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Electric Fields of straight wire

  1. Jan 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A very long, straight wire has charge per unit length 1.47×10^10

    At what distance from the wire is the electric field magnitude equal to 2.57 N/C


    2. Relevant equations

    E = lambda / (2*pi*E_o*r)

    E_o = 8.85*10^-9


    3. The attempt at a solution

    2*pi*E_o*E / lambda = r

    Is this correct so far?
     
  2. jcsd
  3. Jan 26, 2007 #2

    Kurdt

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    Looks fine you just need to plug the numbers in.
     
  4. Jan 26, 2007 #3
    r = 2*pi*(8.85*10^-9)*(2.57) / (1.47*10^-10)

    I'm not sure if lambda is represented correctly and is E just 2.57 or should it be 10^(something)
     
  5. Jan 26, 2007 #4

    Kurdt

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    E will just be 2.57 as stated in the problem. Why are you worried about lambda?
     
  6. Jan 26, 2007 #5
    I wasn't sure if lambda = 1.47*10^-10
     
  7. Jan 26, 2007 #6

    Kurdt

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    probably more likely to be x10-10 than the other way round.
     
  8. Jan 26, 2007 #7
    Thats what I had before 1.47 x 10^-10
     
  9. Jan 26, 2007 #8

    Kurdt

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    What is it in the question?
     
  10. Jan 26, 2007 #9
    The original question was if lambda = 1.47 x 10^-10.

    You said yes, so I just wanted to make sure.
     
  11. Jan 26, 2007 #10
    2*pi*(8.85*10^-9)*(2.57) / (1.47*10^-10) = r

    This evaluated to:
    r = 972

    The online program says I'm off by an additive constant??
     
  12. Jan 26, 2007 #11

    Kurdt

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    I've just noticed you have E on top of the fraction and lambda below. You need to swap these two so the equation is:

    [tex] r=\frac{2k\lambda}{E} [/tex]

    Like I said in a previous thread, try manipulating equations with just their symbols until the very last moment. Its a lot easier to spot problems that way.

    EDIT: Sorry [tex] k=\frac{1}{4\pi \epsilon_0} [/tex]
     
    Last edited: Jan 26, 2007
  13. Jan 26, 2007 #12
    2* (1/4*pi* 8.85*10^-9) * (1.47*10^-10) / 2.57

    = 1.03×10−3

    Still says I'm off by a additive constant.
     
  14. Jan 26, 2007 #13

    Kurdt

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    I think you've made a mistake in the calculation as i get a different answer. Try it again you're 3 orders of magnitude out.
     
  15. Jan 26, 2007 #14

    Kurdt

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    Where did you get that value of epsilon nought from? It should be: 8.85x10-12
     
  16. Jan 26, 2007 #15
    I need to be more careful from paper to online input. I have so much scratch work, some how I changed the epsilon value =[
     
  17. Jan 26, 2007 #16

    Kurdt

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