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Electric Field Of A Cable

  1. Feb 18, 2014 #1

    Kot

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    1. The problem statement, all variables and given/known data
    A long coaxial cable of length L consists of a cylindrical metal sheath of radius b surrounding a conducting wire with outer radius a. The sheath and wire are given the same charge per length +Q/L. Find the electric field in terms of distance r from the central axis of the wire in the regions 0 < r < a, a < r < b, and r > b.


    2. Relevant equations
    Gauss' Law ∫EdA = Qino


    3. The attempt at a solution
    I am not really sure how to do this problem but I think I have an idea. I have to find the electric field in terms of distance r of three different regions 0 < r < a, a < r < b, and r > b. Since the cable and metal sheath have the same charge per length, wouldn't they both have the same charge and cancel out? Could someone explain how I can find the Gaussian surface for the first region?
     
  2. jcsd
  3. Feb 18, 2014 #2

    tiny-tim

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    Hi Kot! :smile:
    For r > b, yes

    So the electric field there is … ? :wink:
    (for r < a) Why don't you just use a cylinder of radius r ?
     
  4. Feb 18, 2014 #3

    Kot

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    For r > b, the electric field wouldn't completely cancel out since I was given an arbitrary a and b.

    For a < r < b, I am still not sure what radius to use. How can i express a radius that is between a and b?
     
  5. Feb 18, 2014 #4

    tiny-tim

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    Hi Kot! :smile:
    But the question says that the charges per length are equal. :confused:
    Just call it r.

    What do you get?​
     
  6. Feb 18, 2014 #5

    Kot

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    It's charge per length is equal but doesn't the length of a and b also contribute to the electric field?

    For a < r < b, if I have a Guassian surface (cylinder in this case) then I get:

    EA = ρV / εo

    E2∏rL = ρ∏r2L / εo

    E = ρ / 2 εo
     
  7. Feb 18, 2014 #6

    tiny-tim

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    charge per length is what it says

    the radius (a or b) has nothing to do with it
    what is ρ ? :confused:

    you seem to be calculating a volume

    the charge is a surface charge, on two cylindrical shells
     
  8. Feb 18, 2014 #7

    Kot

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    Oops,

    EA = σA / εo

    E(2∏rL) = σ(2∏rL) / εo

    E = σ / εo

    where σ is the surface charge density of the cylinders. Is this correct?
     
  9. Feb 18, 2014 #8

    tiny-tim

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    what is this supposed to be? :confused:

    (and what does 2πr have to do with it?)
     
  10. Feb 18, 2014 #9

    Kot

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    σ(2∏rL) is σA, the surface charge density of the cylinder. Since I need to find the electric field, I used Gauss' Law ∫EdA = Qenclosed / εo . I replaced Qenclosed with σA.
     
  11. Feb 18, 2014 #10

    tiny-tim

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    But the surface charge depends only on L :redface:
    (and now i'm off to bed :zzz:)
     
  12. Feb 18, 2014 #11

    Kot

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    Could I get a hint on how to do this problem?
     
  13. Feb 19, 2014 #12

    tiny-tim

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    (just got up :zzz:)

    If …
    … then what is the total charge inside a cylinder of length L and radius r (a < r < b) ?
     
  14. Feb 19, 2014 #13

    Kot

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    Would the total charge inside a cylinder be Qenclosed = E2∏rLεo ?
     
  15. Feb 19, 2014 #14

    tiny-tim

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    why?? :confused:

    (and what is E ? and what happened to Q ?)
     
  16. Feb 19, 2014 #15

    Kot

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    I used Gauss's Law. The E is the electric field and Q is the total charge enclosed inside the cylinder.
     
  17. Feb 19, 2014 #16

    rude man

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    Start by setting up a Gaussian surface inside the inner wire ( r < a). What is the shape of the surface?
     
  18. Feb 19, 2014 #17

    Kot

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    The Guassian surface is a cylinder. Since it is symmetrical I don't have to do the integral, I can use EA = Qenclosed / εo. The area of the cylinder would be 2∏rL.
     
  19. Feb 19, 2014 #18

    rude man

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    Right. But what is Qenclosed?
     
  20. Feb 19, 2014 #19

    Kot

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    If I let the Guassian surface (cylinder) to be the same length as the cable, Qenclosed would be the same length as the cable L. I can rewrite Q as λL. Is that right?
     
  21. Feb 19, 2014 #20

    rude man

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    Well, where do you think the charges are located on the inner wire?
     
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