1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook