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

[Adoniram]

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!

 

[PeroK]

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!

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:

$2\theta + sin(2\theta) = \sqrt{\frac{-8C}{D^3}}t$

 

[BvU]

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.
Got the same picture doing a simple numeric integration with excel:

(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.