Application of Coulomb's law to positron velocities

1. Jul 11, 2012

Aeonace32

I have an idea in which I need to manipulate a positron using an electrical field. However, in order for the problem to work, I need to make sure that the positron's kinetic energy is under 511 keV. to do this, I used Coulomb's law (F=$\frac{kQ1Q2}{r^2}$) to obtain the force. Because F = ma, I divided both sides by mass to get acceleration. I know that integrating acceleration with respect to time yields velocity. However, in this case, acceleration is not a function of time, but rather a function of distance. My goal is to obtain an equation which can give me the velocity at different distances so that I can plug the right velocity into the kinetic energy formula. Is integrating the way to solve this, and if so, how do I do it correctly?

2. Jul 11, 2012

Morgoth

$\frac{dF(x(t))}{dt}=\frac{dF(x(t))}{dx} \frac{dx(t)}{dt}$

3. Jul 11, 2012

Aeonace32

Sorry, but could you elaborate a little?

4. Jul 11, 2012

Morgoth

$a=\frac{dv}{dt}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}$

what you write is this:

$\frac{dv}{dr} \frac{dr}{dt}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}$

dr/dt is the velocity so:

$v \frac{dv}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}$

5. Jul 11, 2012

Staff: Mentor

You can directly get the energy by integrating the force over a distance (or using the Coulomb potential), you do not have to use any time-dependence at all for this.
As a bonus, you avoid relativistic effects in this calculation, which can influence the acceleration.

Are you sure this is the real requirement?

6. Jul 11, 2012

Aeonace32

Sorry if I'm starting to sound incompetent, but what would I do about the $\frac{dv}{dr}$

7. Jul 11, 2012

Aeonace32

@ mfb:
Thanks. The reason that I need to make sure it is under 511 keV is because if it is any higher than that, the positron-electron annihilation will start to produce heavy particles instead of energy.

8. Jul 11, 2012

Morgoth

$v \frac{dv}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}$

$\frac{1}{2} \frac{dv^{2}}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}$

$\frac{d}{dr} ( \frac{1}{2} v^{2}+\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r} )=0$

$\frac{1}{2} v^{2}+\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r}=const$

9. Jul 11, 2012

Aeonace32

Thanks for the information.

10. Jul 12, 2012

Staff: Mentor

If you already have positrons, they can annihilate with electrons - and the cross-section is smaller for higher energy. Very high-energetic positrons might produce gamma rays of >1 MeV somewhere, which could do pair-creation, but as you can see this already requires more than 1 MeV of energy.

11. Jul 12, 2012

Aeonace32

I see. However, the main focus is not the amount of energy generated, but rather the efficiency of matter converting into energy.