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 27.1 cm, and that the proton will approach the plates at a speed of 15.0 km/s, what electric field strength should the plates be designed to provide, if the proton must be deflected vertically by 1.20 ⋅ 10−3 rad?1.20 · 10−3 rad?   N/C (b) What speed does the proton have after exiting the electric field?  m/s (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)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)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 rad1.20 · 10−3 rad , what deflection will kaons with the same momentum as the protons experience? (You may enter your calculation using scientific notation.)  rad

(a) The deflection of the proton is given by the formula θ = Fd / (mv^2), where F is the force on the proton, d is the distance over which the force is applied (the length of the plates), m is the mass of the proton, and v is the speed of the proton. The force on the proton is given by F = Eq, where E is the electric field strength and q is the charge of the proton. Substituting this into the first equation gives θ = Eqd / (mv^2). Solving for E gives E = θmv^2 / (qd). Substituting the given values gives E = (1.20 x 10^-3 rad)(1.67 x 10^-27 kg)(15.0 x 10^3 m/s)^2 / ((1.60 x 10^-19 C)(0.271 m)) = 1.12 x 10^6 N/C.

(b) The speed of the proton after exiting the electric field is the same as its initial speed, because the electric field only changes the direction of

(a) The deflection of the proton is given by the formula θ = Fd / (mv^2), where F is the force on the proton, d is the distance over which the force is applied (the length of the plates), m is the mass of the proton, and v is the speed of the proton. The force on the proton is given by F = Eq, where E is the electric field strength and q is the charge of the proton. Substituting this into the deflection formula gives θ = Eqd / (mv^2). Solving for E gives E = θmv^2 / (qd). Substituting the given values (θ = 1.20 x 10^-3 rad, m = 1.67 x 10^-27 kg, v = 15.0 x 10^3 m/s, q = 1.60 x 10^-19 C, d = 27.1 x 10^-2 m) gives E = 1.20 x 10^-3 rad * 1.67 x 10^-27 kg * (15.0 x 10^3 m/s)^2 / (1.60 x 10^-19 C * 27.1 x 10^-2 m) = 1.18 x 10^6 N/C.

(b) The speed of the proton after exiting the electric field is the same as its initial speed, because the electric field only changes the direction of the proton's velocity, not its magnitude. So the speed is 15.0 km/s.

(c) The deflection of the kaons is given by the same formula as the protons, but with the mass of the kaons instead of the protons. So θ = Eqd / (mv^2). Substituting the given values (E = 1.18 x 10^6 N/C, q = 1.60 x 10^-19 C, d = 27.1 x 10^-2 m, m = 8.81 x 10^-28 kg, v = 15.0 x 10^3 m/s) gives θ = 1.18 x 10^6 N/C * 1.60 x 10^-19 C * 27.1 x 10^-2 m / (8.81 x 10^-28 kg * (15.0 x 10^3 m/s)^2) = 2.68 x 10^-3 rad.