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Point of equilibrium between charges

  1. Jun 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Take two charges, one positive with charge q+, and another with charge q- are at a distance d away from each other. Under what conditions is there a point of equilibrium and, if it exists, where would it be located? Show mathematically. (Hint: use the quadratic equation and comment about the discriminant.)

    2. Relevant equations

    Coulomb's Law: Fe=keq1q2/r2

    3. The attempt at a solution

    For equilibrium to exist (with the positive charge a distance x from the origin and the negative charge therefore a distance (d-x) from the origin), F+=F-, so ke*(q1*q2)/x2=ke*(q1*q2)/(d-x)2.

    I simplify this to (d-x)2=x2

    Simplifying more, d-x=x, and x=d/2, meaning that there's equilibrium halfway between the charges. Though I'm VERY unsure of this. Could someone tell me if I'm headed the right direction?
     
  2. jcsd
  3. Jun 26, 2012 #2
    Hi pilotguy! Welcome to PF! :smile:

    q1 and q2 are not the same for both the equalities. Let the test charge be q' . On one side is the repulsive force, and on the other side is an attractive force.
     
  4. Jun 26, 2012 #3
    So does that mean that it's Frep=ke*(q'*q+)/x^2 and Fattr=ke*(q'*q-)/(d-x)^2?

    Then equate them and solve for x?
     
  5. Jun 26, 2012 #4
    Do you think the equilibrium point will lie halfway?
    Think about it, if you place a charge between the charges q+ and q-, there will never be a situation where Frep = Fattr. Can you see why is this so?
    Make a diagram and place a charge midway and draw the direction of forces.
     
    Last edited: Jun 26, 2012
  6. Jun 26, 2012 #5
    Oh, I think I see what you're saying. If the + charge is on the left, and the - charge is on the right, with a positive test charge between them, the test charge will be repelled from the positive charge and attracted toward the negative charge, meaning that both forces would act to the right, so there wouldn't be equilibrium.

    Am I on the right track?
     
  7. Jun 26, 2012 #6
    Yes! There won't be an equilibrium position in between the charges. Think outside the box(charges)? :wink:
     
  8. Jun 26, 2012 #7
    So there's no equilibrium between them; does that mean that there's only equilibrium at +/- infinity? That would reduce the force to essentially zero. I'm a little lost.
     
  9. Jun 26, 2012 #8
    Seems so. This is basically an electric dipole. Field due to dipole is zero at infinity.
     
  10. Jun 26, 2012 #9
    Yep! :approve:
     
  11. Jun 26, 2012 #10
    I see. Any guess as to what was my professor trying to get at with his hint about the discriminant in the quadratic equation?
     
  12. Jun 26, 2012 #11
    Yes. You would get an negative sign under the square root, while solving the quadratic(imaginary roots)
     
  13. Jun 26, 2012 #12
    Perfect! Thanks for the help!
     
  14. Jun 26, 2012 #13

    ehild

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    Are those charges q+ and q- equal in magnitude?

    ehild
     
  15. Jun 26, 2012 #14
    I'm not sure; the question is a little ambiguous on that. Does it make a difference to the solution from earlier in the thread?
     
  16. Jun 26, 2012 #15

    ehild

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    Yes, it makes a difference. There can be a neutral point outside the segment d. Try q+=12 mC and q-=-3 mC, and d=1 m, for example.

    ehild
     
  17. Jun 27, 2012 #16
    I think it is to be interpreted as a '+q' charge and a '-q' charge, making them equal in magnitude.
     
  18. Jun 27, 2012 #17

    ehild

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    Then it would have been written as +q and -q, instead of writing "one positive with charge q+, and another with charge q-" and the problem were not be a real problem with discriminant and everything. Try to solve it with charges of different magnitudes. It is a good practice. :smile:

    ehild
     
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