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Potential difference between two cylinders

  1. May 13, 2012 #1
    I need help with the attached exercise. Problem is not that I am not able to do it. You can find the field of an infinite cylinder with Gauss' law and then use that to find the potential difference.
    My solutions notes however do it in a slightly different way, still arriving at the correct results. They use the expression for the potential of an infinite line:

    V(r) = −1/(4[itex]\pi[/itex][itex]\epsilon[/itex]0)(2Q/L) ln(r/R0)

    And simply use that to calculate the potential difference. Can someone explain to me why this would yield the same result?
     

    Attached Files:

  2. jcsd
  3. May 13, 2012 #2

    tiny-tim

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    hi aaaa202!
    an infinite uniformly charged line and an infinite uniformly charged cylinder clearly both have cylindrically symmetric potentials (and fields)

    so any cylinder larger than both will be an equipotential surface, and so from Gauss' law, if the enclosed charge-per-length is the same, so will be the fields (and the potentials) :wink:
     
  4. May 14, 2012 #3
    Hmm okay. So is it because you mightaswell view the field of an infinite cylinder with charge Q as that of an infinite line placed along the center of the cylinder with the same charge?
     
  5. May 14, 2012 #4

    tiny-tim

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    outside the cylinder, yes! :smile:
     
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