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Application of Coulomb's law to positron velocities

  1. Jul 11, 2012 #1
    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=[itex]\frac{kQ1Q2}{r^2}[/itex]) 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?
     
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  3. Jul 11, 2012 #2
    [itex]\frac{dF(x(t))}{dt}=\frac{dF(x(t))}{dx} \frac{dx(t)}{dt}[/itex]
     
  4. Jul 11, 2012 #3
    Sorry, but could you elaborate a little?
     
  5. Jul 11, 2012 #4
    [itex]a=\frac{dv}{dt}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}[/itex]

    what you write is this:

    [itex]\frac{dv}{dr} \frac{dr}{dt}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}[/itex]

    dr/dt is the velocity so:

    [itex] v \frac{dv}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}[/itex]
     
  6. Jul 11, 2012 #5

    mfb

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    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?
     
  7. Jul 11, 2012 #6
    Sorry if I'm starting to sound incompetent, but what would I do about the [itex]\frac{dv}{dr}[/itex]
     
  8. Jul 11, 2012 #7
    @ 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.
     
  9. Jul 11, 2012 #8
    [itex] v \frac{dv}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}[/itex]

    [itex] \frac{1}{2} \frac{dv^{2}}{dr}=\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r^{2}}[/itex]

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

    [itex]\frac{1}{2} v^{2}+\frac{k}{m}\frac{{Q}_{1}{Q}_{2}}{r}=const [/itex]
     
  10. Jul 11, 2012 #9
    Thanks for the information.
     
  11. Jul 12, 2012 #10

    mfb

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    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.
     
  12. Jul 12, 2012 #11
    I see. However, the main focus is not the amount of energy generated, but rather the efficiency of matter converting into energy.
     
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