Stuck on this question!! Do I use conservation of energy? Newton\'s second law?
ID: 1462039 • Letter: S
Question
Stuck on this question!! Do I use conservation of energy? Newton's second law? But I dont have the velocity....
20-78 In an oscilloscope, an electron is accelerated from rest through a potential difference of 1.14 X 10' V. It then passes through deflecting plates that are 2.00 cm apart, across which there is a potential difference of 412 V (Figure 20-43). The electron enters the plate region with a velocity that is perpendicular to the electric field of the plates. When the electron leaves the plate region, what distance will it have moved toward the positive plate? 2.00 cm AV- 412V Electron 4.00 cm Figure 20-43 Question 20-78Explanation / Answer
kinetic energy gained by the electron = q*1.14*10^3
(1/2)*m*v^2 = q*1.14*10^3
v = sqrt(2*q*1.14*10^3/m)
= sqrt(2*1.6*10^-19*1.14*10^3/(9.1*10^-31))
= 2*10^7 m/s
Electric fiedl between lates, E = delta_V/d
= 412/0.02
= 20600 N/c
electric force on electron upward, F = q*E
m*a = q*E
a = q*E/m
= 1.6*10^-19*20600/(9.1*10^-31)
= 3.62*10^15 m/s^2
time taken to cross the plates, t = x/v
= 0.04/(2*10^7)
= 2*10^-9 s
distancetravelled towards posistive plate in this time, y = 0.5*a*t^2
= 0.5*3.62*10^15*(2*10^-9)^2
= 0.00724 m or 0.724 cm or 7.24 mm