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Net electrostatic force : four particles form a square

  1. Jan 12, 2006 #1
    I apologize in advance - I am completely clueless about this one. I thought I had it figured out, but it turns out that I think I don't know where to even start.
    Four particles form a square. The charges are q1 = q4 = Q and q2 = q3 = q. What is Q/q if the net electrostatic force on particles 1 and 3 is zero?
    [​IMG]
    See, I thought it would work out to be 1 since I thought q and Q must be the same for the force to be zero. Apparently I am mistaken. So... show me start to finish pretty please??? :biggrin:
     
  2. jcsd
  3. Jan 12, 2006 #2

    daniel_i_l

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    Gold Member

    Your right that Q and q have to have oppisite sighns for the forces to cancel out. But you have to show some work first and eaven the we almost never don't give full answers.
    Hint: Just draw out the force components and show when the total on each axis is zero.
     
  4. Jan 12, 2006 #3
    But I thought this was one of those that you don't really do work - it's logic. Say Q is 2. q has to be -2. Q/q=-1? But that's not right either...

    I tried figuring it out this way:
    The distance from particle 1 to 4 is sqrt(2a^2), so the force there is k(2Q/(sqrt(2a^2)^2) which simplifies to k(Q/a).
    The combined forces of 2 and 3 will pull 1 in at a 45 degree angle to 2 or 3, which is the same direction 4 is pushing away at.
    So, for 1: F=k(Q*q/a^2), except we have to take the 45 degrees into account, so F=k(2*Q*q*cos(45)/a^2).
    Set them equal, and we have:
    k(Q/a)=k(2*Q*q*cos(45)/a^2)
    Q=2*Q*q*cos(45)/a
    q=a/2cos(45)

    Then for particle 3:
    F=k(2q/(sqrt(2a^2)^2)=k(q/a)
    F=k(2*Q*q*cos(45)/a^2)
    k(q/a)=k(2*Q*q*cos(45)/a^2)
    Q=a/2cos(45)

    Q=a/2cos(45)=q, so Q/q is 1. ???
     
  5. Jan 13, 2006 #4

    daniel_i_l

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    Gold Member

    It should be:
    k(Q^2/(sqrt(2a^2)^2).
     
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