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

[sami23]

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

(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

 

[ehild]

(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]

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

u = 7*10^6 m/s

tan $$\varphi$$ = v/u = 0.215

$$\varphi$$ = 12.15 degrees.

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

 

[ehild]

Yes.

ehild

 

[paranormal]

= 1.939 cm

Thank you for providing the given parameters. Based on the information provided, the deflection of the electron in a CRT can be calculated using the equations of motion.

(a) The distance below the axis that the electron will reach at the end of the plates can be calculated using the equation y = ut + (1/2)at^2, where y is the vertical distance, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get y = (0)(8.57*10^-9) + (1/2)(1.76*10^14)(8.57*10^-9)^2 = 6.465*10^-3 m or 6.465 mm.

(b) The angle at which the electron will be moving as it leaves the plates can be calculated using the equation tanθ = y/S, where θ is the angle with the axis, y is the vertical distance, and S is the horizontal distance. Plugging in the values, we get θ = tan^-1(6.465*10^-3/0.06) = 6.15 degrees.

(c) To calculate the distance below the axis where the electron will strike the fluorescent screen, we can use the equation tanθ = y/(S+S'), where S' is the additional horizontal distance from the end of the plates to the screen. Plugging in the values, we get y = (0.06+0.12)tan6.15 degrees = 0.01939 m or 1.939 cm.

It is important to note that these calculations assume ideal conditions and do not take into account factors such as air resistance or the curvature of the CRT screen. The actual deflection of the electron may vary slightly in real-world scenarios.