Electric field strength and particle deflection

In summary, a proton entering an electrostatic separator between two metal plates will be deflected vertically by 1.19⋅10−3rad.
  • #1
castrodisastro
82
0

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
[itex]\phi[/itex]=0.00119rad
mp=1.67*10-27kg
qp=1.067*10-19C



Homework Equations


F=ma
F=Eq
vy=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=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=[itex]\phi[/itex]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=2*0.037312N/C*(1.6067*10-19)*(0.00119rad)/(1.67*10-27kg)

(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 [itex]\phi[/itex] = 0.00265 rad and solve for s?

Thank you any help is appreciated.
 
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  • #2
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...
 
  • #3
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:
(vf,y)2=vi,yt+(1/2)at2=vi,yt+(2Eqp/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.
 
  • #4
BvU said:
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
 
  • #5




Your approach for parts A and B seems correct. However, for part C, you cannot simply equate the momenta of the proton and kaon since they have different masses. Instead, you need to use the conservation of momentum and energy to calculate the velocity of the kaon after exiting the electric field. You can then use this velocity to find the deflection angle of the kaon using the same method as in part A.
 

Related to Electric field strength and particle deflection

1. What is electric field strength?

Electric field strength is a measure of the force experienced by a charged particle in an electric field. It is represented by the symbol E and is measured in units of volts per meter (V/m).

2. How is electric field strength related to particle deflection?

Electric field strength is directly proportional to the amount of deflection experienced by a charged particle. This means that a stronger electric field will cause a greater degree of deflection in a charged particle.

3. How is electric field strength calculated?

Electric field strength is calculated by dividing the force acting on a charged particle by the magnitude of the charge of the particle. Mathematically, it can be represented as E = F/q, where E is the electric field strength, F is the force, and q is the charge of the particle.

4. Can electric field strength be negative?

Yes, electric field strength can be negative. This means that the direction of the electric field is opposite to the direction of the force acting on a charged particle. It is important to note that the magnitude of the electric field strength is always positive.

5. How does distance affect electric field strength?

In general, electric field strength decreases as the distance from the source of the field increases. This is because the force acting on a charged particle decreases as the distance between the particle and the source of the field increases. The relationship between electric field strength and distance is inversely proportional.

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