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Electric Field and an Uniformly Charge Plastic Pipe

  1. May 22, 2006 #1
    A plastic pipe has an inner radius of a = 35.00 cm and an outer radius of b = 71.00 cm. Electric charge is uniformly distributed over the region a < r < b. The charge density in this region is 30.00 C/m3

    A)Calculate the magnitude of the electric field at r = 0.44 m.

    B)Calculate the magnitude of the electric field at r = 1.59 m.



    I need some help on where to begin. I know the e-field for cylindrical sym. is E=2kQin/rL

    I also know that Qin=charge density * Volume.

    How do I find the volume so I can find the Qin in both A and B?
     
  2. jcsd
  3. May 22, 2006 #2

    siddharth

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    Hint: Gauss's Law

    Have you figured out what gaussian surface to use? Once you do that, you'll be able to calculate the charge enclosed.
     
  4. May 22, 2006 #3
    I dont understand
     
  5. May 22, 2006 #4

    siddharth

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    What don't you understand?
     
  6. May 22, 2006 #5
    I don't understand those formulas, we learned E=2kQin/rL. Was my approach the wrong way of doing it. Finding Qin and then using E=2kQin/rL to get the answer.
     
  7. May 23, 2006 #6

    siddharth

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    Yes, but more important than the formula is the underlying principle. Do you know how you got that formula?
     
  8. Feb 14, 2007 #7
    Sorry to revive this old thread, but it is the exact same problem I am working on.

    What is the formula used to solve the equation? I thought all I would have to do is E = lambda/2pi*epsilon*r^2 but I was wrong.

    In the previous posts, in the equation E=2kQin/rL, what is L? Also, how do I find the volume so I can find the Qin in both A and B? Thank you.
     
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