Peak Electron Number Density of a Beam

[mickyfitz13]

Homework Statement

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(\frac{\pi r^2}{2a^2})​

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.

Homework Equations

∇.E=p/ε₀
∫ ∇.E dV = ∫ E dS
V=∫E dl
p=dQ/dV

The Attempt at a Solution

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

∇.E=p₀/ε₀ cos(\frac{\pi r^2}{2a^2})

Then substituting this into the second equation gives

∫ p₀/ε₀ cos(\frac{\pi r^2}{2a^2}) dV = ∫ E dS

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

Q₀/ε₀ cos(\frac{\pi r^2}{2a^2}) = ∫ E dS

and since this is a radial beam then dS=4\pir² which gives;

E=Q₀/4\pir²ε₀ cos(\frac{\pi r^2}{2a^2})

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?

 

[mickyfitz13]

Does anyone have any advice to point me in the right direction?

 

[mfb]

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.

 

[mickyfitz13]

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.

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!