Please help, i think i should use the equaiton 1/2m(v)^2=q(of e-)(v), but then h
ID: 2000894 • Letter: P
Question
Please help, i think i should use the equaiton 1/2m(v)^2=q(of e-)(v), but then how would the distance term fall out?????
Consider the setup below, which is a view of an electron trajectory through a CRT. Only the vertical deflection plates are shown; the horizontal deflection plates are not shown for simplicity. The x-axis is along the originating electron beam direction, and the y-axis is along the direction of the vertical electric field between the plates. Using the information below, calculate the angle at which the electrons have been deflected relative to the +x- axis when they emerge from the vertical plates. Show your work and use extra paper as needed. Note: Electron mass: 9-11x10-31 kg; electron charge:-e =-1.602x10-19 C 1. E-500 N/Cj Eo(375/d) N/C i Steps to calculate : Electrons are boiled off the heat filament shown at left. Some of them emerge through a pinhole, which creates a beam of low energy ( v·0) electrons. The electrons are first accelerated through an electric field E. The magnitude ofthe electric field in this region, Eo, is given as (375/d) N/C where d is the distance (in meters) over which this electric field is applied. For example, if d in the figure is 2.00 cm, the electric field in this region is (375/0.02) 18750 N/C. In this setup, we don't know what "d" is but it should drop out of your calculations. Calculate the velocity ·Explanation / Answer
Here, 1/2*mv2 = qE*d
=> v2 = 2qE*d/m
= 2*1.6*10-19*375/9.11*10-31
=> v = 11.47*106 m/sec -------------> horizontal velocity
Vertical velocity = (1.6*10-19*500/9.11*10-31) * (0.01/11.47*106)
= 87.81*1012 * 8.718*10-10
= 765.52*102 m/sec
=> angle of deflection =tan-1(765.52*102/11.47*106)
= 0.382 degrees