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Power transported down a coaxial cable

  1. May 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A long coaxial cable carries current I. The current flows down the surface of the inner cylinder (of radius a) and back along the outer cylinder (of radius b). Calculate the power transported down the wire.

    2. Relevant equations

    The Poynting vector is given by [itex]\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}[/itex]. From there I should have no trouble finishing off the problem.

    3. The attempt at a solution

    I found the magnetic field no problem with Ampère's law. It is [itex]\mathbf{B}(s) = \frac{\mu_0I}{2\pi s}\hat{\boldsymbol{\phi}}[/itex] for a < s < b and B(s) = 0 otherwise.

    My trouble is with finding the electric field. There are two possible sources of electric fields. One is time-varying magnetic fields. But the magnetic field here is time-independent. The second is static charges. But both the inner and outer cylinder are electrically neutral overall. (Their positive charges happen to be stationary while their negative charges are in motion, but they still should have equal amounts of each.) So by my calculations, E = 0.

    But that would mean no Poynting vector, and that would mean no power, and that's probably not the answer to this question.
     
  2. jcsd
  3. May 9, 2014 #2
    You can find the electric field by using Gauss's Law in integral form.
     
  4. May 9, 2014 #3
    If I could find somewhere to put my Gaussian surface such that the total enclosed charge was nonzero, I'd be very happy to.
     
  5. May 9, 2014 #4
    In Griffiths, you are given a diagram, right? It would be easier if I can refer to one.
     
  6. May 9, 2014 #5
    Oh, is this problem in Griffiths? Do you by chance know where in the book?
     
  7. May 9, 2014 #6
    A Gaussian surface outside the cable will enclose a net charge of zero, and hence no electric field. Inside the inner cylinder, no charge will be enclosed, so no electric field there either. Can you see what to do now?
     
  8. May 9, 2014 #7
    But a Gaussian surface enclosing only the inner cylinder but not the outer cylinder encloses a net charge of zero too. The electrons are all moving and the protons are all stationary, but there are still an equal number of both. That was my point in the original post.
     
  9. May 9, 2014 #8

    TSny

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    There will be a potential difference V between the inner and outer conductors in order to make the current flow. This potential difference would be supplied by some external source. So, there will be an electric field between the conductors due to surface charges on the conductors. I'm not exactly sure what the question is asking. Maybe you are meant to express the electric field in terms of V and then show that the power calculated using the Poynting vector equals the usual expression IV.
     
  10. May 9, 2014 #9
    This is probably not an introductory physics problem, by the way.

    I found a diagram, but I can't get the image to show, so go here: http://tinypic.com/r/2lw7zsx/8
    We only need to worry about the region between r = a and r = b as before so we’ll use the Gaussian surface defined in that diagram. Now, find the electric field between the plates.
     
    Last edited: May 9, 2014
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