A proton enters the gap between a pair of metal plates (an electrostatic separat

Question

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 13.5 cm, and that the proton will approach the plates at a speed of 13.9 km/s, what electric field strength should the plates be designed to provide, if the proton must be deflected vertically by 1.49 10-3 rad?

3.2e-6 N/C i got this answer but it is WRONG


(b) What speed does the proton have after exiting the electric field?

13.9 km/s i got it and it is RIGHT


(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 10-28 kg), compared to the mass of the proton, which is 938 MeV/c2 (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 1.20 10-3 rad, what deflection will kaons with the same momentum as the protons experience?

Explanation / Answer

let field be E hence force =E*q a= E*q/m Vy =at =E*(q/m) *t where t=0.135/13900 = 9.71*10^(-6) hence Vy = E*1.6*10^(-19)/[1.66*10^(-27)] *9.71*10^(-6) Vy =E*936 tan 1.49 10-3 =Vx/Vy hence Vy =9328852m/s E =Vy/936 =9328852/936 =9966.7 N/C ...other answer you have calculated right..

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