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Homework Help: Electric Potential Energy Question

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Two electrons start at rest with a separation of 5.0*10-12m. Once released, the electrons accelerate away from each other. Calculate the speed of each electron when they are a very large distance apart.

    2. Relevant equations

    3. The attempt at a solution
    Okay so I know that:

    • The Electric force is in the same direction as the displacement, which means kinetic energy is increasing and electric energy is decreasing (ΔEK=-ΔEE)
    • As they move to a really far distance, the change in electric energy is equal to 0 - EE1 as anything divided by infinity is equal to zero
    mass of electron = 9.109*10-31 Kg
    q = -1.60*10-19 C
    k = 8.99*109
    v1 = 0 m/s
    r1 = 5.0*10-12 m

    My process:

    Is this correct?

    Hopefully that shows it well, if not here is a picture (I write small, probably can't make out much):

    EDIT: Looks like pic is a no show


  2. jcsd
  3. Nov 20, 2014 #2


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    Looks to me as if you are moving only one charge to infinity. But the electrons are supposed to move away from each other.
  4. Nov 20, 2014 #3
    You are absolutely correct, as I discovered today walking into class. The conservation of energy comes from the kinetic energy of not just one but both electrons MOVING AWAY from each other, as you just said. Thanks a lot anyways :)
  5. Nov 20, 2014 #4


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    The last equation is wrong. The potential energy of a system of two point charges is kq1q2/d, where d is the separation between them.
    The electrons interact and there is no external force. So the centre of mass of the system stays in rest The kinetic energy of the system is the sum of the kinetic energies of both electrons, not the difference between them! Initially, d=5.0*10-12m
  6. Nov 21, 2014 #5


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    If you look at how he is using it, this is not the problem, r1 and r2 correspond to the distance between the electrons at two different times. The problem, as BvU noted, is in the setup and treatment of the kinetic energy.

    Mystiq: The expression for the kinetic energy difference is wrong not only because you forgot about one of the electrons. The difference should be proporional to v1^2 - v2^2, not (v1-v2)^2. In this case, one of the velocities was zero so it did not make a difference, but it is something you will generally need to consider.
  7. Nov 21, 2014 #6


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    Sorry. I see now that r1 and r2 were distances at two different times.
    The two-electron system can be replaced by a single particle with reduced mass, μ=m/2 if the particles are identical. The velocity of that "imaginary" particle is equal to the relative velocity u=va-vb of the real particles, while the CoM of the system is stationary. As they are identical va=-vb=v and u=2v. The kinetic energy is then KE=1/2 μ u^2, and the change of the kinetic energy is ΔKE=1/2 μ(u22-u12).
    As the particles were in rest initially, u1=0, so the change of the KE is ΔKE=1/2 μu22.
    Last edited: Nov 21, 2014
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