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Resistance between coaxial cylinders

  1. Mar 3, 2006 #1
    Two coaxial cylinders, inner radius a and outer radius b are separated by a material of conductivity given by [itex]\sigma (r) = k/r[/itex] for some constant 'k'. Find the resistance between the cylinders.

    Here the conductivity is a function of position and the charge density is not zero in the resistive medium, and the electric field [itex]\vec E[/itex] does not go as 1/r. So how is this problem solved?
     
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  3. Mar 3, 2006 #2

    Meir Achuz

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    Assuming steady state, div j=0 so j=j_0/r and E=j/\sigma=constant.
     
  4. Mar 4, 2006 #3
    The charge density is not zero here, so [itex]\vec \nabla \cdot \vec E = {1\over \sigma}\vec \nabla \cdot \vec J = 0[/itex] will not hold good.
     
  5. Mar 4, 2006 #4

    Meir Achuz

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    Div E is NOT zero. Div j is zero because rho is constant in time.
    Your equation above should be
    [itex]\vec \nabla \cdot \vec E = \vec \nabla \cdot [\vec J/\sigma] ,[/itex]
    but Div j does =0.
    Do you want to argue or to do the problem?
     
    Last edited: Mar 4, 2006
  6. Mar 5, 2006 #5
    Ok, here goes...
    I need to evaluate the current and the voltage in order to find the resistance.
    [tex]\vec \nabla \cdot \vec E = \vec \nabla \cdot [\vec J/\sigma][/tex]

    Integrating over a volume dV.
    [tex]\int \vec \nabla \cdot \vec E dV= \int \vec \nabla \cdot [\vec J/\sigma] dV[/tex]

    By Divergence theorem,

    [tex]\int \vec E \cdot d\vec a = \int (\vec J /\sigma)\cdot d\vec a[/tex]

    [tex]{Q_{enc}\over \epsilon_0} = \int (\vec J /\sigma)\cdot d\vec a[/tex]

    I'm getting stuck here. I don't know how to proceed.

    Anyway, I found the current.
    But by Ohm's law, [itex]\vec J = \sigma \vec E[/itex]
    Current I is given by;
    [tex]I = \int \vec J \cdot d\vec a = \int {k\over r}\vec E \cdot d\vec a[/tex]

    But, I need [itex]\vec E[/itex] to compute the voltage. How do I do that?
     
    Last edited: Mar 5, 2006
  7. Mar 5, 2006 #6

    Meir Achuz

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    Read my first post.
     
  8. Mar 6, 2006 #7
    [tex]\vec \nabla \cdot \vec J = 0[/tex]
    [tex]J = {J_0\over r}[/tex]
    [tex]E = {J\over \sigma} = {J_0\over r} {r\over k} = {J_0\over k}[/tex]

    [tex]I_{enc} = \int \vec J \cdot d\vec a = \int_a^b {J_0\over r} 2\pi r dr = J_0 2\pi (b-a)[/tex]

    [tex]V_a - V_b = -\int_b^a \vec E \cdot d\vec r = {J_0\over k}(b-a)[/tex]

    Resistance R will be:
    [tex]R = \frac{V_a - V_b}{I} = \left({J_0\over k}(b-a)\right)\left({1\over (J_0 2\pi (b-a)}\right)[/tex]
    [tex]R = {1\over 2\pi k}[/tex]

    Thank you for your time!
     
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