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Potential of Concentric Cylindrical Insulator and Conducting

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data
    An infinitely long solid insulating cylinder of radius a = 4.3 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 28 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 10.1 cm, and outer radius c = 12.1 cm. The conducting shell has a linear charge density λ = -0.34μC/m.

    1) What is Ey(R), the y-component of the electric field at point R, located a distance d = 48 cm from the origin along the y-axis as shown?

    2. Relevant equations

    ρ = Qin/V
    V = 2πRL (Disregard L if I take it at 1m)
    λ = Qout/L (L @ 1m) so λ = Q
    Gauss - ∫E⋅dA = Qenclosed/∈0
    Agauss = 2πRa < May be area of confusion

    3. The attempt at a solution

    Qin = 2π(.043)(28x10-6 = 7.56495X10-6
    Qout = λ = -.34x10-6
    Qnet = 7.22495x10-6
    E = (Qnet4πk)/2πRd
    = 270935.816619 N/C

    I think I might be having trouble determining where to place the outside of the Gaussian surface. I'm assuming the different fields are at:
    -outside of outer conductor
    -with in conductor = 0
    -at inner surface of conducting shell
    -between conductor and inner insulator
    -surface of inner insulator
    -and within the inner insulator
    I think I remember seeing something different, but I'm not sure. I blasted through a similar problem last night, and somehow finished the section on Gauss' law but now I seem to have forgotten everything.
     

    Attached Files:

  2. jcsd
  3. Jan 30, 2017 #2

    BvU

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    Did you already do an exercise like this but with spheres ? If you place your Gaussian surface as a cylinder around the z axis through point R, what does the theorem tell you about ##\vec E## ?
     
  4. Jan 31, 2017 #3

    kuruman

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    It is indeed an area of confusion. Your expression for the area is incorrect and does not even have the correct dimensions. In fact, it is the circumference of a circle of radius Ra. You need the total area through which electric field lines emerge.
     
  5. Jan 31, 2017 #4

    BvU

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    I think the poster realizes that, from
     
  6. Jan 31, 2017 #5

    BvU

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    I don't see this one in your relevant equations. I recognize the 2πRd and the 4πk but I don't see a Q there, only a ##\lambda## :rolleyes:
     
  7. Jan 31, 2017 #6

    kuruman

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    Right you are. I saw OP's earlier "V = 2πRL (Disregard L if I take it at 1m)" and I took that to be the area despite the symbol "V" defining it. I now see that this is meant as a volume, except that it's incorrect.
     
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