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Two Electorns in a Vacuum

  1. Jul 12, 2010 #1
    Five years ago I posted this question. At the time it was difficult for me to understand almost all of what I was trying to learn and was very frustrated. I put it down and studied other stuff including math. I just looked at this question again and am finding myself still unable to derive any answers.

    I'm having some trouble with a pretty basic question and I'm not sure what I'm missing that will correct my thinking. If I have two electrons in a vacuum and they are set at arbitrary origins at a correspondingly arbitrary fixed distance between them with initial velocity 0 how do I find time as a function of distance?

    I'm looking at this like so:
    [tex]a(r) = k_{e} \frac{q^{2}}{m \cdot r^{2}}[/tex]

    where a is acceleration, r is the distance between the two electrons, [tex] k_{e} [/tex] is the Coulomb Constant, m is two times the electron mass and q is the charge on one electron. Both electrons are allowed to move freely!

    I feel that plotting acceleration as a function of distance would be useful but I'm not seeing how to integrate in time? What am I missing about the mathematics which is also probably rather elementary and is preventing me from logically thinking this through?
  2. jcsd
  3. Jul 12, 2010 #2
    For these kind of problems, you use the energy conservation.
    [tex] E=\frac{1}{2}m\left(\frac{dx}{dt}\right)^{2} + V(x) \ \ \rightarrow \ \ \frac{dx}{dt} = \pm \sqrt{2[E-V(x)]/m}[/tex]
    [tex]\rightarrow \ \ dt=\pm \frac{dx}{\sqrt{2[E-V(x)]/m}} [/tex]

    [tex]\pm[/tex] in the expression is determined by the initial condition of the problem. (for example, whether they approaching or moving away from each other?)
  4. Jul 13, 2010 #3
    Thank you for redirecting my approach!
  5. Jul 13, 2010 #4
    While weejee is absolutely correct in everything he did, We can solve this problem
    from first principles, just the way you asked. [and it includes a common physics-trick,
    which is probably good to see used.] The energy is sometimes called a FIRST INTEGRAL
    of the motion. meaning we get it when we integrate the equations of motion 1 time.

    start from a = (kq^2)/(mr^2)
    recognize a = dv/dt = d^2r/dt^2
    [the trick]
    [tex] \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} [/tex]

    So that
    [tex] v\frac{dv}{dr} = \frac{k q^2}{m r^2} [/tex]
    [tex] \int_{v_0}^v v dv = \frac{k q^2}{m} \int_{r_0}^r r^{-2} dr [/tex]
    [tex] \frac{v^2}{2} - \frac{v_0^2}{2} = \frac{k q^2}{m}
    \left( \frac{1}{r_0} - \frac{1}{r} \right) [/tex]

    Now this can be rearranged a-la weejee.

    of course, we really wouldn't want to do this integral from scratch everytime
    for each different force F that we run into. If the force is conservative (i.e. the
    gradient of some potential) we can do this one time to arive at the law of
    conservation of energy -- and from then on just use conservation of energy.
  6. Jul 13, 2010 #5
    That's awesome ..... I wish I had come up with this myself with all the time I spent trying to figure it out ..... but I'm glad to get it out of my head, it's been stuck there for a while. Thanks again!
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