WiteyPLUS My PLUS Halliday, Fundamentals of Physics, 10e Physics for Engineers (

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WiteyPLUS My PLUS Halliday, Fundamentals of Physics, 10e Physics for Engineers (Physics for Engineers) Assignment Gradebook ORION Downloadable eTextbook Chapter 22, Problem 040 An electron with a speed of 5.74 x 10° cm/s in the positive direction of an x axis enters an electric field of magnitude 1.05 x 10 N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is 5.05 mm long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region? (a) Number 0.08924 Units (b) Number 1.55E-8 Units (c) Number Units This answer has no units Click if you would like to Show Work for this question: Open Show Work By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.

Explanation / Answer

Given

electron speed is v = 5.74*10^8 cm/s = 5.74*10^6 m/s

electric field E = 1.05*10^3 N/C ,

a)

as the electron is moving in the electric field , the electric force acts on it  

F = E*q,

here the work done on the electron is equal to the change in kinetic energy

W = F*s = (k.e2-k.e1)

==> s = d(k.e)/W

==> s = (0.5*m*v^2)/(E*q)

===> s = (0.5*9.1*10^-31*(5.74*10^6)^2)/(1.05*10^3*1.6*10^-19)

===> s = 0.0892331 m

so the distance covered by the electron before stopping is s = 0.0892331 m

b) using equations of motions  

V = u-at

v =0 m/s, u = 5.74*10^6 m/s , a = F/m = E*q/m = (1.05*10^3*1.6*10^-19)/(9.1*10^-31) m/s2 = 1.8462*10^14 m/s2

t = u/a = (5.74*10^6)/(1.8462*10^14) s

t = 3.1091*10^-8 s

c) now the electric field is confined to a region of 5.05*10^-3 m only

so

the initial k.e = 0.5*m*v^2 = 0.5*9.1*10^-31*(5.74*10^8)^2 J = 1.4991*10^-13 J

the work done in 5.05*10^-3 m is equal to the k.e lost

so work done W = E*q*d

W = 1.05*10^3*1.6*10^-19*5.05*10^-3 J = 8.484*10^-19 J

the fraction of energy lost is  

W/k.e = 8.484*10^-19)/(1.4991*10^-13)

W/k.e = 5.6594*10^-6

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