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Homework Help: Capacitance of coaxial cable

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data

    http://smg.photobucket.com/albums/v68/jumpboyb/?action=view&current=scan0001.jpg" [Broken]
    Pretty much looking for a general equation for capacitance of a coaxial cable per meter length.

    -a diagram (click link)
    -two radii
    -dielectric material
    -outer conductor is grounded (V = 0V)
    2. Relevant equations
    C = Q/V

    3. The attempt at a solution
    used the second equation but won't work since its for a parallel plate capacitor.

    thats it! I'm lost on this one
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 8, 2008 #2
    I think the best way to approach this problem is with Gauss's law. You need to set an arbitrary charge (it will vanish in the end.). Use a cylinder as your Gaussian surface. Hope this helps
  4. Dec 8, 2008 #3
    Still stuck. Here's what I did with Gauss' Law in mind. (btw, the answer is (2pi[tex]\epsilon[/tex])/ln(R2/R1) and thats not pi to the epsilon not power btw.)

    So, with Gauss, you have integral of E dot dA which equals Q/[tex]\epsilon[/tex].

    Capacitance - Q=VC...so subsitute that back into the gauss equation so you have....

    EA = VC/epsilon [tex]\Rightarrow[/tex] C= EA[epsilon]/V

    Electric potential is the negative rate of change of a electric field...ie E = -dV/ds = -V/r where r = R2 - R1


    and yea....nothing. I see I need to integrate something, but idk where or what logic to use for it.
  5. Dec 8, 2008 #4


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    Homework Helper

    To get V you must integrate the expression for E with respect to dr. You can't just use E = -V/r which is true only for the simplest 1D case.
  6. Dec 8, 2008 #5
    Right, try adding the integral of E dot dl or dr (whatever you guys call it.) into your arsenal.
  7. Oct 19, 2010 #6
    the expression you want to integrate is dV = -[Q/(2Pi*R)] with limits from R1 to R2 (inner to outter). This is obtained through gauss's law, saying that EA = Qin/Episilon, solving for E and substituting for the original integral of dV = -[E dot ds]. The area "A" is 2*Pi*r because after the length is taken into acount, it will account for the surface area.
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