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Beginner's emt question

  1. Feb 18, 2006 #1
    I have fixed electron at origin, and one other with initial coords [tex](x_0,y_0,z_0)[/tex]. I'm trying to get v(t), ignoring all relativistic effects including magnetism to keep it simple (hoping to extend solution into relativistic one). I've so far derieved v(x,y,z) only, but that's not what I need!

    Any ideas?
     
  2. jcsd
  3. Feb 18, 2006 #2
    hey, that's not a homework question!
     
  4. Feb 18, 2006 #3

    nrqed

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    This is not very clear. let me try to understand. First, this other electron, is it released from rest? And what are you trying to find exactly, its speed as a function of time, right? And the other electron is not allowed to move away from the origin, correct? So the second electron (if released from rest) will move away from the origin, and you want to find its speed as a function of time. Am I understanding correctly?
    It`s easy to find the speed of the second electron as a function of its position, but it`s a much more difficult question to find its speed as a function of time. The reason is that the force on it will depend on its position so the acceleration is not constant.
    Do you feel at ease with calculus? Because it is required to solve this question. To find the velocity as a function of time, you will have to integrate the acceleration.

    what is your level of math?
     
  5. Feb 18, 2006 #4
    Yup, exactly. I gotta find better words next time, I guess ^-^' It was just a question I've "invented" for myself, by the way.

    I'm pretty fluent in calculus, I'm just stuck with the physics part. I can write various equations, with no t in! (except for dt, of course, which I cannot relate with x,y,z. I tried this, for instance:

    [tex]\int \vec{F} dt = m \vec{v}[/tex]
    But what do I do with dt?!?
    So I solved this:
    [tex]m \frac{d\vec{v}}{dt} \frac{d\vec{r}}{d\vec{r}} = m \frac{v dv}{d\vec{r}} = \vec{F}[/tex]
    where F is the coloumb force, expressed in terms of x,y,z... Am I hopeless?
     
    Last edited: Feb 18, 2006
  6. Feb 18, 2006 #5

    nrqed

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    Well, F contains the position which is itself a an unknown function of time. So this is a dead end. One way is to set up F= m a which is a differential equation and solve that for the position as a function of time. Then you can differentiate to get v(t).

    First, let me point out that there is no need to work with vectors here. This is really a one dimensional problem since the electron will be moving along a straight line. Imagine aligning your x axis with the line connecting the two electrons and then the motion is only along the x axis.

    what you wrote is correct and can be integrated but it simply leads to conservation of energy. What it will give you is the velocity (and speed) as a function of the position (not as a function of time) and this is not difficult, as I mentioned earlier.

    Pat
     
  7. Feb 18, 2006 #6
    Aha! So sweet! Well, I've already solved it for one direction last year (in gravity field, an object going up thru sky).
    Well, what if electron's not at rest initially, but has [tex]\vec{v_0}[/tex]. Kinda generalization!
     
  8. Feb 18, 2006 #7

    nrqed

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    Ah, ok. Then you are done since the force has the same r dependence. (except that the force is repulsive, here...but if you have solved the case of gravity you are basically done).

    if there is some initial velocity then the motions perpendicular to the x axis (using the x axis aligned with the two charges) will be simply constant velocity motions. So y(t) and z(t) are trivial. So you get a parametric representation of the motion, (x(t), y(t), z(t)). You can eliminate t to find the shape of the motion. For example, if tehre is no motion along z, you could eliminate t to get y(x) and see what the trajectory actually looks like.

    Have fun :-)

    Pat
     
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