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Electric potential

  1. Sep 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Initial, 3 charges q are separated r at vertices of the equilateral triangle. Then they move freely to infinity.
    Find the velocity of charge when they are in infinity.
    2. Relevant equations
    If we seem 1 charge is affected by potential of 2 charges, we have:
    so ##v=2q\sqrt{\dfrac{k}{mr}}##
    If we seem that they are a system. The enery of system charge is:
    And potential enegy will change to dynamic enegy.
    $$ \frac{3kq^2}{r}=\frac{3}2mv^2$$
    so ##v=\sqrt{2}q\sqrt{\dfrac{k}{mr}}##
    3. The attempt at a solution
    Where is wrong?
    Thanks for helping
  2. jcsd
  3. Sep 10, 2016 #2
    Do you have the answer for this problem? It's the first or the last one?
  4. Sep 10, 2016 #3


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    You seem to be certain that your answer is wrong, so you must have a book answer at hand. Right ?

    In your first attempt yo release one charge and find an expression for ##v##.
    If you then release the second charge, what ## v## woul it get ?
    And the third will of course be left in place, so ##v = 0##.
    What do you think of that ?

    In your second attempt you have ##

    How did you get that ?
  5. Sep 10, 2016 #4
    The first is right.
    Energy of system charge is:
    $$W=\frac {1}2\Sigma q_iV_i$$
    3 charge move to infinity with the same velocity in 2 cases.
  6. Sep 10, 2016 #5


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    Doesn't your ##W## have to be divided over three charges ?

    So what is my mistake in post #3 ?
  7. Sep 10, 2016 #6
    Yes W is energy of System.
    Your mistake is 3 charge will together move. Not 1 charge, after that 2 and 3.
  8. Sep 10, 2016 #7


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    I agree. So I find it strange you say that 'first is right'.
  9. Sep 10, 2016 #8
    But I dont understand why the last is wrong.
  10. Sep 10, 2016 #9
    The potential of 1 charge at initial V= q(V2+V3). After that they are in infinity so initial V=0 . So V=KE . That is the first equation
  11. Sep 10, 2016 #10


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    No, BvU was not making that mistake. BvU was pointing out that your first method made that mistake.
  12. Sep 10, 2016 #11
    But the second is wrong. The first right
  13. Sep 10, 2016 #12


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    I agree with BVU and haruspex that the first solution in the OP is wrong while the second is correct.

    The mistake in the first solution is trying to ascribe a definite amount of potential energy to just one of the charges of the system. The system as a whole has potential energy. The individual charges do not have potential energy. You get the right answer by using conservation of energy for the system: ##KE^{sys}_{final} + PE^{sys}_{final} = KE^{sys}_{initial} + PE^{sys}_{initial}##. This will handle all cases: (1) Releasing all three charges, (2) Releasing only 1 charge, (3) Releasing 2 charges.

    Sometimes we get the right answer by thinking in an incorrect way. For example, drop a book from a height h and find the speed of the book when it reaches the floor. We get the right answer in this case by saying that "the book" has potential energy mgh. But really, that potential energy does not "belong" to the book, it belongs to the book-earth system.

    Likewise, if you have only two identical charges initially separated by r0, the potential energy of the system is V0 = kq2/r0. You can release just one charge or you can release both chrages. If you release one charge while holding the other in place, you will get the right answer whether or not you think of the potential energy V0 as belonging to the system or to just the charge that is released. But if you release both charges, you will get the wrong answer if you try to set up conservation of energy for each charge separately and claim that each charge intially had the potential energy V0.
  14. Sep 10, 2016 #13
    Thanks. I understand. I have a question:
    We seem they are a system. They have a center of mass. The PE when they run to infinity will change to KE center of mass. (They not rotate around center of mass so Rotate KE=0) velocity of center of mass always equals 0 so KE=0. where PE will change ?Where is wrong?
  15. Sep 10, 2016 #14


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    There are no external forces acting on the system of three particles once the particles are released simultaneously. So, the CM will remain at rest. The PE is converted to the total KE of the three particles as measured relative to the CM frame.
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