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Insulating Cylindrical Shell

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data A long cylindrical insulating shell has an inner radius of a = 1.38 m and an outer radius of b = 1.61 m. The shell has a constant charge density of 4.80 C/m3.
    1. What is the magnitude of the electric field at a distance of 1.87 m from the axis?
    2. What is the magnitude of the electric field at a distance of 1.51 m from the axis?
    3. If we take the potential at the axis to be zero, what is the electric potential at the outer radius of the shell?

    2. Relevant equations
    [tex] E=\frac{kq}{r^2}[/tex]
    [tex] \oint \vec{E}\vec{dr}= \frac{Q_{Enclosed}}{\epsilon_{o}}[/tex]

    3. The attempt at a solution
    For 1 i use [tex] E=\frac{kq}{r^2}[/tex] but i am confused on finding q because i am given density of c/m^3. I am assuming an infinity long cyclinder?[tex]\vec{}[/tex]
     

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  3. Sep 15, 2007 #2

    learningphysics

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    careful.. the integral should be E.da, not E.dr

    Take a cylindrical section of arbitrary length L...
     
  4. Sep 15, 2007 #3
    You are correct dA.
    So:
    [tex] dE= \frac{kdQ}{r^2}, dQ=pdV \longrightarrow dQ= 4.80 C/m^3 * 2\pi r l dr[/tex]
     
  5. Sep 15, 2007 #4
    Then:
    [tex] E= \int \frac{k 4.80* 2 \pi r l}{r^2}dr[/tex]
    With he boundries a=a b=b
     
  6. Sep 15, 2007 #5

    learningphysics

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    No... use gauss' law... what is your gaussian shell... what is the charge enclosed...
     
  7. Sep 15, 2007 #6
    Oh of course..I would pick a cyclinder as my G surface.
     
  8. Sep 15, 2007 #7

    learningphysics

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    exactly. can you write out the integral using gauss law for the first part? we need the field 1.87 m away.
     
    Last edited: Sep 15, 2007
  9. Sep 15, 2007 #8
    Ok, thanks. i got 1 & 2.
     
  10. Sep 15, 2007 #9
    how about 3?
    I know [tex] V=\frac{kq}{r}[/tex]
     
  11. Sep 15, 2007 #10

    learningphysics

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    Nah... I don't think that'll work...

    The voltage at r minus the voltage at 0 is:

    [tex]-\int_0^r{\vec{E}\cdot\vec{dr}[/tex]

    You need to integrate this from 0 to b=1.61...
     
  12. Sep 15, 2007 #11
    But then i get a ln(r/0), with other stuff infront. ln is -inf?
     
  13. Sep 15, 2007 #12

    learningphysics

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    That doesn't look right... the integral from 0 to 1.38m is 0... because the field is 0 for this part... so you just need the integral from 1.38 to 1.61.
     
  14. Sep 16, 2007 #13
    so does:
    [tex] V=\frac{-p (b^2-a^2)}{2 \epsilon_{o}}ln(\frac{b}{a})[/tex]
    sound on the right track?
     
  15. Sep 16, 2007 #14

    learningphysics

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    Hmm... I'm getting something different. Can you show how you got that integral?
     
  16. Sep 16, 2007 #15
    Well how should I do it?
     
  17. Sep 16, 2007 #16

    learningphysics

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    Get the field inside the insulator. Then do [tex]-\int_{1.38}^{1.61}\vec{E}\cdot\vec{dr}[/tex]

    What do you get for the field in terms of r?

    Use gauss' law. [tex]4.80*{\pi}r^2h = 2{\pi}rh*E[/tex]
     
    Last edited: Sep 16, 2007
  18. Sep 16, 2007 #17
    mmm.. i Think it should be [tex]E= \frac{-p(b^2-a^2)}{2r \epsilon_{o}}[/tex]
    I choose [tex] -p(b^2-a^2)=Q_{enclosed}[/tex] because i had to integrate to find the charge enclosed.
     
  19. Sep 16, 2007 #18

    learningphysics

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    But the charge enclosed changes with r. You're integrating from r = 1.38 to 1.61... as you increase r, the charge enclosed becomes bigger.
     
  20. Sep 16, 2007 #19
    so b is replaced with r
     
  21. Sep 16, 2007 #20

    learningphysics

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    Ah... yes, you're right... I made a mistake in my post. I should have written:

    [tex]4.80*{\pi}(r^2 - a^2)h = 2{\pi}rh*E[/tex]

    So that gives what you have by replacing b with r.
     
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