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Guass' Law (Conducting Cylinder)

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Two coaxial cylindrical conductors are shown in perspective and cross-section above. The inner cylinder has radius a = 2 cm, length L = 10 m and carries a total charge of Qinner = + 8 nC (1 nC = 10-9 C). The outer cylinder has an inner radius b = 6 cm, outer radius c = 7 cm, length L = 10 m and carries a total charge of Qouter = - 16 nC (1 nC = 10-9 C). What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r = 4 cm from the origin and makes an angle of 30o with the x-axis?

    2. Relevant equations
    E∫dA=qenclosed


    3. The attempt at a solution

    Okay so I found the charge per unit length (λ) which is Lguassσ. λ=0.8 nC. Then I performed the integral and solved for E to get E=qenclo/(2*Pi*ε*r*L). I wasn't sure how to get the x component though but figured it must be just E mag * Cos[theta] but this is wrong. Can anyone help me? I hope I showed enough work to get a response. I can double check my units and everything but I am pretty sure I converted everything to the proper units.
     
  2. jcsd
  3. Feb 3, 2012 #2

    vela

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    If ##\lambda## is charge per unit length, it should have units of C/m.

    What values did you use in the formula?

    Please post the figure the problem is referring to.
     
  4. Feb 3, 2012 #3
    coaxial_1.jpg

    Opps I meant qenclosed =0.8nC

    r=0.04m
    1/(ε*Pi*2)=8.85*10^-12 N*m^2/C^2
    L=10m
     
  5. Feb 3, 2012 #4

    vela

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    Is that what you really meant? ##\epsilon_0 = 8.85\times10^{-12}~\text{N m}^2\text{/C}^2## by itself.
     
  6. Feb 3, 2012 #5
    1/(8.85*10^-12 (Pi)*2)=1.79836*10^10
     
    Last edited: Feb 3, 2012
  7. Feb 3, 2012 #6

    vela

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    Your work looks okay to me. What answer did you get?
     
  8. Feb 3, 2012 #7
    35.8N/C but this is wrong...
     
  9. Feb 3, 2012 #8

    vela

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    Did you multiply by cos 30 since you only want the x-component?
     
  10. Feb 3, 2012 #9
    Yes and still I had the wrong answer.
     
  11. Feb 3, 2012 #10
    Well I multiplied 35.8 N/C by Cos(30deg)
     
  12. Feb 3, 2012 #11

    vela

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    What result did you get?
     
  13. Feb 4, 2012 #12
    31.0028 n/c
     
  14. Feb 4, 2012 #13

    vela

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    It might just be rounding error or the number of significant figures. I get 31.1 N/C to three sig figs.
     
  15. Feb 4, 2012 #14
    I actually entered the formula directly into the online hw thing. I just tried 31.1 N/C and still didn't work. I really dislike the way this hw is set up. Waste of time!
     
  16. Feb 4, 2012 #15

    gneill

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    The OP needs to check his value for linear charge density; it could be that he's using a value that's 10x too small.
     
  17. Feb 4, 2012 #16

    vela

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    Yup, that's it. I grabbed 0.8 nC from one of your earlier posts, but the original problem said 8 nC.
     
  18. Feb 5, 2012 #17
    Yeah thanks, I ended up figuring it out that it should be 8 not .8. This is very frusterating because the online hw assignment said that is was 0.8. So I wasted a lot of time due to that.
     
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