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Capacitance of Parallel cylinders

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Two long, cylindrical conductors of radii [itex]a_1[/itex] and [itex]a_2[/itex] are parallel and separated by a distance [itex]d[/itex], which is large compared with either radius. Show that the capacitance per unit length is given approximately by
    [tex]
    C=\pi\epsilon_0 \left(\ln\frac{d}{a}\right)^{-1}
    [/tex]
    where [itex]a[/itex] is the geometric mean of the two radii.
    Approximately what gauge wire (state diameter in millimeters) would be necessary to make a two-wire transmission line with a capacitance of [itex]1.2\times10^{-11}[/itex] F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?


    2. Relevant equations
    [tex]
    C=\frac{Q}{\phi}
    [/tex]
    [tex]
    \phi=-\int \vec{E}\cdot d\vec{l}
    [/tex]


    3. The attempt at a solution
    Honestly, I'm pretty lost on this... I'm not sure how to construct this from scratch, my only work is from backwards, and I feel shady about it...

    Here it is:
    We know that [itex]C=Q/\phi[/itex] and we are given the capacitance per unit length [itex]C'=C/l[/itex], so I did this
    [tex]
    \phi=\frac{q}{C' l}=\frac{\lambda}{\pi\epsilon_0}[\ln(d)-\ln(\sqrt{a_1 a_2})]
    [/tex]
    with [itex]\lambda[/itex] the charge per unit length, which can be written as
    [tex]
    \phi=-\int_{d}^{\sqrt{a_1 a_2}} \frac{\lambda}{\pi\epsilon_0 }\frac{1}{r}dr
    [/tex]
    Then the electric field would be
    [tex]
    \vec{E}=\frac{\lambda}{\pi\epsilon_0 r}\hat{r}
    [/tex]
    Now if I were to continue with this (and I'm not committed to this :) ), how could I manage to write this as a superposition of two cylindrical electric fields from two different conductors with equal and opposite charges...?


    Can someone give me a hand here?

    Thanks in advance,
     
  2. jcsd
  3. Sep 5, 2011 #2
    Find the electric field of the cylinders by using Gauss' Law, and use that to find the potential difference between the two cylinders. Then you can substitute your result into the formula for capacitance.
     
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