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Electric field strength and particle deflection

  1. Feb 19, 2014 #1
    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
    [itex]\phi[/itex]=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=[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.
     
  2. jcsd
  3. Feb 20, 2014 #2

    BvU

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    Hello Castro,

    Let's start with A). You state
    But where does that come from ? and how can x = v2 ? Dimensionally I don't follow you. Please explain a little...
     
  4. Feb 20, 2014 #3

    BvU

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    Science Advisor
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    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.
     
  5. Feb 20, 2014 #4
    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
     
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