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Find the force of the rod on the point charge

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A rod of length L and total charge Q is a distance D from a point charge q which lies along the
    perpendicular bisector of the rod. Find the force of the rod on the point charge


    2. Relevant equations

    E = ∫ dE

    dq = λdx

    F = qE
    3. The attempt at a solution

    I'm trying to integrate this does the second integral of j go to zero?


    [tex](kQ/L) [i\hat D \int _0L dy/(D^2+y^2)^(3/2) -j\hat \int _0^L ydy/ (D^2+y^2)^(3/2)][/tex]

    sorry for the equation I'm still learning to use latex
     
  2. jcsd
  3. Apr 25, 2012 #2

    collinsmark

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    Gold Member

    The second integral should go to zero due to symmetry. But I don't think you're setting up the limits of integration correctly. The problem statement says the test charge q "lies along the perpendicular bisector of the rod."

    In other words, set up the system in your mind such that the test charge is on one of the axis, a distance D from the origin, and the the rod lies on a different (perpendicular) axis such that the center of the rod is at the origin. Then integrate along the rod from -L/2 to L/2.
    See the link in my signature for a primer in [itex] \LaTeX [/itex] btw.
     
    Last edited: Apr 26, 2012
  4. Apr 25, 2012 #3

    collinsmark

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    By the way, as a [itex] \LaTeX [/itex] example, I think the equation you were trying to type in, with limits of integration corrected, is (Right lick on it to see the LaTeX source):

    [tex] \vec F = \frac{kQq}{L} \left[\int_{-L/2}^{L/2} {\frac{D \ dy}{\left(D^2+y^2 \right)^{\frac{3}{2}}} \hat \imath} + \int_{-L/2}^{L/2} {\frac{y \ dy}{\left(D^2+y^2 \right)^{\frac{3}{2}}} \hat \jmath} \right][/tex]
     
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