A proton and an \"positron\" (identical to an electron, except positively charge
ID: 1595509 • Letter: A
Question
A proton and an "positron" (identical to an electron, except positively charged) are brought 6 mu m apart and released from rest. What is the initial potential energy stored by this system? U_sys = In all of the previous problems on this homework, the system's collective potential energy has been assigned to a single charge which is moveable. In this problem, we'll let BOTH charges move. The collective kinetic energy of the charges (K_sys) will be drawn from U_sys. However, since both charges are moving, they will need to "share" this energy. Once in motion, what percentage of K_sys does each charge have at any given moment? proton: K = % of K_sys positron: K = % of K_sys (In this scenario, how valid is our usual approximation to assign the entire U_sys to one particle? Which particle should it be assigned to?) What is the speed of each charge after they have repelled a long distance apart? proton: v_f = m/s positron: v_f = m/sExplanation / Answer
a. initial PE of two charge system = kq1q2/r = 8.98*10^9 *(1.6*10^-19)^2/6*10^-6 = 3.84*10^-23 J
b. at some random distance x m apart
PE-final = k(e)^2/d [ where e is charge on proton]
KE-final = PE initial - PE final ( form conservation of energy)
now, as both particles always experience dsame force, so the acceleration ratio of proton and psoitron, a/A = m/M [ M is mass of proton, m is mas of positron]
so, for initial velocity of zero for both particles, final velocity ratio of proton amd positron = v/V = a/A = m/M
KE ratio = (v/V)^2 = (m/M)^2
let toatal KE os sys be Ksys
(KE proton + KE positron) /KE positron = [(9.1*10^-31/1.6*10^27)^2 + 1]
KE positron/KE sys = 0.9994
% = 99.4
KE proton/LE sys % = 100 - 99.4 = 0.6 %
c. KE final = PE initial = 3.84*10^-23 J
Vf ( positron) = sqroot(2*0.9994 3.84*10^-23 J/9.1*10^-31 kg) = 9159.10643 m/s
Vf ( proton) = sqroot(2*0.006*3.84*10^-23/1.6*10^-27 kg) = 16.97 m/s