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Forces between 3 charged particles

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    What is the force F_vec on the 1 nC charge?
    Give answer as magnitude in Newtons.


    2. Relevant equations
    F = K qq/r^2
    K = 1/4(pi)epsilon_knot; = 8991804694 N m^2/C

    3. The attempt at a solution
    A. Convert it all. cm to m. nC to C.
    B. Vectors for Forces. The two x-component forces from the two 2 nC charges are equal and opposite (both repelling the 1 nC charge) so they cancel.
    So F(x_net) = 0 Newtons
    C. Similarly the y-component forces are the equal and in the same direction. So I merely found one and doubled it.
    To obtain y, I used the ole: y = sqroot[ (.01m)^2 - (.01m/2)^2) = .00866m

    F = Fnet_y = 2F_y = [2 * (8.99 Nm^2/C^) * (2E-9 C)(1E-9 C)] / (.00866m)^2
    F = 4.80 * 10^-4 N

    where did i go wrong? thanks
     
  2. jcsd
  3. Sep 22, 2007 #2
    F = Fnet_y = 2F_y = [2 * (8.99 Nm^2/C^) * (2E-9 C)(1E-9 C)] / (.00866m)^2

    means that two charges are separated by 0.00866 m, according to definition of this formula. But are they really separated by that distance?

    You are saying here that move both 2 nC charges to the midpoint of the bottom line!
     
  4. Sep 22, 2007 #3
    cool i got it now. thanks
     
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