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Electric field generated by a charged wire

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A charge per unit length +λ is uniformly distributed along the positive y-axis from y = 0 to y = +a. A charge per unit length −λ is uniformly distributed along the negative y-axis from y = 0 to y = −a. Write an expression for the electric field at a point on the x-axis a distance x from the origin. (Use the following as necessary: k, λ, x, and a.)


    2. Relevant equations

    E = k|q|/r2

    I derived this equation by treating the ends of the wire as point charges.
    Ey = 2kqa/(y2+x2)3/2

    3. The attempt at a solution

    I quickly determined that the field vectors x and k would be 0.

    For the y vector:

    dEy = 2ak(dq)/(y2+x2)3/2

    dq/dy = Q/2a

    ∫dEy = (4akQ/2a)∫dy/(y2+x2)3/2
    = 2kQy/[x2(y2+x2)1/2]


    2kλy/[x2(y2+x2)1/2]


    The limits of integration are from 0 to a.


    I am pretty sure this is not correct, and have been working at it for a while, so I would appreciate some help.
     
  2. jcsd
  3. Feb 9, 2013 #2

    haruspex

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    On the RHS, part of the expression is valid for any y in the range, but part of it is only valid at the endpoints. You later treat it as though it's valid over all y.
     
  4. Feb 9, 2013 #3
    I think I understand what you are talking about. Here is my revised process.

    Ey = kqy/[(x2+y2)3/2]

    dq = λdy

    dEy = kλydy/[(x2+y2)3/2]

    The limits of integration are from -a to a. I choose to integrate from 0 to a and multiply the result by two.

    2∫dEy = 2kλ∫ydy/[(x2+y2)3/2]
    = -2kλ/[(x2+y2)1/2]
    =-2kλ/[(x2+a2)1/2]
     
  5. Feb 9, 2013 #4

    haruspex

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    Better. But it's a definite integral. y should have turned into a.
     
  6. Feb 9, 2013 #5
    =-2kλ/[(x2+a2)1/2]

    How much better? Am I on the right track or am I already there?
     
  7. Feb 9, 2013 #6
    Actually, I just input that answer to webassign and it didn't accept it. I am kind of frustrated. lol
     
  8. Feb 9, 2013 #7

    haruspex

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    ... and a definite integral has two bounds.
     
  9. Feb 9, 2013 #8
    ok. New answer for limits of integration 0 to a:

    -2kλ/[(a2+x2)1/2]+2kλ/x
     
  10. Feb 9, 2013 #9
    Here is an attachment showing all of my work.
     

    Attached Files:

  11. Feb 9, 2013 #10

    haruspex

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    That's what I get.
     
  12. Feb 9, 2013 #11
    Yes, this worked. Thanks for your help.
     
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