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Homework Help: Electrical energy question

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data

    two electrons seperated by a large distance are fired directly at eachother. The closest approach in this head on collision is 4e-14 m. One electron starts with twice the speed of the other. Assuming there is no deflection from the original path, determine the initial speed of the electron.

    2. Relevant equations

    Ee = (k q1 q2) / r

    ek = 0.5 m v^2

    3. The attempt at a solution

    see here:


    Answer from book is:
    5.3e7 m/s
  2. jcsd
  3. Jul 20, 2009 #2
    bump please, im sorry i have a test tom. im taking in alot of info today.
  4. Jul 20, 2009 #3


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    Staff Emeritus
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    Homework Helper

    I see a problem with your solution:

    When the electrons are at their closest approach, their speeds (and hence KE) are not zero. However, you can use conservation of momentum to get their speed at that point.
  5. Jul 20, 2009 #4
    ok in that case what is a better way to go about it?

    i dont think he wanted us to use momentum.

    the only other euation i got with velocity is:
    q/m = v / (B *r)

    q= charge (got that)
    m = mass (got that)

    v = velocity, what im looking for

    b = (dont have that and cant determine without a current)

    and r i do have

    Ill ask him in the morn before the test, and will post back his soln.
    Last edited: Jul 20, 2009
  6. Jul 20, 2009 #5


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    Staff: Mentor

    Oh wow, I didn't see that RB. Thanks. If their initial velocities were equal, that would be different. But the unequal initial velocities....
  7. Jul 20, 2009 #6


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    You can go about it the same way you did, using conservation of energy.

    The difference is, you need to account for the kinetic energy when they are at their closest approach ... you seemed to assume it was zero, but it isn't.

    An alternative approach is to transform to a frame of reference in which the initial speeds are equal, and solve the problem in that reference frame -- still using conservation of energy. Finally, transform back to the lab frame to find the velocities.
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