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Posted

I'm having some trouble with a pretty basic question and I'm not sure what I'm missing that will correct my thinking.

 

If I have two electrons in a vacuum and they are set at arbitrary origins at a correspondingly arbitrary fixed distance between them with initial velocity 0 how do I find time as a function of distance?

 

I'm looking at this like this:

[math]a® = k_{e} \frac{q^{2}}{m \cdot r^{2}}[/math]

 

where a is acceleration, r is the distance between the two electrons, [math] k_{e} [/math] is the Coulomb Constant, m is two times the electron mass and q is the charge on one electron.

 

It is all well and fine to plot acceleration as a function of distance but I'm not seeing how to integrate in time?? What am I missing about the mathematics which is also probably rather elementary which is preventing me from logically thinking this through?

Posted

I posted this question on a site I had asked the same question 5 years ago ..... and I figured I might as well post the very sensible reply! Pretty obvious now that I look at it, I should probably find a book on introductory physics.

 

For these kind of problems' date=' you use the energy conservation.

[math'] E=\frac{1}{2}m\left(\frac{dx}{dt}\right)^{2} + V(x) \ \ \rightarrow \ \ \frac{dx}{dt} = \pm \sqrt{2[E-V(x)]/m}[/math]

[math]\rightarrow \ \ dt=\pm \frac{dx}{\sqrt{2[E-V(x)]/m}} [/math]

 

[math]\pm[/math] in the expression is determined by the initial condition of the problem. (for example, whether they approaching or moving away from each other?)

Posted

And from my starting point ...

 

[math]a® = k_{e} \frac{q^{2}}{m \cdot r^{2}}[/math]

 

 

[math]a = \frac{dv}{dt} = \frac{d^{2}r}{dt^{2}}[/math]

 

[math] \frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr} [/math]

 

 

[math] v\frac{dv}{dr} = \frac{k q^2}{m r^2} [/math]

 

or

 

[math] \int_{v_0}^v v dv = \frac{k q^2}{m} \int_{r_0}^r r^{-2} dr [/math]

 

or

 

[math] \frac{v^2}{2} - \frac{v_0^2}{2} = \frac{k q^2}{m}

\left( \frac{1}{r_0} - \frac{1}{r} \right) [/math]

 

which would then be rearranged into the form of post #2

 

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