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Force equalibrium

  1. Jan 16, 2006 #1
    Three point charges are arranged along the x-axis. Charge q_1=−4.05 nC is located at x_1 = 0.195 m, and charge q_2 = 2.80 nC is at x_2=−0.340 m A positive point charge q_3 is located at the origin.Where along the x-axis can q_3 be placed and the net force on it be zero, other than the trivial answers of x= \pm \infty?
  2. jcsd
  3. Jan 16, 2006 #2
    show some work.

    how do you think it should be solved?
  4. Jan 17, 2006 #3
    [tex]\frac{4.05* q_3}{(-0.34-x)^2} =\frac{-2.8 * q_3}{(0.195+x)^2} [\tex]

    I dont know why it doesnt work
    Last edited: Jan 17, 2006
  5. Jan 18, 2006 #4
    may anyone?
  6. Jan 18, 2006 #5
    draw the diagram:


    if x is some value, then to find the distance, you have to use the distances as (x-x2) and (x1-x)
  7. Jan 18, 2006 #6
    how do you know if x will land in the interval between [tex] q_1 and q_2 [\tex]
    the problem i get is when i assume the q3 will at equalibrium, x become imaginary when i take the square root

    2.8/x^2 == -4.05 / (0.535+x)^2

    i have shifted the system to the right. Then the negative sign became PAIN under the square root
    Last edited: Jan 18, 2006
  8. Jan 18, 2006 #7

    i'm sorry, i didn't read the question carefully enough. as far as i can tell, it has to be either left of q_2 or right of q_1. just from eyeballing the figure, it looks like it will be to the left of q_2 because the smaller repelling charge needs to be closer to cancel out the larger attracting charge.
  9. Jan 18, 2006 #8
    nvm i got it.
    Thank you everyone
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