) Positive 4(-) Timer Notes Evaluate e. Feedback-Print A positive charge of 4.60
ID: 1575605 • Letter: #
Question
) Positive 4(-) Timer Notes Evaluate e. Feedback-Print A positive charge of 4.60 uC is fixed in place. From a distance of 3.60 cm a particle of mass 5.10 q and charge +3.90 uC is fired with an initial speed of 72.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away? The particle comes to rest when it has no kinetic energy. The change in kinetic energy will be equal to the increase in electrical potential energy. Note that the particle has some potential energy at the starting point ubmit Answer Tries 1/20 Previous Tries Post Discussion Send FeedbackExplanation / Answer
from conservation of energu
k1+u1 = k2+u2
u2 = k1+u1
k2 = 0 (rest)
u2 = 1/2mv1^2+kq1q2/r1 = 1/2*5.1*10^-3*72^2+9*10^9*4.6*3.9*10^-12/0.036
u2 = 13.219+4.485 = 17.704
kq1q2/r1 = 17.702
r2 = 9*10^9*4.6*3.9*10^-12/17.702 = 0.912*10^-2 m = 0.912 cm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.