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Force exerted by a charged rod on a point charge

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the electric force exerted on a negative point charge –Q located a distance d away from a line segment of charge, extending from –L/2 to +L/2 along the x-axis. The line segment has a non-uniform charge distribution, lambda(x)=lambda_0*x .


    2. Relevant equations
    F= q1q2/4pi(epsilon_0)r^2


    3. The attempt at a solution


    Fy=0

    Fx= -Q/4*pi*epsilon_0[the integral from 0 to -L/2 of(lambda_0*x/(d+L/2+x)^2) + the integral from 0 to L/2 of(lambda_0*x/(d+L/2-x)^2)]

    Notice the -x in the distance in the second integral. This is just my attempt at the question. I have no clue whether it is right or wrong. I feel the need to express the equation into 2 different integrals because you have to show the distance from the origin. Well at least I think so. Any thoughts would be very helpful!
     
  2. jcsd
  3. Sep 27, 2011 #2

    SammyS

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    Is charge, -Q, located on the x-axis, or is it located a distance d from the center of the rod on the y-axis?
     
  4. Sep 27, 2011 #3
    -Q located some distance d from the end of the rod in the positive direction ON the x-axis.
    so Fy=0
     
  5. Sep 28, 2011 #4
    hi split the integral in two parts. since the line charge depends on x coordinate, the charge would be negative for x < 0 and positive for x > 0
     
  6. Sep 28, 2011 #5
    That's what I've done but I'm not sure if I did it correctly.
     
  7. Sep 28, 2011 #6

    SammyS

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    It appears that you have the rod of length, L, centered at the origin. One end of the rod is at x = L/2. The charge is at distance, d, from L/2 in the positive x-axis, thus it's located at x = (L/2) + d .

    There is no need to break up the integral the way you did. The distance from any point at x on the rod to the charge is (L/2) + d - x . Algebra will take care of x being negative.

    Also: By switching the limits of integration, like you did in your first integral, you may introduce an unintended change of sign in that integral.
     
  8. Sep 28, 2011 #7
    Thanks a lot, I figured that out about 10 minutes ago but it's nice having it confirmed!
     
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