Question

Cathode ray tube question

Answer 1

Diagram: http://puu.sh/bERqv/67e68b7551.png, hey I'm having a hard time finding delta y2 is it possible someone can show me how to get such, I got delta y1 using F = qE and uniform motion and constant acceleration. But, I'm having a hard time finding delta y2. I know \[x=x_0+v_xt~~~~y=y_0+v_yt\] and I see I have the velocities at t = 0. But I have no idea how to relate this with electrons at the end of plates.

Answer 2

Any help is much appreciated

Answer 3

I can show how I got \[\Delta y_1\] if that helps

Answer 4

delta y1 = q E L^2/2m v ^2 ?

Answer 5

Yup

Answer 6

I know how to get that, just having a hard time figuring out how to get delta y2

Answer 7

They move in the region between the plates that can be described as a combination of uniform motion in horizontal direction and motion with constant acceleration in the vertical direction.

Answer 8

That should be d^2 instead of L^2

Answer 9

yea i didn't look at pic.
delta y = Eq d^2 / 2mv^2.
For y 2

\[\tan \theta = \frac{ V _{y} }{ V _{x} } = \frac{ y _{2} }{ L }\].

Answer 10

thus you will get Y 2

Answer 11

Vy is the final velocity

Answer 12

I'm sort of confused, I see the final answer says \[y_2 = \frac{ eE_0dL }{m v_0^2 }\]

Answer 13

e=q

Answer 14

yea i know w8

Answer 15

By the way, thank you so much for helping, I really do appreciate it.

Answer 16

\[V _{y} final= at = \frac{ q E }{ m } \frac{ d }{ V _{x} }\]
and \[V _{x} will-stay-as-a-symbol \]
\[Y _{2 }= \frac{ L V _{y} }{ V _{x} }=substitute-with-V _{y}= \frac{ qELd }{ mv _{x}^{2} }\]

Answer 17

Ah, I see, thank you so much!

Answer 18

OK we have finished.
I am so slow :(

Answer 19

Lol, no you did good, I spent a lot of time earlier trying to figure it out, got no where though haha.