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Distance dependent force

  1. Oct 23, 2004 #1
    I’m trying to solve the following problem for a long time:
    There are two charged particles, one doesn’t move, the other is being accelerated and is initially at rest, I would use the particle that doesn’t move as the referential, this referential has only one dimension, how do I find the position as a function of time.
    I tried this approach, by Coloumb’s law the force in the second particle is F=(k*q*Q)/r^2, for simplification I put a C in place of k*q*Q, now I use Newton’s second law, I get a=C/(m*x^2), then I try to put velocity as a function of time, so the work done on the particle as a function of the distance is found by solving the integral of the force as a function of the distance between x and x0(this is the initial position), I got W=(c/x)-(c/x0), knowing that the particle starts at rest, its kinetic energy must be equal to the work done by the electrical force so W=(mv^2)/2, I put velocity as a function of the position, that is v=√((2c/(xm))-(2c/(x0m))), now the idea of all this stuff was to use the relation a=dv/dt to get the position as a function of time, this is what I don’t know how to do and would appreciate someone’s help in solving the problem this way or a different propose for a solution.
     
  2. jcsd
  3. Oct 23, 2004 #2

    jcsd

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    You know that from the defintion of accelartion:

    [tex]\frac{d^2r}{dt^2} = \frac{C}{mr^2}[/tex]

    multiply both sides by [itex]2\frac{dr}{dt}[/itex], integrate with respect to r and take the square root:

    [tex]\frac{dr}{dt} = \sqrt{\frac{-2C}{mr} + A}[/tex]

    Where A is a constant.

    Then it's a case of seperating the variables to solve the equation.
     
  4. Oct 24, 2004 #3
    “Then it's a case of seperating the variables to solve the equation.”
    It is this part that I don’t know how to do, could you solve that, please.
     
  5. Oct 24, 2004 #4

    HallsofIvy

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    "Separating the variables" means exactly what it says:
    [tex]\frac{dr}{dt} = \sqrt{\frac{-2C}{mr} + A}[/tex]
    Now get the variable r on one side, the variable t on the other:

    [tex]dr= \sqrt{\frac{-2C}{mr}+A}dt[/tex]
    [tex]\frac{dt}{\sqrt{\frac{-2C}{mr}+A}}= dt[/tex]

    and integrate both sides.
     
  6. Oct 25, 2004 #5
    Could you explain to me how you passed from [tex]dr= \sqrt{\frac{-2C}{mr}+A}dt[/tex]
    to [tex]\frac{dt}{\sqrt{\frac{-2C}{mr}+A}}= dt[/tex]
    , when you talk abut integrating the last equation e have no problem integrating the right side, but on the left side there is a dt so the independent variable is t, and I am to solve the integral between two instants t, but on the function that is being integrated there is no t, this doesn’t makes sense to me.
    I’m sorry if I am doing trivial questions, but I know few about calculus, thanks for your help and patience.
     
  7. Oct 25, 2004 #6

    arildno

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    It's quite simple; HallsofIvy made a typo.
    On the left-hand side, it should be dr not dt
     
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