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A problem about Coulomb's Law

  1. Feb 4, 2005 #1
    two charges, -Q and -3Q are a distance r apart. These two charges are free to move but do not because there is a third charge nearby. What must be the charge and placement of the third charge for the first 2 to be equilbrium?

    The answer is:
    The charge is 0.4Q
    and the distance is 0.37l from -Q toward -3Q

    Can someone explain to me?
     
    Last edited: Feb 5, 2005
  2. jcsd
  3. Feb 4, 2005 #2
    You're gonna have to use a bunch of Coulomb's law expressions and equal them. The repulsion of the two charges is gonna half to be equal to the attraction they feel toward the third charge. In the denominator (distance squared), use x for the distance between the -Q charge and the third charge and (r-x) for the distance between the -3Q charge and the third charge (so U only have 1 unknown).
     
  4. Feb 5, 2005 #3
    I still dont quite understand thsi, can you explain it more clearly?
    Like the "attraction" and "replusion" that you are talking about, do they mean force? So i can just use F = (k*Q(I)*Q(II)) / r^2 to find right?
     
  5. Feb 5, 2005 #4
    I re-read the question several time and I find the word "equilibrium" very misleading.
    Does it mean that the third charge will balance the charge between -Q and -3Q? Or does it make the net force of both -Q and -3Q become zero?
     
  6. Feb 5, 2005 #5
    I agree. It will not be an equilibrium, a small deviation of one of the charges will result in an unstable situation.
     
  7. Feb 5, 2005 #6
    The word "equilibrium", in this case, simply means that the charges won't be moving, that their attaction for one charge will be cancelled out either by their attraction for another charge on the other side (opposite direction) or by their repulsion for another charge on the same side. Let me know if you still don't get it. :smile:
     
  8. Feb 5, 2005 #7
    I still dont understand the problem
    I have 3 variables but only 2 equations.
    And I want to know whether or not the net force of -Q and -3Q is 0?
     
  9. Feb 5, 2005 #8
    The distance is 0.371 what? r?
     
  10. Feb 5, 2005 #9
    yes, that was a typo
    it should be 0.37r
     
  11. Feb 6, 2005 #10

    s_a

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    To keep the two negative charges -Q and -3Q in equilibrium, you will need a positive charge somewhere in between (say having charge +nQ, and a distance pr from the charge -Q) to keep them together. You need to find n and p (we know n>0 and 0<p<1).

    Firstly the forces exerted on the -Q charged particle from the +nQ and the -3Q need to be equal in magnitude:
    -> -Q * nQ / (pr)^2 = Q * 3Q / r^2 (permittivity cancels out)
    -> n/p^2 = 3
    -> n = 3p^2 .. (A)

    Secondly, the forces exerted on the -3Q charged particle from the +nQ and -Q need to be equal in magnitude too:
    -> 3Q * nQ / ((1-p)r)^2 = 3Q * Q / r^2
    -> n = (1-p)^2 .. (B)

    Thirdly, the forces exerted on the +nQ charged particle from the -Q and -3Q need to be equal in magnitude as well. You don't need to bother with Coulomb's Law here though, since Newton's 3rd Law (for every action there's an equal opposite reaction) guarantees the net force on the +nQ charged particle is zero provided the above two conditions above are met. If you do apply Coulomb's Law, you'll get nothing new.

    solving A and B simultaneously,

    3p^2 = (1-p)^2
    +- p*sqrt(3) = 1 - p
    p(1 +- sqrt(3)) = 1
    p = 1/(1 +- sqrt(3))
    = (-1 +- sqrt(3))/2

    The positive root is p = 0.37, so the distance of +nQ from the charge -Q towards -3Q is 0.37r. The negative root of p is meaningless since the positively charged particle MUST be between the two negatively charged particles. Also n=0.4 from the above equations A and B, so the positive charge has magnitude 0.4Q.
     
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