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Coaxial metal cylinders

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Two long, coaxial metal cylinders are separated by a material of conductivity sigma and dielectric constant epsilon. The radius of the inner cylinder is a, the radius of outer cylinder is b, and the length of both is L.

    Suppose that the inner conductor is held at a potential Vo with respect to the outer one. What is the current density J(r) between the two?

    2. Relevant equations
    J(r) = I/A = sigma*E
    C=Q/V
    R=s/(A*sigma)

    3. The attempt at a solution
    First attempt:
    I found resistance of the dielectric to be ln(b/a)/(2pi*L*sigma). As a function of radius r, R(r) = ln(r/a)/(2pi*L*sigma).

    Current = V/R = Vo / (ln(b/a)/(2piLsigma)) = Vo*2pi*L*sigma / ln(r/a) as a function of r.
    Then divide by Area=2pi*L*r to get current density J(r) = Vo*sigma / (r*ln(r/a)).


    Second attempt:
    Vo = Q/C. I found capacitance of the configuration to be 2pi*epsilon_o*L*epsilon / ln(b/a) using the expression for resistance. Then Q = Vo*2pi*L*epsilon*epsilon_o / ln(b/a).
    EA = Q/epsilon_o, so E = Vo*epsilon / (r*ln(b/a)). Using J(r) = E*sigma, I get J(r) = sigma*Vo*epsilon / rln(b/a).

    The difference in the two is the presence of epsilon (dielectric constant) in the numerator, and b rather than r in the ln term of the denominator. Are either of these approaches correct?
     
  2. jcsd
  3. Oct 20, 2016 #2

    rude man

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    Can bound charges support current flow?
    Better stick with method 1.
     
  4. Oct 20, 2016 #3

    haruspex

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    Looks reasonable (but I've not checked it in detail).
    That I don't understand. The same total current has to flow through any given radius, no?
     
  5. Oct 20, 2016 #4
    Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?
     
  6. Oct 20, 2016 #5

    rude man

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    Sounds right, looks right ... I'm buying!
     
  7. Oct 20, 2016 #6

    haruspex

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    Looks right to me too.
     
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