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Homework Help: Electric Charge homework

  1. Jan 13, 2008 #1
    [SOLVED] Electric Charge

    1. The problem statement, all variables and given/known data
    Two point charges q and 4 q are distance L apart and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium. What is the magnitude of the third charge?

    2. Relevant equations
    F = kq1q2 / r^2
    k ~= 8.988x10^9

    3. The attempt at a solution
    Not a clue. All I've figured out is that for this to work in perfect equilibrium, the third charge must be directly in between the first two.

    Not looking for an all-out solution from anyone (Although that'd be cool too), but more looking for just a push in the right direction.

    In theory, it'd involve drawing free body diagrams for the three charges, and then working with the charge q3 to balance the equilibrium... but I can't grasp my head around that.
     
  2. jcsd
  3. Jan 13, 2008 #2

    mjsd

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    yes! just need to go ahread and write down some simultaneous equations and solve them... obviously the variables will be the charge for the unknown/3rd particle, and its distances from each one.
     
  4. Jan 13, 2008 #3
    Particle 3:
    Net Force = 0 = F1 + F2
    -F1 = F2

    -kq1q3/(r13)^2 = kq2q3/(r23)^2
    The problem here is that I don't know the distance between 1 and 3, or between 2 and 3.
    Also, the q3 cancels out on each side. /boggle
     
  5. Jan 13, 2008 #4

    mjsd

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    also need equations for particle 1 and 2, all particles must have net force add to zero, since particle 3 is in between the two, the the two distances (13, 23) should add to L which gives u another equation... I haven't solved this, but that's how you go about it.
     
  6. Jan 13, 2008 #5
    Particle 1:
    Net Force = 0 = F2 + F3
    -F2 = F3

    -kq1q2 / L^2 = kq1q3 / (r13)^2
    -4q / L^2 = q3 / (r13)^2

    Particle 2:
    Net Force = 0 = F1 + F3
    -F3 = F1 (Doing it the other way, since the force would be in the other direction)

    -kq2q3 / (r23)^2 = kq1q2 / L^2
    -q3 / (r23)^2 = q / L^2

    L = r13 + r23
    r13 = L - r23

    -4q / L^2 = q3 / (L - r23)^2
    Unsure where to go from here. Could simplify the above line a little, but then I'd still have two unknowns: q3 and r23.

    Wait, I have an idea..
    -Sqrt((q3*L^2) / -4q) + L = r23
    Which means:
    r13 = L - r23
    r13 = Sqrt((q3*L^2) / -4q)

    Then from the particle 3 formula... plugging in r13 and r23...
    -q3/(r13)^2 = 4q3/(r23)^2
    -q3/(Sqrt((q3*L^2) / -4q))^2 = 4q3/(-Sqrt((q3*L^2) / -4q) + L)^2
    -1/(Sqrt((q3*L^2) / -4q))^2 = 4/(-Sqrt((q3*L^2) / -4q) + L)^2
    -((q3*L^2) / -4q) = ((-Sqrt((q3*L^2) / -4q) + L)^2) / 4
    q3*L^2/q = ((-Sqrt((q3*L^2) / -4q) + L)^2)

    Just checking, am I on the right path? Computing the right side is going to be a HUGE mess, I don't want to do it if I'm already doing things wrong.
     
  7. Jan 13, 2008 #6

    mjsd

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    I haven't checked the last bit but once you have got those equations it is just a matter of solving them.... and method of substition seems to be the way to go.. so you are on right track
     
  8. Jan 13, 2008 #7
    Sigh... here goes nothing.

    q3*L^2/q = (-Sqrt((q3*L^2) / -4q) + L)^2
    q3*L^2/q = 2L^2 - q3*L^2/4q3
    q3*L^2/q = 2L^2 - L^2/4
    q3/q = 2 - 1/4
    q3 = 2q - 1/4q
    q3 = 8/4q - 1/4q
    q3 = 7/4q
    q3 = 1.75q

    Trying this result...
    Incorrect.
     
  9. Jan 13, 2008 #8
    I just wanna point out that I'm almost completely certain that the mistake is mine somewhere along the line. Just saying that the answer I came up with happened to be incorrect, not saying that your advice is off or that. I do appreciate you helping me at this ungodly hour.
     
  10. Jan 13, 2008 #9

    mda

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    I can't see where you've gone wrong, but it is.
    It may be easier to first solve for L concentrating on the added particle.
     
  11. Jan 14, 2008 #10

    mjsd

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    ungodly hour? :rofl:

    Couldn't you see my Avatar?
    :rolleyes:
     
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