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Electric Fields: Thin wire in conducting cylinder

  1. Sep 19, 2014 #1
    Hello PF, I am having a bit of difficulty understanding this question.

    1. The problem statement, all variables and given/known data
    "A thin wire with linear charge density λ is surrounded by a conducting cylindrical shell."

    (There is a hollow cylinder with a wire though it)

    "If the electric field must be zero inside a conductor, is the electric field due to the wire shielded from extending beyond the conducting shell?Find the electric field as a function of distance r from the thin wire."


    2. Relevant equations

    EA = Qenc0

    Q = P/V

    surface area (A) = 2πrl

    3. The attempt at a solution

    I'm having more conceptual difficulty with this question than mathematical. There is a charged wire in a conductor, but the inside of the conductor must have an E field of zero (at equilibrium). How does this happen? How does the charge on the conductor rearrange or move to cancel the E field? Do I assume the charge can't leave the wire?

    To find the E(r) I would just use a Gaussian surface at different intervals (I know it's zero until r>R of cylinder). After that the Q enclosed is just constant from the surface of the cylinder shell.

    Thanks for the help.
     
  2. jcsd
  3. Sep 19, 2014 #2

    gneill

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    Staff: Mentor

    Hi wannabee_engi, Welcome to Physics Forums.

    The electric field inside a conductor is zero because the free like charges repel each other to the periphery and arrange themselves for lowest energy state. Effectively this means that all the fields from the charges balance each other and cancel inside the conductor. Likewise charges in a conductor will rearrange themselves to cancel an externally applied field. Consider: if a field existed inside a conductor (where there are mobile charges), the potential difference caused by that field would cause the mobile charges to flow as a current. Current stops when the charges can go no further (can't escape the surface), and they "pile up" and create their own field to dissuade further charges from approaching. That is, they cancel the external field within the space of the conductor.

    Now, this can happen inside a conductor where there is a path for charges to follow. In you wire+cylinder case there is free space between the wire and cylinder where charges cannot flow. So you can expect the charges on the wire to live on the surface of the wire, and their electric field to extend across the gap to the inner surface of the cylinder. So there will be a field in that gap. That field in turn will cause charges to rearrange themselves in the cylinder (which is also a conductor).

    What you have is what is called a cylindrical capacitor. A web search will turn up details about the charge distribution and fields.
     
  4. Sep 19, 2014 #3
    Ok, I understand the idea of what you are saying. The charge on the wire won't go through the space between it and the cylinder, but the charge on the shell will move to cancel the field of the charge on the surface of the wire. It's tough to conceptualize this stuff from the textbook. Thanks for the response.
     
  5. Sep 20, 2014 #4

    rude man

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    Homework Helper
    Gold Member

    I don't see why there wouldn't be fringing effects at the ends of the shielded wire, assuming the ends of the shield are not sealed. Finding the axial E field close to either end would seem a tough assignment. Maybe I'm misinterpreting the question.
     
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