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Homework Help: Strength of Electric Field

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Halliday and Resnick edition 7 chapter 22 number 29

    In Fig 22-45, positive charge q = 7.81 pC is spread uniformly along a thin nonconducting rod of length L = .145m. The y distance between the point and rod is R = .06m. What is the magnitude and direction of the electric field produced at point P.

    I will attempt to describe the figure:

    There is a thin nonconducting rod in the x dimension with length L and a point P is directly above the center of the rod in the y dimension (R = .06m).

    2. Relevant equations

    E = (1/4[tex]\pi[/tex][tex]\epsilon[/tex])(q/r[tex]^{2}[/tex])

    3. The attempt at a solution

    [tex]\int[/tex]dE = 1/4[tex]\pi[/tex][tex]\epsilon[/tex][tex]\int[/tex][tex]\lambda[/tex]Rdx/(R[tex]^{2}[/tex]+x[tex]^{2}[/tex])[tex]^{3/2}[/tex]
    I then took lamba and R out of the integral since they are constants
    (4 is not rasied to ([tex]\pi[/tex][tex]\epsilon[/tex])

    After taking the integral I got:


    When I plug in for the varibles I do not get the correct answer, I get 7.48 n/C but the correct answer is 12.4 N/C

    I plug in L for x and R for R and for [tex]\lambda[/tex] I use q/L
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 4, 2009 #2


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    Homework Helper

    Hi glennpagano44! :smile:

    (have a lambda: λ and an epsilon: ε and a pi: π and a square-root: √ :wink:)
    No … how can you still have an x when you've just eliminated x by integrating over it?

    And where did that extra R come from? :confused:
  4. Feb 4, 2009 #3
    I went back to class today and I figured it out thanks alot. I just integrated it wrong, I had to do some trig subsitutions. (that is where the extra R came from)
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