# Peak Electron Number Density of a Beam

1. Feb 16, 2013

### mickyfitz13

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($\frac{\pi r^2}{2a^2}$)​

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.

2. Relevant equations

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

3. The attempt at a solution

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

∇.E=p00 cos($\frac{\pi r^2}{2a^2}$)

Then substituting this into the second equation gives

∫ p00 cos($\frac{\pi r^2}{2a^2}$) dV = ∫ E dS

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

Q00 cos($\frac{\pi r^2}{2a^2}$) = ∫ E dS

and since this is a radial beam then dS=4$\pi$r2 which gives;

E=Q0/4$\pi$r2ε0 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?

2. Feb 17, 2013

### mickyfitz13

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

3. Feb 17, 2013

### 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.

4. Feb 17, 2013

### mickyfitz13

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!