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Uniform rod problem

  1. Sep 6, 2007 #1
    _______-^y
    ________l
    ________l
    ________l
    ________l
    ________________
    l________________l--------> x

    Sorry for the bad picture, but suppose there's a uniform rod of charge has a length L and a net charge +Q, find the force that this rod exert on a point charge q placed at (0,y).

    Ok, I found by symmetry that the F(x) is 0.
    [tex]\lambda[/tex]=dQ/dl

    So, F(y) = [tex]\int[/tex] k[tex]\lambda[/tex]qcos[tex]\Theta[/tex](dl)/ (l^2 + y^2) , [-L/2, L/2]

    And since y/r=cos[tex]\Theta[/tex], r= ysec[tex]\Theta[/tex], where r = (l^2 + y^2)

    Then it boils down to 2kqy[tex]\lambda[/tex][tex]\int[/tex] (dl)/ (y^3(sec[tex]\Theta[/tex])^3) , [0, L/2]

    Then I don't know how to carry on without using the integral table....
    But it should be something like this and i don't know why:
    F(y) = 2kqy[tex]\lambda[/tex][tex]\int[/tex](y(sec[tex]\Theta[/tex])^2d[tex]\Theta[/tex])/ (y^3(sec[tex]\Theta[/tex])^3)
    and gives 2kqy[tex]\lambda[/tex][tex]\int[/tex](cos[tex]\Theta[/tex]) d[tex]\Theta[/tex]
    ............. ans= kqQ/ y[tex]\sqrt{y^2 + (L/2)^2}[/tex]

    Anyways, my question is, without using the integral table, how do people actually move on to that next step?

    Thanks
     
    Last edited: Sep 7, 2007
  2. jcsd
  3. Sep 7, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Get rid of those angles! You should have an expression for dF in terms of x & y (I use x as the coordinate of the line charge element):
    [tex]dF = \frac{kq\lambda}{(y^2 + x^2)}\cos\theta \; dx = \frac{kq \lambda x}{(y^2 + x^2)^{3/2}} \; dx[/tex]

    You should have no problem integrating this without a table.
     
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