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Doubt about potential energy

  1. Sep 5, 2015 #1
    I'm trying to solve an exercise that asks what is the final speed of two positive particles after they are very distants each other. Then I've used the conservation of energy to solve it. The initial energy of the system is the potential energy = Q².k/d The final energy is the 2 knetic energy of them. So Q².k/d = Ke1 + Ke2
    My doubt is why use the same equation for the two particles in the init? Why not 2.Q².k/d since both of them have acumulated same amount of energy? (sorry my bad english)
     
  2. jcsd
  3. Sep 5, 2015 #2

    mfb

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    Q2 k / d is the total potential energy in the system. It is not a property of an individual particle, so there is no reason to double it.
     
  4. Sep 5, 2015 #3
    yes I know that. But the thing is why only one equation Q²K/d is the total potential energy....?
     
  5. Sep 5, 2015 #4

    mfb

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    k is defined that way. The definition is arbitrary, but this is the easiest one as you don't need additional prefactors.
     
  6. Sep 5, 2015 #5
    But since we are adding the energy of each charge to get the total energy of the system, I guess it's wrong to use Q².k/d becous it's the equation for only one charge. Am I right?
     
  7. Sep 5, 2015 #6

    mfb

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    That is a questionable approach, but if you do it both charges account for 1/2 of Q2k/d.
     
  8. Sep 5, 2015 #7
    Think of it this way. Say these two pieces were already at infinitely far away. To get one piece to a certain distance d to the other, work of the magnitude q^2k/d has to be done on that one particle. The same amount of work is needed to move the first piece near the second. Therefore the total work done to set the system is q^2k/d, which is the potential energy due to the electric field being a conservative force, and thus no matter how the two positive pieces got there it has that potential energy.. That is why the total kinetic energy can only be equal to q^2k/d
     
  9. Sep 5, 2015 #8
    Same thing. Doesnt matter the way you used to calculate the potential energy. The fact is that Q².k/d is the acumulated energy on one charge.... so when we added the total energy of the system composed by the two charges, if would be 2.Q².k/d (sorry my bad english)
     
  10. Sep 5, 2015 #9
    No, you are not. A single charge by itself has no potential energy.
    Only the interaction between the two charges "produces" some potential energy.

    I suppose you are confusing the potential of the field of one charge with the potential energy of the system of two charges.
    If you were to calculate the potential of the field of the two charges you will indeed add the potential produced by each individual charge.
     
  11. Sep 5, 2015 #10
    I tried to translate the exercise for you see the situation. (sorry my bad translation)

    http://[ATTACH=full]199884[/ATTACH]
     
    Last edited by a moderator: May 7, 2017
  12. Sep 6, 2015 #11

    mfb

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    For (a), you'll also have to take conservation of momentum into account. Apart from that, everything else said above applies.
     
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