distinct for 3D, since 3D is pretty special and not like most other n-dimensions.

    And, similar for coulomb's law: the geometry is tripping me up. The 3D radial dependence of the charges have to be modified for 2D and 1D... and is there geometry in the prefactor?

    Thanks very much!
  2. jcsd
  3. Sep 19, 2010 #2
    Unless you are using crazy units it is not del dot E = 4*pi*rho, it is del dot E = rho/epsilon zero. TO be honest, I am not quite sure what the question is. When you go to two dimension you can think of it as just x and y, so the z component is zero. That should pretty much answer your questions except for the curl being a vector in 2d. To be honest, the curl isn't really a vector anyway, even in 3d, it is just convenient to assign it a vector value because you can dot other vectors into it and the math works out. It is a bookkeeping device. If it helps, think of it still as a vector, but pointing in the z direction, and then all you need to know is the magnitude of it because you would just dot it with the normal of the surface (z hat) to do calculations anyway. Yes the normal of the 2d plane is z, the dimension is still there whether or not the question says it is 2d or not. In other words, there is no 2d, it is all a 3d problem, it just so happens that your problem has a symmetry along one direction (it is a line of charge extending in the z direction or something similar) and this allows you to only look at the 2d (x,y) part of it (because the z part is zero due to a symmetry/geometry thing).
  4. Sep 20, 2010 #3
    Thanks Prologue! No, not crazy units really. I should have mentioned I was using gaussian units. The equations are much more compact that way.

    So, I figured it all out. The cross product is in fact a scalar, which you can get by a few methods, some of which involve some shading dealings with det's that I don't approve of. But, all said and done, the cross product of a gradient is always zero (E field is neg deriv. of potential). As for the dot product, it becomes 2*pi*rho because of geometry. Coulomb's law in 2D is just as you'd expect from geometry manipulation. But Coulomb's in 1D is pretty interesting, although not surprising once you find it: it has no radial dependence. pretty fun stuff.
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