# Beam of particles shot through slits

1. Jan 28, 2014

### Queequeg

1. The problem statement, all variables and given/known data
A beam of positively charged particles with radius $R$ (current $I$) is shot from a region of potential $-V$ through a potential difference of $+V$ to a grounded plate. The beam passes through a slit on this plate then passes through another region of length $L$ meters and passes through a slit on a second plate before hitting a target. The proton density is $p$ protons per cubic meter.

a. What is the speed, $v$ of the particles when the beam reaches the first slit?

b. What is the electric field as a function of the radius within and outside the beam?

c. What is the electric potential as a function of the radius if the potential at the center of the beam is $0$?

d. Due to particle interactions, some of beam diverges over the distance $L$. If the radial field is constant and $1/2$ the maximum field, estimate how far a proton would move radially due to this field over the distance $L$.

2. Relevant equations

$U=qV$
$K=\frac{1}{2}mv^2$
$\rho = \frac{Q}{Volume}$
$E=\frac{Q}{4\pi ε_0 R^2}$
$V =\frac{Q}{4 \pi ε_0 R}$

3. The attempt at a solution

a. The initial electric potential energy of the particles is converted to kinetic energy as the particles reach the first plate, so $qΔV=\frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2qV}{m}}$

b. The volume of the beam over the distance $L$ is $V= (\pi)(R^2)(L)$ so there are $p\pi R^2 L$ protons and the total charge is$(1.6*10^{-19})p\pi R^2 L = Q$. So the electric field within the beam at a distance $r<R$ is $E=\frac{Qr}{4\pi ε_0 R^3}$ and for $r>R$, $E=\frac{Q}{4\pi ε_0 r^2}$

c.For $r<R$, the electric potential is $V=\frac{Q}{4\pi ε_0 R}$ and for $r>R, V=\frac{Q}{4\pi ε_0 r}$

d. No idea

2. Jan 29, 2014

### BvU

Don't understand the step from b) answer to c) answer for r<R. Wasn't E the the divergence of V ?In the OP they state that V=0 at r = 0.
Your V for r<R comes from what assumption about the charge distribution in the cross section of the beam ?
Very quick reply, may need to edit some more...

The equations you give are for a point charge. The beam is probably more like a line charge!

Last edited: Jan 29, 2014
3. Jan 29, 2014

### lightgrav

a) is ok.

b) you need the Area of the cylinder shell, length L and radius r.

in c), they wanted V =0 in the beam center (not at infinity) , so you integrate E from inside, outward.

d) how long (time) do the protons take to travel L?
where is the maximum E-field? (r=R) ... presume that ½ this value will provide constant acceleration outward.

4. Jan 29, 2014

### Queequeg

Ok so this beam is essentially a cylinder of charge? But the circular faces aren't pierced since it's continuous.

For part B, with $r$ being the distance I use Gauss' law to calculate the electric field for $r<R$ so the enclosed charge $q$ is then $Q\frac{\pi r^2 L}{\pi R^2 L} = Q\frac{r^2}{R^2}$, so using Gauss' law, $E(2\pi rL)= \frac{Qr^2}{ε_0 R^2} \rightarrow E = \frac{Qr}{2\pi ε_0 R^2 L}$ for $r<R$. Thus for $R<r, E(2\pi rL) = \frac{Q}{ε_0} \rightarrow E = \frac{Q}{2\pi ε_0 rL}$ Where $Q = \pi R^2 L(p)(1.6*10^{-19})$ and the field is maximum when $r=R$

Last edited: Jan 29, 2014
5. Jan 29, 2014

### lightgrav

b) the E-field is radial outward, so NONE of it pierces either end-cap (r+L) →L; otherwise ok.
c) straightforward integral, now that E is right.
d) trying to use non-constant acceleration as the beam expands would be a nightmare.
so just use constant-acceleration approx (ok, if they're going fast)

Last edited: Jan 29, 2014
6. Jan 29, 2014

### Queequeg

so for part C, I would integrate E from $-\frac{L}{2}$ to $\frac{L}{2}$. Not going to show all the work but does this integral look correct: $\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{Q}{2\pi ε_0 rL}\,dr$ I'm just not sure if I'm using the correct equation for the electric field. since the field is zero at the center, I used the electric field for r>R

Then for part D, I calculated the acceleration from $v^2 = 2aL \rightarrow a = \frac{qV}{mL}$. to calculate the distance a proton moves radially, do I need to relate this to the work done by the electric force?

Last edited: Jan 29, 2014
7. Jan 29, 2014

### lightgrav

c) E is RADIAL, so integrate from r=0, outward. The center of the beam (r=0) has V=0.
what happened to the r on top? how did it arrive on the bottom ... there used to be a R² down there.

8. Jan 29, 2014

### Queequeg

I see what you mean by radial but I have two different equations for electric field. So I need to integrate each one from 0 to R (electric field in the beam) and then from R to infinity (electric field outside beam)?

9. Jan 29, 2014

### lightgrav

c) yes, integrate the Qr/R² from r=0 to r=R, then add to that the integral of Q/r from r=R.
d) no, qV/mL was the acceleration along the beam, before the grounded plate.
here we want the acceleration outward (radially) of a typical proton (say, at r=R/2),
while the beam drifts (constant speed, found in (a)) from ground-plate to beam-dump.

10. Jan 29, 2014

### Queequeg

OK so my two integrals are $\int_0^R \frac{Qr}{2\pi ε_0 R^2 L} \,dr = \frac{Q}{4 \pi ε_0 L}$ and $\int_R^r \frac{Q}{2\pi ε_0 rL} \,dr = \frac{Q}{2\pi ε_0 L} \ln{\frac{r}{R}}$The second one somehow doesnt seem right though... since r > R, shouldnt potential decrease the farther you get and shouldn't my first equation imply V=0 at the center?

For part D ok so the radial acceleration would be from $\frac{v^2}{r}$?

Last edited: Jan 29, 2014
11. Jan 29, 2014

### lightgrav

c) looks good. notice that a uniform field (|| plate capacitor) has V increase _linearly_ with distance.
d) the protons repel each other, by the E-field you calculated in b) ... F = eE causes mproton a

12. Jan 29, 2014

### Queequeg

Great, and for D, the displacement would then be calculated using $v^2 = 2 a \Delta x$ and the velocity from a)? Thanks so much for your help.

13. Jan 30, 2014

### lightgrav

no, the velocity you calculated in part (a) was along the length L; use it to find time-of-flight.
part (d) wants Δr = ½ a t²