Xittenn Posted July 13, 2010 Posted July 13, 2010 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?
Xittenn Posted July 13, 2010 Author Posted July 13, 2010 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?)
Xittenn Posted July 13, 2010 Author Posted July 13, 2010 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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