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Gauss' Law for line of charge

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A long line of charge with density λ (C/m) is surrounded by a concentric cylindrical conducting shell of inner radius R1 and an outer radius of R2. The shell carries a net charge of -2λ (C/m). Use Gauss' Law to determine the electric field as a function of the distance 'r' from the line of charge for

    a) 0 < r < R1

    b) R1 < r < R2

    c) r > R2

    And

    d) What is the surface charge density σ (C/m2) on the inner and outer surfaces of the conductor?



    2. Relevant equations

    E = λ/2∏ε0R

    E = Qencl0



    3. The attempt at a solution

    I don't quite see a way to find the Electric field of those three stated points.

    To start I thought about finding the electric field only for the middle line of charge, then finding the field for the shell and adding them together. I really don't know where to begin this problem though..

    I know the first equation applies for the simple long ling of charge, but how can i find E of the shell?

    Also, to choose a gaussian surface, could I just simply enclose a section of both the middle line and the shell, with the outer edge of the gaussian cylinder at R2 in order to get the E outside both?

    If so, how can I find the bounds for the other two points?
     
  2. jcsd
  3. Feb 4, 2013 #2

    SammyS

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    Choosing a Gaussian surface is a good place to start.
    What are the necessary considerations for choosing a Gaussian surface?​

    Choose a Gaussian surface appropriate for each region to be considered. Start with 0 < r < R1 and work your way out.
     
  4. Feb 4, 2013 #3
    Hm,

    Thanks for the reply. Sorry I am very lost for this...

    could I simply say that the radius of the gaussian cylinder is R1 - r, for the first region?

    I guess to clarify, I dont know if the edge of the gaussian cylinder can be anywhere within that region.

    For the region, r > R2, the cylinder should rest with its edge at R2 I think.

    But for the other two, I don't see immediately where the cylinder could go...

    Possibly for R1 < r < R2, we could say that edge should be at R1+((R2-R1)/2) to put us right in between the outer shell?
     
  5. Feb 4, 2013 #4

    SammyS

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    The important question, which I asked earlier was:
    "What are the necessary considerations for choosing a Gaussian surface?"​

    Without knowing the answer to this question, it's futile to try to set-up a Gaussian surface for the purpose of calculating the electric field.

    Gauss's Law itself, merely gives the flux through a closed surface. In order to get the electric field from that, we need the following for how the electric field is related to our surface.
    1. For at least a portion of our surface, we need the electric field to be uniform (i.e. constant) and oriented to the field in a uniform way. (Usually we want the field to be perpendicular to the surface.) This requires that there is a lot of symmetry regarding the charge distribution.

    2. For any portion of our surface over which we can be sure that the field is uniform, we need to have the surface be parallel with the field. This assures that no flux passes through this portion of the surface.

    3. If any portion of our surface falls in a region in which the electric field is zero, then we no that no flux passes through this region. This is also acceptable.​

    From the charge distribution in this problem, it's pretty clear that each of your Gaussian surfaces should right-circular-cylinders of finite height. The axis of each of these cylinders should lie along the central line of charge.
    Do you understand why this is the case?​
     
  6. Feb 5, 2013 #5
    Because the field lines must be perpendicular to the gaussian surface at all times, yes? I realize the axis must go through the center line of charge so that the surface of the cylinder will actually fulfill that requirement. But I don't know how to choose the necessary radii from the axis outwards.
     
  7. Feb 5, 2013 #6

    SammyS

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    In each part, pick some arbitrary height for the cylinder. You can even pick a height of 1, assuming that the line of charge is much longer than 1 unit of length .

    For each part, the radius should be as follows:
    a) 0 < r < R1
    For this the only charge within the cylinder is a portion of the line charge equal in length to the height of the cylinder.​

    b) R1 < r < R2
    This should tell you how much charge resides on the inner surface of the conductor. From that, you can infer how much charge resides on the exterior surface of the conductor.

    c) r > R2



    Furthermore, in part b) you know that E = 0 inside conducting material .
     
  8. Feb 6, 2013 #7
    So really what we get for part a is:

    E = Qenclo

    where Qencl = λx

    where x is any distance on the infinite line?

    So

    E = λx/εo?

    One thing though, would the field created by the cylindrical shell change the field, even though we are only looking at enclosed charge to get the field?

    Also, I don't see why E would be 0 inside the shell.

    and for part b) how does this tell me how much charge is on the inner surface?
     
  9. Feb 6, 2013 #8

    SammyS

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    No.

    Qencl is correct.

    But E∙A = Qencl0 .

    What is the area of the portion of your cylinder which has flux passing through it?

    (Your cylinder has radius, r, and height, x .)
     
  10. Feb 6, 2013 #9
    Ah, I see...

    In this case, the x cancels were left with

    E = λ/2∏r

    For any r in the defined region.


    Then for a region r > R2

    How would I deal with finding Qencl with two charge densities?

    Also, wouldn't I need to know the charge density of the shell in m3 as the shell has volume?

    I'm looking at it as Q of the line, plus the Q of the shell.
     
  11. Feb 6, 2013 #10

    SammyS

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    Not for all regions. It's only true for r < R1

    What is the electric field in the region, R1 < r < R2 ? (That's in the conducting material itself.)
     
  12. Feb 6, 2013 #11
    Within the shell, it is zero, but there would still be charge on the surface of the shell, which would be within the gaussian cylinder.
     
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