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Use integration to find the total force on a point HELP!

  1. Jan 13, 2009 #1
    http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/newerlib/Graphics/Gtype51/prob19a.gif[/URL]

    1. A charge Q = 8.10 ×10-4 C is distributed uniformly along a rod of length 2L, extending from y = -11.4 cm to y = +11.4 cm, as shown in the diagram 'on your assignment above. A charge q = 1.80 ×10-6 C, and the same sign as Q, is placed at (D,0), where D = 17.5 cm.
    Use integration to compute the total force on q in the x-direction




    It is already proven that :
    The magnitude of the force on charge q due to the small segment dy is
    dF=(kqQ/2Lr2)dy.




    3. Given whats already proven that for a small dy:
    dF = (kqQ/2Lr2)dy

    so i integrated both sides and i got:
    ∫dF = k*q*Q / 2L ∫ 1/r^2 dy

    since r^2 = D^2+y^2

    therefore the function becomes:
    ∫dF = k*q*Q / 2L ∫ 1/D^2+y^2 dy

    now what should i do next???? and the equation i derived, was it right at the first place???


    THANKS!
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 13, 2009 #2

    Andrew Mason

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    Last edited by a moderator: Apr 24, 2017
  4. Jan 13, 2009 #3
    oh, i see...
    so now since only the x-component is considered then
    cos theta = D/r
    and the function becomes:
    dF=(kqQ/2Lr^2)dy * D/ r ... is that correct???

    then further integrating the above gives:

    F= k*q*Q*D^2/2L* ∫1/(D^2+y^2)^-2/3 8 dy??? -- did i do this step correct??
     
  5. Jan 14, 2009 #4

    Andrew Mason

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    Not quite.

    [tex]F = \frac{kqQD}{2L}\int_{-L}^{L}\frac{1}{r^3} dy = \frac{kqQD}{2L}\int_{-L}^{L}\frac{1}{(\sqrt{D^2 + y^2})^3} dy =\frac{kqQD}{2L}\int_{-L}^{L}(D^2 + y^2)^{-\frac{3}{2}}dy[/tex]

    Good luck working out that integral

    AM
     
  6. Jan 14, 2009 #5
    thanks, i solved it :)
     
  7. Jan 15, 2009 #6
    Last edited by a moderator: May 3, 2017
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