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Charge inside and outside coaxial cable

  1. Sep 22, 2009 #1
    I'm attempting to teach myself some electrostatics, so that I'll be prepared when I head back to school full time.

    1. The problem statement, all variables and given/known data
    Determine the electric field inside the inner conductor, between the conductors, and outside the conductors of a coaxial cable when the inner conductor is negatively charged with a linear charge density of [tex]-\lambda[/tex] and the outer conductor is positively charged with a linear charge density of [tex]\lambda[/tex].


    2. Relevant equations

    [tex]\iint \vec{E} \cdot dS = \frac{q_0}{\epsilon_0}[/tex]

    [tex]\lambda = \frac{q}{l} = 2\pi r \sigma[/tex]


    3. The attempt at a solution

    Using cylindrical symmetry, I can dispense with the integral and say that [tex] \vec{E}(2\pi r l) = \frac{2 \pi r l \sigma}{4\pi \epsilon_0}[/tex]. I construct my first Gaussian cylinder inside both the positively charged conductor and the negatively charged conductor. There is no charge enclosed by this Gaussian surface, so the electric field must be zero. I construct my second Gaussian surface outside the negatively charged conductor, but inside the positively charged conductor. The positively charged conductor doesn't contribute anything since it is outside the surface, but the negatively charged conductor, using the equation above, contributes an electric field of [tex]-\frac{\sigma}{4 \pi \epsilon_0}[/tex]. I construct a third Gaussian surface outside both the positively charged conductor and the negatively charged conductor. The net field here is the sum of both the positive flux pointing outward from the positive conductor, and the negative flux pointing inward towards the negative conductor, so the field is 0 outside both conductors.

    Does that line of reasoning look correct?

    edit: I'm getting confused in this problem by the two variants of Gauss' law I'm seeing - one I'm looking at has the right hand side as being [tex]4\pi k_e q_s[/tex]. That's why I'm getting mixed up with the 4 pi. I'll have to do this over.
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 23, 2009 #2
    Note that, in this equation,

    [tex] \vec{E}(2\pi r l) = \frac{2 \pi r l \sigma}{4\pi \epsilon_0}\ ,[/tex]

    the 'r' on the left would be the radius of the Gaussian surface. The 'r' on the right would be the radius of the inner conductor.
     
  4. Sep 23, 2009 #3
    Thank you for the clarification. So for the field between the two conductors, I get [tex]E = \frac{R \sigma}{r \epsilon_0}[/tex], where R is the radius of the inner conductor, and r is the radius of the gaussian surface. This is the same as [tex]2k_e\frac{\lambda}{r}[/tex], where lambda is the linear charge density and [tex]k_e = \frac{1}{4\pi\epsilon_0}[/tex].
     
  5. Sep 23, 2009 #4
    I don't know what units you are using in the class, so I can't comment on the factor of 1/4pi.

    The inner conductor is negatively charged. You need to fix the sign.

    Earlier, you had this

    [tex] \vec{E}(2\pi r l) = \frac{2 \pi r l \sigma}{4\pi \epsilon_0} \ .[/tex]

    A little nit-picking, but note that the left side is vector valued and should be a scalar.
     
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