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.