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Electromagnetism: Gauss's Law

  1. Sep 28, 2011 #1
    Not sure if this is advanced. Highly doubt it but oh well

    1. The problem statement, all variables and given/known data
    Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρo r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

    2. Relevant equations
    Gauss's Law
    q=ρ δτ

    3. The attempt at a solution
    (a) E inside cylinder
    I sketched a Gaussian surface inside of the cylinder.
    I believe that E is parallel to ds ( E⃗ ||ds⃗ )
    So, gauss's law becomes E∮ds = q/ϵ for the side

    I believe the integral of ds is 2π r L (L being the length of the cylinder even though it is infinite.
    And q = ρo r π r2 L
    derived from q=ρ δτ

    So we have E (2π r L) = ρo r π r2 L /ϵ
    Simplifying to E = ρo r2/ 2ϵ

    Is this correct for (a)?
    And for (b) would it be the same idea but with a gaussian surface outside of R?

    Thanks!
     
  2. jcsd
  3. Sep 29, 2011 #2

    Doc Al

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    Staff: Mentor

    It's a bit difficult to read your notation. Consider using Latex for your equations.

    (For example: [itex]q = \int \rho dv[/itex].)

    Show again how you integrated to find the total charge within your gaussian surface.
     
  4. Sep 30, 2011 #3

    collinsmark

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    Homework Helper
    Gold Member

    So close! :cry: But as Doc Al says, it's kind of hard to understand your notation. By "r2" do you mean r2? If so, try the final simplification once more. I think you forgot to cancel something out.
    Yes. The trick is to just be careful about determining q. When outside the cylinder, is the total charge q within the Gaussian surface a function of r or a function of R?
     
  5. Sep 30, 2011 #4
    Yes I meant r squared. The thing is when i reposted this problem the pasting messed up a couple of things.
     
  6. Sep 30, 2011 #5

    collinsmark

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    Right. I think I figured that out when I looked at the other thread. Check that thread. I left a hint for you there.
     
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