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Surface charges and surface current on conductors carrying steady currents

  1. Aug 2, 2010 #1


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    It is known that on a wire carrying steady current there are surface charges (and hence electric field outside the wire, but let's forget about it). This surface charges play an important role to maintain a uniform electric filed along the whole wire. There are only few geometries for which one can calculate the surface charges distribution. Let us focus on a long cylindrical wire of radius [tex]a[/tex] carrying a steady current [tex]I[/tex]. Current returns along a perfectly conducting grounded coaxial cylinder radius [tex]b[/tex] (see, for example, Prob. 7.57 in Griffiths). Ok, in this special case we have the surface charge density to be a linear function of [tex]z[/tex]:
    [tex]\sigma(z)=\frac{\varepsilon_0 I \lambda}{\pi a^3 \ln{a/b}}z[/tex]
    where [tex]z[/tex] is measured along the axis of the cylinder, [tex]k = const[/tex],[tex]\lambda[/tex] is the resistivity (I renamed [tex]\rho[/tex] in the Book not to be confused with the volume charge density [tex]\rho[/tex]).

    How about the surface current [tex]\mathbf{K}[/tex] (in Griffiths's notation)?

    I mean the charges on the surface certainly move, but in such a way that the spatial distribution of the charges does not change: [tex]\nabla \cdot \mathbf{j}=0[/tex]. This state is stable: if somewhere we have less charges than required, the condition [tex]\nabla \cdot \mathbf{j}=0[/tex] is violated in such a way to correct this issue.

    But we can not write for the surface current just
    [tex]\mathbf{K}=\sigma v \hat{\mathbf{z}}[/tex]

    because [tex]\sigma[/tex] changes along the wire and we will get [tex]\mathbf{K}[/tex] which is not a constant---surely, a nonsense.

    Any ideas?
  2. jcsd
  3. Aug 3, 2010 #2


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    I think there will be something more difficult than [tex]\sigma v[/tex], but the problem is WHAT? Maybe Griffiths can explain... I'll send an e-mail to him.
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