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The Electric Field Produced by a Finite Charged Wire XD

  1. Jul 27, 2009 #1
    http://img22.imageshack.us/img22/1958/physicsl.jpg [Broken]

    My attempt,
    After all the integrals I've got the final equation

    (1/4piE0) * |q| / ( d * sqrt ( d^2 + (L/2)^2) )

    i'm not too sure how to express it how the question asks,

    please help!

    the bit i'm confused on, is the ( d * sqrt ( d^2 + (L/2)^2) )

    can it also be seen as (d * d) + (l/2)
    or is it d * (d+(l/2))

    bah
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 27, 2009 #2
    d * (sqrt ( d^2 + (l/2)^2) ) = d*(L/2)?
     
  4. Jul 27, 2009 #3
    no because that's the thing if it becomes infinitely long and i'm dealing with a finite length
     
  5. Jul 27, 2009 #4

    Cyosis

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    Not sure what you're trying to do here. The question simply asks you to give the answer in terms of L, lambda, d and k. You're missing a k and a lambda, use the definition of k and lambda to get them into your expression.
     
  6. Jul 27, 2009 #5
    yes I am not a total retard as i was able to make the integral to the final equation.
    anyway lol, I just don't know how to convert the

    |q| / ( d * sqrt ( d^2 + (L/2)^2) )


    part of the equation, in terms of lambda and d

    -
     
  7. Jul 27, 2009 #6

    Cyosis

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    What is the definition of k and what is the definition of lambda?
     
  8. Jul 27, 2009 #7

    k=1/4piEo
    lambda=q/L

    i don't know how to make lambda from q / d*sqrt(d^2+(L/2)^2)
     
  9. Jul 27, 2009 #8

    Cyosis

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    So if [itex]\lambda=q/l[/itex] then q=....? Note that the l you're using here is not the same l as in your problem, but a general symbol for length. Perhaps it is wise to show us your integration.
     
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