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Differential Gauss's Law & Coaxial Cable

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    A long coaxial cable consists of an inner solid cylinder, radius a, and an outer thin coaxial cylindrical shell, radius b. The outer shell carries a uniform surface charge density σ.
    Find the uniform volume charge density ρ that the inner cylinder must have in order that the whole cable (inner + outer) is neutral.

    2. Relevant equations
    Acylinder = 2πbl

    Vcylinder = πa2l

    ρ = Q/V

    ∇∙E = ρ/ϵ0

    Qenc = ∫Vρdτ

    3. The attempt at a solution
    I feel like I have all the relevant equations, yet, I still do not know how to start this. Any hints? Very much appreciated.
     
  2. jcsd
  3. Sep 18, 2012 #2
    Hi Nekoteko,

    The problem is asking you to find ρ such that the total charge in the inner cylinder is equal to the total charge on the outer cylinder.

    Another way of saying this is: For what ρ does the electric field in the region outside of both cylinders equal zero?

    Try using Gauss' Law to find the electric field in the region outside of both cylinders.
     
  4. Sep 19, 2012 #3

    vela

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    This is a very basic review problem for an upper-division electromagnetics course. If you have no idea how to start this, you must not have tried very hard. You need to show a decent effort before you receive help here.
     
  5. Sep 19, 2012 #4
    lol Is there a reason you're so reluctant to help?

    E_M_C, that's really all the hint I needed! It's so simple, not sure why I didn't see that yesterday. Thanks. :)
     
  6. Sep 19, 2012 #5

    vela

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    Yes, read the forum rules, the ones you agreed to when you registered here. Did you really need a hint to understand what the word neutral meant?
     
  7. Sep 19, 2012 #6
    Lol I didn't ask for help to piss you off. What do you want me to do if I can't figure something out myself? I understand the forum rules, but if I needed help, I needed help, I don't care how stupid you think my question was (it was). Besides, not like I was asking anybody to solve it for me, just a hint is all I needed. And neutral isn't the problem I had w/ this question. Relax your internet authority, lol.
     
  8. Sep 19, 2012 #7
    There's no need for that kind of behavior.

    You're very welcome :)
     
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