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Motion of a charged particle

  1. Jun 25, 2007 #1
    In one dimension, the motion of a charged particle (q1) will be [assume q2 is stationary]

    [tex]\frac{d^2 x}{d t^2} =\frac{Kq_1q_2}{m x^2}[/tex]

    Is there a solution to this differential equation?
  2. jcsd
  3. Jun 25, 2007 #2


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    How about
    is anything you want.
  4. Jun 25, 2007 #3
    Thanks, that seems to work. I wonder why maple couldn't give me that answer.
  5. Jun 25, 2007 #4


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    well... it's a pretty particular solution I gave you... it only works when the "total energy" is zero, i.e., when
    [tex]\frac{1}{2}mv^2 + \frac{Kq1q2}{x} = 0[/tex].

    If you rewrite A and B in terms of x(0) and v(0) you will see that the condition on A means that
    1/2mv(0)^2+Kq1q2/x(0)=0... but it is also easy to show that 1/2mv^2 + Kq1q2/x is a constant in time thus it is always zero.

    In general the solution is hard, but using the constants of the motion we can write
    to find t(x) and then invert to find x(t)...
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