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Equation of motion: Help with DiffEq (2nd order non linear)

  1. Aug 20, 2015 #1
    I am trying to solve the differential equation that will give me the equation of motion of a point charge under the influence of another point charge's electric field.

    Say point charge A is free to move, and it currently a distance D away from point charge B. Point charge B is fixed in space.

    Say charge A has q = +q, and charge B has q = -q. The two charges will attract.

    Ignoring all other influences (gravity, etc), charge A should experience a force F = qE, where E is the field due to charge B, or:
    F = -(k q^2)/r^2
    where k = 1/4πε (imagine that ε is the permittivity of free space; I'm using the available symbols)

    Solving for equations of motion, I use:
    ma = -(k q^2)/r^2

    or
    a = -(k q^2)/(m r^2)

    Putting it another way, r → r[t], a → r''[t]
    Then I get:
    r[t]2 r''[t] = -(k q^2)/m

    Or
    r[t]2 r''[t] = C

    How do I solve that differential equation? It is a 2nd order non-linear diff eq... The Mathematica answer I get is very complicated, but I'm hoping someone can help me out with this one.

    Also, I can use the following initial conditions:
    r[0]=D
    r'[0]=0

    Thanks!
     
  2. jcsd
  3. Aug 20, 2015 #2

    PeroK

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    You can do it in three stages:

    1) First, multiply by ##r'## (integrating factor) to get:

    ##\frac{d}{dt}(r'^2) = \frac{d}{dt}(\frac{-2C}{r})##

    2) (The key trick): Let ##r = Dcos^2\theta##

    This leads to:

    ##(cos^2\theta)\theta ' = \sqrt{\frac{-C}{2D^3}}##

    3) Integrate to get:

    ##2\theta + sin(2\theta) = \sqrt{\frac{-8C}{D^3}}t##
     
  4. Aug 20, 2015 #3

    BvU

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    So no simple solution. I typed x'' = -1/x^2 here and it came back with this picture:

    http://www4c.wolframalpha.com/Calculate/MSP/MSP12181g1a4ceci2e2i16500005a8e262bb5i0g8e3?MSPStoreType=image/gif&s=10&w=388.&h=101. [Broken]
    Got the same picture doing a simple numeric integration with excel:

    upload_2015-8-20_22-29-40.png

    (x on the left axis, v and a on the right axis)


    Basically the mobile charge just "falls" towards the fixed charge in the same way a small mass falls towards a planet: initially slowly (the change in attractive force is small) and then accelerating faster and faster.
     
    Last edited by a moderator: May 7, 2017
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