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Coulomb's Law, what charge will make the two charges in static eq.

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data
    In the figure below the charge in the middle is
    Q = -3.7 nC. For what charge
    q1 will charge q2
    be in static equilibrium?



    2. Relevant equations
    F = (K*q1*q2)/(r^2)


    3. The attempt at a solution

    I'm not sure where to go after acknowledging that Fq1->q = -Fq2->q
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2013 #2

    SammyS

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    attachment.php?attachmentid=56732&d=1363310960.png


    Let's see ...

    q1 is twice the distance from q2 that Q is from q2, and we have an inverse square law.
     
  4. Mar 14, 2013 #3
    So that means the force is (3.7^2) = ~14 nC?
     
  5. Mar 14, 2013 #4

    SammyS

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    The unit for force is not Coulombs .

    Why would you square the charge anyway ?
     
  6. Mar 14, 2013 #5
    I misunderstood the inverse square law.

    F=q1=q2=k*q1*q2/r^2 = k*q1*Q/(r/2)^2

    (9x10^9)*(q1*q2)/400 = (9x10^9)*q1*(3.7)/100

    q1*q2/400 = q1*0.037
    q2 = 14.8 N = F

    Is that correct?
     
  7. Mar 14, 2013 #6

    SammyS

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    Doubling distance reduces force to 1/4 , so to compensate for that, charge must be 4 times what is would be at Q . (Of course with opposite sign.)

    So, is 14.8 = (4)(3.7) ?

    ... Yes. So you're O.K.
     
  8. Mar 14, 2013 #7
    Oh! Wow, that's a lot simpler than I was trying to make it. Thanks!
     
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