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Gaussian cylinder in the finite case

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider two long coaxial metal cylindrical tubes, with radii a and b and length L. (You may assume a,b<<L. Also a<b.) Suppose the inner cylinder is given a charge +Q and the outer cylinder a charge -Q.

    Using Gauss' Law, compute the electric field for all r between a and b.



    2. Relevant equations

    Gauss' Law, the usual Gaussian surface stuff.

    3. The attempt at a solution

    As the textbook recommends, I put my Gaussian cylinder far from the ends of the tube. So for a<r<b the enclosed charge is

    [tex]Q_{\text{enc}}=\frac{Q\lambda}{L},[/tex]

    where lambda << L is the length of the Gaussian cylinder. So by Gauss' law,

    [tex]2\pi r\lambda|\textbf{E}|\approx\oint_{\text{surface}} \textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}= \frac{Q\lambda}{L\epsilon_0},[/tex]

    (What about the ends???) Therefore in cylindrical coordinates

    [tex]\textbf{E}=\frac{Q}{2\pi L\epsilon_0 r}\boldsymbol{\hat r}[/tex]

    My questions:

    (1) Why do we get to ignore the ends? Or do we get to ignore the ends?

    (2) I'm not really sure what the restrictions are for the Gaussian cylinder when the case is finite as opposed to infinite. If we don't put it at the very center of the wire(s)/tube(s)/whatever, then doesn't that ruin symmetry? And even if we just put it far from the edges, then it seems like we can't say anything about those edges, i.e. we can only calculate the electric field at points far from the edges of the wire(s)/tube(s).

    I guess I just don't understand the physical interpretation behind the math, and it's giving me trouble. Any help would be much appreciated!
     
  2. jcsd
  3. Feb 29, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi hatsoff! :smile:
    in an exam question, "long" always means that you can treat it as infinite :wink:
    you won't be asked about that

    but yes, you're right … the field lines will be bent slightly out of shape, and difficult to calculate :smile:
     
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