Electric field strength and particle deflection

[castrodisastro]

Homework Statement

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
m_p=1.67*10^-27kg
q_p=1.067*10^-19C

Homework Equations

F=ma
F=Eq
v_y=v₀t+(1/2)at²

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=m_pa_p=Eq_p
(m_pa_p)/q_p=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=\phix = 0.00119*(18300)² = 398519.1m

to find our acceleration we take s/t²

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

From our (m_pa_p)/q_p=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

(v_f,y)²=v_i,yt+(1/2)at²=v_i,yt+(2Eq_p/m)s)

since initial y velocity is zero...

(v_f,y)²=(2Eq_p/m)s

(v_f,y)²=2*0.037312N/C*(1.6067*10^-19)*(0.00119rad)/(1.67*10^-27kg)

(v_f,y)²=8543.66m/s

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

v=((v_x)²+(v_y)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

m_pv_p=m_pv_k

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.

 

[BvU]

Hello Castro,

Let's start with A). You state

s=ϕx = 0.00119*(18300)² = 398519.1m

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

 

[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:

(v_f,y)²=v_i,yt+(1/2)at²=v_i,yt+(2Eq_p/m)s)

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

 

[castrodisastro]

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...)

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