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Peak Electron Number Density of a Beam

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider what happens as a parallel beam of non-relativistic electrons with beam radius a is injected as a current I into a region of vacuum after acceleration through a voltage difference V. If the radial profile of the beam can be described as

    p=p0 cos([itex]\frac{\pi r^2}{2a^2}[/itex])​

    where ρ is the charge density, find an expression for ρ0 in terms of parameters a, V and I and hence determine the peak electron number density N0 in the beam.

    2. Relevant equations

    ∫ ∇.E dV = ∫ E dS
    V=∫E dl

    3. The attempt at a solution

    My attempt was first substituting in the charge density into Gauss's equation to give

    ∇.E=p00 cos([itex]\frac{\pi r^2}{2a^2}[/itex])

    Then substituting this into the second equation gives

    ∫ p00 cos([itex]\frac{\pi r^2}{2a^2}[/itex]) dV = ∫ E dS

    Using p=dQ/dV i got rid of the dV to give;

    Q00 cos([itex]\frac{\pi r^2}{2a^2}[/itex]) = ∫ E dS

    and since this is a radial beam then dS=4[itex]\pi[/itex]r2 which gives;

    E=Q0/4[itex]\pi[/itex]r2ε0 cos([itex]\frac{\pi r^2}{2a^2}[/itex])

    Using the potential equation V=∫E dl we can find the potential, however i haven't got that far to calculate due to the fact it feels like i'm going in the wrong direction. What do you's think?
  2. jcsd
  3. Feb 17, 2013 #2
    Does anyone have any advice to point me in the right direction?
  4. Feb 17, 2013 #3


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    Staff: Mentor

    The charge density looks strange. That cos() expression does not vanish for large r, so you magically convert a well-defined beam into something that spreads out in the whole volume, with positive and negative densities?
    Should we assume that this expression is valid for r<a only?

    In that case: Current is the same before and after acceleration. If you integrate your density over the beam profile, this number is related to current via the velocity of the electrons. Current/Charge conservation could be enough to solve the problem.
  5. Feb 17, 2013 #4
    I think it would safe assume that it is for r<a. I think you're right with the theory, i'll try it out anyways and see how it goes from there, thanks!
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