Deflection in a CRT


by sami23
Tags: deflection
sami23
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#1
May21-10, 12:26 AM
P: 76
1. The problem statement, all variables and given/known data
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


(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
Accleration 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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ehild
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#2
May21-10, 01:02 AM
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Quote Quote by sami23 View Post
(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
sami23
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#3
May21-10, 08:52 AM
P: 76
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?

ehild
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#4
May21-10, 09:59 AM
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P: 9,817

Deflection in a CRT


Yes.

ehild


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