# Electric field strength and particle deflection

1. Feb 19, 2014

### castrodisastro

1. The problem statement, all variables and given/known data
A proton enters the gap between a pair of metal plates (an electrostatic separator) that produces a uniform, vertical electric field between them. Ignore the effect of gravity on the proton.

a) Assuming that the length of the plates is 11.1cm , and that the proton will approach the plates at a speed of 18.3km/s , what electric field strength should the plates be designed to provide, if the proton must be deflected vertically by 1.19⋅10−3rad?

b) What speed does the proton have after exiting the electric field?

c) Suppose the proton is one in a beam of protons that has been contaminated with positively charged kaons, particles whose mass is 494 MeV/c2 (8.81⋅10−28kg) , compared to the mass of the proton, which is 938 MeV/c2 (1.67⋅10−27kg) . The kaons have +1e charge, just like the protons. If the electrostatic separator is designed to give the protons a deflection of 2.65⋅10−3rad, what deflection will kaons with the same momentum as the protons experience?

d=11.1cm=0.111m
v0=18.3km/s=18300m/s
$\phi$=0.00119rad
mp=1.67*10-27kg
qp=1.067*10-19C

2. Relevant equations
F=ma
F=Eq
vy=v0t+(1/2)at2

3. The attempt at a solution

Part A

The process was to equate the force of the proton to the electric field strength times the charge on the proton.

F=mpap=Eqp
(mpap)/qp=E

First we find the time it takes for the proton to travel between the length of the plates.

v=x/t → t=x/v = 6.0656*10-6s

the deflection s

s=$\phi$x = 0.00119*(18300)2 = 398519.1m

to find our acceleration we take s/t2

ap=(1.321*10-4m)/(3.679*10-11s2)=3.5905*106m/s2

From our (mpap)/qp=E

(1.67*10-27kg)(3.5905*106m/s2)/(1.6067*10-19C) = E

E=0.037312 N/C

Part B

I used the motion equation to get the vertical acceleration

(vf,y)2=vi,yt+(1/2)at2=vi,yt+(2Eqp/m)s)

since initial y velocity is zero...

(vf,y)2=(2Eqp/m)s

(vf,y)2=8543.66m/s

To get the total magnitude of the velocity in the x and y direction...

v=((vx)2+(vy)2)(1/2)

v=18300.233m/s

Part C

This is where I don't know how to proceed. If I am still dealing with the same basic setup since the problem states that our setup has been contaminated to include kaons and not a different situation, the only values that change are mass and velocities.

Since the problem asks what the deflection would be if the kaon had the same momentum as the proton, I tried

mpvp=mpvk

and I solved for the velocity required of the kaon to have the same momentum as the proton.

My question is if it is the same process as in part a except that I just change the values for mass, velocity, set $\phi$ = 0.00265 rad and solve for s?

Thank you any help is appreciated.

2. Feb 20, 2014

### BvU

Hello Castro,

But where does that come from ? and how can x = v2 ? Dimensionally I don't follow you. Please explain a little...

3. Feb 20, 2014

### BvU

We come to part B). You can check your work somewhat more thoroughly: Can your vertical speed really be consolidated with a deflection angle of 0.00119 radians ?

What is the s ? can you imagine two plates of 11.1 cm long being placed 400 km apart ? (this applies to part A as well, so: back to the drawing board...)

Then:
is a dimensional mess: m2/s2 on the left, m/s plus m in the middle and m/s plus Nm on the right.

4. Feb 20, 2014

### castrodisastro

You are right. I will start again from the beginning but this time more thoroughly. Some of these typos came from transferring my work onto the computer. I honestly should've triple checked when going from paper to keyboard. Sorry about that, I will post my new work as soon as a can