C) starting with the three particle system, find the change in electric potentia
ID: 1596088 • Letter: C
Question
C) starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place) D) use conservation of energy to calculate the speed of the alpha particle at infinity E) if the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity My Notes o Ask Your Teacher 1 points l Previous Answers sercP9 16 P019 10 (a) Calculate the electric potential energy associated with this configuration. 363e-14 Ju t (o, y) (3.17, 3.17) fm. Calculate the electric potential energy associated with this 2e, mass 6.64 x 10 kg) is now (b) An alpha particle (charge configuration 224.77 check each step carefully.) Your response differs significantly from the correct answer Rework your solution from the beginning and (c) Starting with the three particle system, find the change in electric potential energy rthe alpha particle is allowed to escape toinnnity whee the two protons (d) use conservation of energy to calculate the speed of the alpha particle at infinity. (e) the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity. If Need Help? LReadIL
Explanation / Answer
Part B :
total Potential energy due to system of three charges
Unet = U12 + U23 + U31
since q1 =q2
where Each U = Kq1q2/r = Kq^2/r
here q1= q2 = 1.6 *10^-19C
q3 = 3.2 *10^-19C
Unet = Ua +U31 *2
r31^2 = (3.17^2 + 3.17^2)
r31 = 4.48 cm
Unet = 3.63*10^-14 + 2 * ((9*10^9 * 1.6 *10^-19* 3.2*10^-19)/(4.48*10^-15))
Unet = 2.42 *10^-13 J
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part C:
change in PE = 2.42 *10^-13 - 3.63 *10^-14
Change in PE = 2.057*10^-13 J
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part D :
use change in PE = KE
0.5 mv^2 = 2.057 *10^-13
v^2 = 2* 2.057 *10^-13/(6.67 *10^-27)
v = 7.85 *10^6 m/s
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oart E:
use change in PE = totaa KE
0.5 mv^2 + 0.5 mv^2 = 2.057 *10^-13
v^2 = 2.057*10^-13/(6.67*10^-27)
v = 5.55*10^6 m/s
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