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Coaxial cable

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi, I'm having issues with a question about a coaxial cable.
    This is the question:

    A coaxial cable consists of a long straight cylindrical wire of radius R1, surrounded by a coaxial cylindrical shell of radius R2. For a situation where the inner cylinder carries constant volume charge density ρ and the shell carries constant surface charge density σ, find the electric field within the wire, between wire and shell, and outside the cable.

    2. Relevant equations

    Gauss's Law and any other continuous charge distribution equations I guess...

    3. The attempt at a solution

    I have a number of issues with this question. Firstly, for the electric field within the inner cylinder I can, using Gauss's law find E=ρr/2ε0. However I thought that the electric field within a conductor was zero.
    Secondly, I wasn't sure if the charge of the outer shell would make a contribution to the field within the inner wire, and if it does, how to include/calculate it.
    Basically, my main issue here is working out the electric field when there are two charged surfaces affecting the field.

    Any help would be most appreciated :)
     
  2. jcsd
  3. Oct 16, 2011 #2
    You are correct that the first E-field (over surface s1) is E = [itex]\frac{\rho*R_1}{2*\epsilon_o}[/itex] and E would be in the radial direction.
    The reason the E-field is not zero is due to the charge density being distributed over the entire volume of the inner wire.

    The second surface would enclose all of the inner wire so using Gauss's Law similar to before:
    [itex]E_r*\epsilon_o*2*\pi*r*h = \int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{R_1}\rho*rdrdzd\theta[/itex] which equals:
    [itex]E_r*\epsilon_o*2*\pi*r*h = \rho*\pi*h*R_1^2[/itex], so E will again be in the radial direction: [itex]E = \frac{\rho*R_1^2}{2*\epsilon_o*r}[/itex]

    For the third Gaussian surface (s3):
    We have to include both the volume charge density which will be the same as before as well as the surface charge density now included in the enclosed surface
    [itex]E*\epsilon_o*2*\pi*r*h = \rho*\pi*R_1^2*h+ \int_{0}^{2\pi}\int_{0}^{R_2}\int_{0}^{h}\sigma*dr*dz*d\theta[/itex] which gives after integration:
    [itex]E = \frac{\rho*R_1^2}{2*\epsilon_o*r}+ \frac{\sigma*R_2}{\epsilon_o*r}[/itex]
    which again is in the r direction.

    Hope this helps.
     
  4. Oct 17, 2011 #3

    rude man

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    You are correct in stating that any static charge on the inner conductor is confined to the surface. So the E field within the inner conductor is zero unless there is also current flow. But this current does not constitute net free charge so there is no E field produced ouside the inner conductor by virtue of current flow alone.

    If there is current flow there is an E field within the conductor, directed in the direction of current: E = i/γ where i = area current density and γ = conductivity.

    It is possible for the wire to simulataneouskly have static surface charge and also carry a current, in which case the static charge again makes the E field inside the conductor = 0 but for r > R1 it's not.

    Just use Gauss' theorem for net free charge contained within your surfaces. I suppose you could pretend the "wire" is a dielectric so that charge density can indeed exist within the inner "wire". In which case you have 3 surfaces with varying net free charge to contend with.

    Like I said, the premise of the question is impossible so I would confront your instructor with that fact. Either that, or he needs to educate me!
     
    Last edited: Oct 17, 2011
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