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Charged cylinders

  1. Sep 10, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    2 long cylinders with radius [tex]a[/tex] and [tex]b[/tex] [tex](a<b)[/tex] have a charge density by unit of length worth [tex]- \lambda[/tex] and [tex]\lambda[/tex] respectively.
    Use Gauss's law to find the electric field in every point of the space.
    Find the potential in all points of the space, assuming that the potential is worth [tex]V_b[/tex] over the cylinder with radius [tex]b[/tex].



    2. The attempt at a solution
    I believe that the electric field inside both cylinders is null.
    For any point [tex]r[/tex] such that [tex]a<r<b[/tex], I believe that only the cylinder with radius [tex]a[/tex] contributes to the electric field. I get that [tex]E=-\frac{2 \lambda k}{r}[/tex] using Gauss's law.

    And outside both cylinder, [tex]E[/tex] is determined by both cylinders, so [tex]E=-\frac{2 \lambda k}{b+R}+\frac{2 \lambda k}{R}[/tex]

    I know I'm completely wrong. I remember a helper at university saying that outside both cylinders, the electric field is null. He also said that the situation is not the one of a capacitor... I don't know why.
    I really need help... Thanks in advance!
     
  2. jcsd
  3. Sep 10, 2009 #2

    mukundpa

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    What is R and b + R ?
     
  4. Sep 10, 2009 #3

    fluidistic

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    First : thanks for helping me!
    Second : Sorry, I should have mentioned it : R would be the distance from the surface of the biggest cylinder to any point outside both cylinders.
    b is the radius of the biggest cylinder, hence b+R is the distance from the center of both cylinders to any point outside them.

    Do you understand what I'm saying? (Sorry for my English)
     
  5. Sep 10, 2009 #4

    mukundpa

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    But in the formula r is the distance from axis of the charge system. Actually this is the radius of Gaussian cylinder (virtual closed surface) considered coaxially to find field at distance r from axis.
     
  6. Sep 10, 2009 #5

    fluidistic

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    Right.
    I don't see the inconstancy. R is not r.
     
  7. Sep 10, 2009 #6

    mukundpa

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    As the distance r of the point (at which we find field outside both cylinders) is taken from the axis of the system it is same for both cylinders
     
  8. Sep 10, 2009 #7

    fluidistic

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    Ah you're right. I see my error! So indeed the electric field outside both cylinders is 0.
    Was I right when I wrote [tex]E=-\frac{2 \lambda k}{r}[/tex] inside the biggest cylinder but outside the smaller one?

    Thank you very much for all!
     
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