lem FULL SCREEN PRINTER VERSION BACK NEXT Chapter 22, Problem 062 Your answer is
ID: 3162004 • Letter: L
Question
lem FULL SCREEN PRINTER VERSION BACK NEXT Chapter 22, Problem 062 Your answer is partially correct. Try again. ua) what is the magnitude of an electron's acceleration in a uniform electric neld of magnitude 1.60 x 106 M/c? (b) How long would the electron take, starting from rest, to attain 112-th the speed of ligbt? (c) How far would it travel time? Units m/s, 2 (a) Number 2.81E17 Units T times (b) Number 8.90E-11 (c) Number 1.112E-3 click if you would like to show question: open sbow work Work for this Question Attempts 1 of 3 used SAVE rom LATER SUBMIT ANSWER copyright o zooo 2017 by John Wiiev 6 sons, inc or related companies Version 4.22.1.2 Division of John sons All Rights ReservedExplanation / Answer
a]
Given:-
E= 1.60*106 N/C
F= ma and F= qE
Hence ma= qE
a= qE/m= 1.6*10-19* 1.60*106 / 9.1*10-31= 0.28* 1018 m/s2
b]
Vi= 0m/s [rest]
Vf=1/12* c= 1/12* 3*108 m/s= 0.25*108 m/s
Vf= Vi+at
t= Vf- Vi / a= 0.25*108 / 0.28* 1018 = 0.893*10-10 s
c]
d= Vi* t+1/2at2
d= 1/2at2 [ Since Vi=0 m/s]
d= 1/2* 0.28* 1018*( 0.893*10-10 ) 2 = 0.22*10 -2 m = 2.2 mm
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