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castrodisastro

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## 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

^{−3}rad?

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/c

^{2}(8.81⋅10

^{−28}kg) , compared to the mass of the proton, which is 938 MeV/c

^{2}(1.67⋅10

^{−27}kg) . 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

^{−3}rad, what deflection will kaons with the same momentum as the protons experience?

**d**=11.1cm=

**0.111m**

**v**=18.3km/s=

_{0}**18300m/s**

**[itex]\phi[/itex]**=

**0.00119rad**

**m**

q

_{p}=1.67*10^{-27}kgq

_{p}=1.067*10^{-19}C## Homework Equations

F=ma

F=Eq

v

_{y}=v

_{0}t+(1/2)at

^{2}

## 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

_{p}a

_{p}=Eq

_{p}

(m

_{p}a

_{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

^{-6}s

the deflection s

s=[itex]\phi[/itex]x = 0.00119*(18300)

^{2}= 398519.1m

to find our acceleration we take s/t

^{2}

a

_{p}=(1.321*10

^{-4}m)/(3.679*10

^{-11}s

^{2})=3.5905*10

^{6}m/s

^{2}

From our

**(m**

_{p}a_{p})/q_{p}=E(1.67*10

^{-27}kg)(3.5905*10

^{6}m/s

^{2})/(1.6067*10

^{-19}C) = E

**E=0.037312 N/C**

**Part B**

I used the motion equation to get the vertical acceleration

(v

_{f,y})

^{2}=v

_{i,y}t+(1/2)at

^{2}=v

_{i,y}t+(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

^{-27}kg)

(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

_{p}v

_{p}=m

_{p}v

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

Thank you any help is appreciated.