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Deflection in a CRT: Problem more than halfway solved, stuckby SuperCass
Tags: electric potential 
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#1
Feb2110, 02:02 PM

P: 60

1. The problem statement, all variables and given/known data
In Fig. 2432 an electron is projected along the axis midway between the deflection plates of a cathoderay tube with an initial speed of 6.40 106 m/s. The uniform electric field between the plates has a magnitude of 1.04 103 V/m and is upward. (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in part (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S? 2. Relevant equations F=Eq F=ma V=W/q V=kq/r 3. The attempt at a solution I got parts ac correct, a being 1.666e16 N downward, b being 1.8473e14 m/s^2 downward, and c being .008118 meters. Part d is giving me trouble. I've tried doing arctangent of .06/.008118, with .06 being the length of the plates and .008118 being how far down it travelled, but that didn't work. What am I doing wrong? I'm also stuck on part e, but I think that's because I don't have part d. Thank you!! 


#2
Feb2110, 02:44 PM

HW Helper
P: 3,394

In (d) get the angle of the velocity vector. The horizontal component is given and you can find the vertical component from the vertical acceleration (you must have found the travel time when doing part c).



#3
Feb2110, 03:52 PM

P: 60

Thanks!
For part e, would I just use tangent of that angle and multiply it by .12 to get the needed value? Because that isn't working for me. What should I do? 


#4
Feb2110, 04:36 PM

PF Gold
P: 2,022

Deflection in a CRT: Problem more than halfway solved, stuck
Hello SuperCass for part e did you remember to add the deflection you calculated in part c?



#5
Feb2110, 04:44 PM

P: 60

*gasp* no!! That worked!! Thank you so much!!



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