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Coaxial Cable

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    The diagram below depicts a cross section of coaxial conductor with an inner wire of diameter and an outer conducting sheath of inside diameter , and some material placed in the space between the two wires. Suppose that you have a coaxial wire with di= 2.85 mm, do= 6.25 mm and mylar ( k= 3.10) is placed in the space between the two wires. If there is a potential of 1 kV between the wires, how much energy is stored in a 10 m piece of cable?

    2. Relevant equations

    [tex] U=\int V dQ[/tex]

    3. The attempt at a solution
    I perform the intergral and come up with a couple equations this one seem the best:
    [tex] U= .5QV[/tex] Right?
     

    Attached Files:

    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 21, 2007 #2

    learningphysics

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    Homework Helper

    How did you get 0.5QV ?

    The way I'd do it is to find the charge per unit length on the inner wire... the charge per unit length on the outer conductor is just - the inner charge per unit length...

    You can get this using Gauss' law... and the voltage = -integral E.dr

    When you find the charge per unit length x... then the total charge is 10*x. The energy stored is 10*x*V.
     
  4. Sep 21, 2007 #3
    You need to know the charge right? Or how charge varies with voltage or something?
     
  5. Apr 22, 2010 #4
    The coax cable has capacitance per meter. You can look the equations up on wiki. Once you find the total capacitance (multiply by length), use E=(CV^2)/2.



    Sterling
     
  6. Jul 14, 2012 #5
    I do not know what I am doing wrong. Here is my work:
    q/(кε) = EA, A = 2*pi*rx
    E = q/(кεA) = q/(2кε*pi*rx)
    V = -integral of Edr from a to b = -(q*ln(b/a))/(2кε*pi*x) = 1000 V
    1000(2кε*pi)/ln(b/a) = q/x
    (q/x)*10*1000 does not give me the answer. Where did I go wrong?
     
  7. Jul 14, 2012 #6
  8. Jul 14, 2012 #7
    Thanks. It turns out my work is right, but the formula for energy is .5qV (i.e. C = q/V, so .5CV^2 = .5qV) , not qV.
     
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