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Conductors and co-axial cables?

  1. Mar 6, 2007 #1
    Conductors and co-axial cables!??

    Say you have a long conducting hollow cylinder with some surface charge density [itex]\rho[/itex] which is inside another hollow (uncharged) cylinder. What would the electric field be outside the outer cylinder?

    I am guessing that the e-field outside the outer cylinder should be zero because the inner cylinder would produce induced charges on the surface of the outer cylinder equal but opposite in magnitude and therefore the total charge enclosed in both of the cylinders would be zero.

    Is this correct?
     
  2. jcsd
  3. Mar 6, 2007 #2

    marcusl

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    What does Gauss' law (in integral form), applied outside the outer cylinder, tell you? Can you spot the mistake in your reasoning?
     
  4. Mar 6, 2007 #3
    If we choose a cylindrical gaussian surface [itex]S[/itex] outside the outer cylinder then

    [tex]\oint_{S} \vec{E}\cdot d\vec{S} = Q_\text{enc} [/tex]

    which is equal to zero since the total charge enclosed inside the surface is 0 (Q on the inner cylinder and -Q on the outer cylinder).
     
  5. Mar 6, 2007 #4

    marcusl

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    Your problem specifies only charge on the inner cylinder so Q enclosed is nonzero.
     
  6. Mar 6, 2007 #5

    Dick

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    If the outer cylinder is conducting, put the gaussian surface inside the outer cylinder.
     
  7. Mar 6, 2007 #6
    But wouldn't there be an induced charge -Q on the outer cylinder due to the inner cylinder?
     
  8. Mar 6, 2007 #7
    But the outer cylinder/conductor is not a shell -- it doesn't have any thickness.
     
  9. Mar 6, 2007 #8

    Dick

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    That's what a gaussian surface inside the outer conductor (is it a conductor) would tell you. That there is surface charge inside it.
     
    Last edited: Mar 6, 2007
  10. Mar 6, 2007 #9

    Dick

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    That's an interesting problem in metaphysics. :smile:
     
  11. Mar 6, 2007 #10
    lol
    Remember what must be true for the outer conductor in a statics situation. Hint: think about the E field.
     
  12. Mar 6, 2007 #11

    marcusl

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    Yes, but since the outer conductor must remain overall neutral there will be an opposite charge on its outer surface.
     
  13. Mar 7, 2007 #12
    Oh I see. So the net charge outside would then be Q (+Q on the inner conductor, -Q on the inner suface of the outer conductor, and +Q on the outer suface of the outer conductor).
     
  14. Mar 7, 2007 #13

    Dick

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    Exactly as you say.
     
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