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Infinite line charge (without Gauß's law)

  1. Sep 20, 2008 #1
    Hi!
    I would like to find the electric field of an infinite line charge without using Gauß's law.
    I already simply integrated over the line charge, which gave me the correct result.
    However, I would like to try yet another approach: the potential equation, which states that:

    [tex] \Delta \phi = \frac{\rho}{\epsilon_0}[/tex],
    for the electric potential.
    For the infinite line charge, we can use the Laplacian for cylindrical coordinates, which is
    [tex]\Delta= {1 \over r} {\partial \over \partial r}
    \left( r {\partial \over \partial r} \right)
    + {1 \over r^2} {\partial^2 \over \partial \theta^2}
    + {\partial^2 \over \partial z^2 }[/tex].

    Because the situation is symmetric, all derivatives with respect to [tex]\theta, z[/tex] vanish.
    The charge density must also be 0 for r>0.
    The equation I have to solve thus looks like:
    [tex]{1 \over r} {\partial \over \partial r} \left( r {\partial \phi \over \partial r} \right) = 0[/tex].

    By integrating twice, I get
    [tex] \phi = \alpha \ln(r) + \beta[/tex]
    which has the correct form I want. I then chose [tex] \phi(1) = 0 \Rightarrow \beta = 0[/tex]

    My problem comes now: How do I get [tex]\alpha[/tex]?

    Thanks for your help!
     
  2. jcsd
  3. Sep 20, 2008 #2

    CompuChip

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    Homework Helper

    I don't quite follow your calculation. If you have an infinite line charge, then why would you use cylindrical coordinates? Isn't it much easier to just put the charge along one of the axes and use Cartesian coordinates?

    [OK, your point probably is, that this would come down to just integrating over the line charge in the end, because first you integrate rho twice to get phi, and then differentiate that to get E]

    Also, I don't really see why you impose the boundary conditions you do. You are putting the line charge along the x-axis (the z-axis seems more convenient, but then again you might as well do it in Cartesian coordinates)? But then why is the charge density zero for r > 0 and why would you want to choose the potential zero at r = 1 ?
     
  4. Sep 20, 2008 #3
    Thanks for your reply. You're right, I should have been more clear in my explanations:
    I'm using cylindrical coordinates because the potential must be cylindrically symmetric, I thought it was kinda the natural choice.
    In my calculation, I put the charge along the z axis, therefore [tex]r^2 = x^2 + y^2[/tex] is the (squared) distance from that axis.

    The charge is only on the z axis, and with my choice of r, the density must be 0 everywhere else.
    I chose [tex]\phi(1) = 0[/tex] so [tex]\beta[/tex] would vanish, just for aesthetical reasons ;).
    Of course, in the end, I want to get the actual electric field using
    [tex] \mathbf E = -\nabla \phi [/tex]

    I hope this has cleared up the situation a bit.
    Thanks for any help!
     
    Last edited: Sep 20, 2008
  5. Sep 20, 2008 #4

    atyy

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    Science Advisor

    Specify the charge density, solve for the potential in the charge density and match solutions at the boundary. Probably alpha is more important than beta, since the potential is unique only up to an additive constant?
     
  6. Sep 20, 2008 #5
    As what I have here is a line charge, there is no "inside" that I could solve the potential equation for.
    I could try to first solve everything for a infinitely long cylinder with radius r and then take the limit as r tends to 0.
     
  7. Sep 21, 2008 #6
    Getting [tex]\alpha[/tex] here is a little messy business. In essence, it's just applying the boundary conditions, so you should integrate both sides of Gauss's law:

    [tex] \int \alpha \nabla^2 \ln r\, da = \int \frac{\rho}{\epsilon_0}\, da = \frac{\lambda}{\epsilon_0}[/tex]

    Now the first integral is a fishy one and you must be really careful about it. What I guess is that it can be solved by means of a 2D version of Gauss's theorem:

    [tex]\int \nabla^2 \ln(r)\, da = \int \boldsymbol{\nabla \cdot} (\nabla \ln(r))\, da = \int \nabla \ln(r) \, dl = \int \frac{1}{r}\, rd\phi = 2\pi[/tex]

    So you would have

    [tex]\alpha = \frac{\lambda}{2\pi \epsilon_0}[/tex]

    The dimensions are alright, just check the 2pi factor using the usual methods.
     
  8. Sep 21, 2008 #7
    Thanks for your help, Irid.
    I think I was trying so hard not to use Gauß's law that I missed the fact that in this case,
    it's impossible without it.

    Thanks!
     
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