What is the deflection of an electron in a CRT with given parameters?

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Homework Help Overview

The problem involves the motion of an electron in a cathode-ray tube (CRT) under the influence of a uniform electric field. The electron is projected with a specified initial speed, and the task includes determining its vertical displacement and angle of deflection as it exits the electric field region.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for vertical displacement and the angle of deflection of the electron as it exits the plates. There are attempts to clarify the correct angle of the velocity vector and how it relates to the initial speed.

Discussion Status

Some participants are exploring the calculations for the angle of deflection, with one participant questioning their result. Another participant confirms the angle derived from the velocity vector, indicating a productive exchange of ideas regarding the interpretation of the problem.

Contextual Notes

There is a focus on the definitions of angles in relation to the velocity vector and the setup of the problem. Participants are also addressing potential misunderstandings about the calculations involved.

sami23
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Homework Statement


An electron with an initial speed of 7x106m/s is projected along the axis midway between the deection plates of a cathode-ray tube. The uniform electric field between the plates has a magnitude of 1000 V/m and is upward. F = 1.6x10-16N and acceleration a = 1.76x1014m/s2
YF-23-63.jpg


(a) How far below the axis has the electron moved when it reaches the end of the plates?
(b) At what angle with the axis is it moving as it leaves the plates?
(c) How far below the axis will it strike the fluorescent screen S?


An electron with an initial speed u = 7x106 m / s
The uniform electric field between the plates E = 1000 V/m
Upward force F = 1.6x10-16N
acceleration a = 1.76x1014m/s^ 2
Horizontal direction :
--------------------------
distance S= 6 cm =0.06 m
time taken to reach end of the plate t = S / u
= 8.57*10^-9 s
In vertical direction :
--------------------------
Initial velocity U = 0
acceleration a = 1.76 * 10^ 14 m / s^ 2
time t= 8.57 * 10 ^ -9 s
distance moved at the end of the plates in vertical direction y = ?
from the relation y = Ut + ( 1/ 2) at ^ 2
= 0 + 6.465* 10 ^ -3
= 6.465 * 10 ^ -3 m

(b). required angle θ = tan -1 ( y / S )
= 6.15 degrees ? I get this wrong


(c). tan θ = Y / ( S + S ' )
where S ' = 12 cm =0.12 m
from this Y = ( S+ S ' ) tan θ
= 0.01939 m
 
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sami23 said:
(b) At what angle with the axis is it moving as it leaves the plates?


(b). required angle θ = tan -1 ( y / S )
= 6.15 degrees ? I get this wrong

You need the angle of the velocity vector.

ehild
 
v = 0 + (1.508*10^6 m/s) = 1.508*10^6 m/s

u = 7*10^6 m/s

tan [tex]\varphi[/tex] = v/u = 0.215

[tex]\varphi[/tex] = 12.15 degrees.

Is this angle where it is moving as it leaves the plates?
 
Yes.

ehild
 

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