## JenniferSmart1 2 years ago question

1. JenniferSmart1

$\Delta y=\Delta y_1+\Delta y_2=\frac 12 a(\frac{L}{v_0})^2+L +D$ $x_1=v_0t$ $t=\frac{x_1}{v_0}=\frac{L}{v_0}$ $\Delta y_1=v_0t+\frac 12 at^2$ $\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2$ $\Delta y_1=\frac 12 a(\frac{L}{v_0})^2+L$ now outside o the acceleration plates $x_2=v_0t$ $t=\frac{x_2}{v_0} =\frac D{v_0}$ $\Delta y_2=v_0(\frac D{v_0})$ Here is where I'm stuck: $ma=qE$ $a=\frac {qE}{m}$ $KE=U$ $\frac 12 mv_0^2=qV_acc$ $v_0^2=\frac {2qV_acc}{m}$ now i Have $\Delta y=\frac{qE}{m}(\frac{Lm}{2qV_acc})^2+L +D$

2. JenniferSmart1

but my equation should look like $\Delta y=\Delta y_1+\Delta y_2=\frac{EL}{2V_{acc}}\left( D+\frac L 2\right)$

3. JenniferSmart1

I probably made a mistake somewhere, because my m's aren't cancelling out....and some other stuff too I guess

4. JenniferSmart1

feel free to write on here fellas:)

5. JenniferSmart1

$y_1=v_yt+\frac 12 at^2$ is there a y velocity within the plates?

6. JenniferSmart1

7. TuringTest

can I type today on OS (testing)

8. JenniferSmart1

9. experimentX

10. TuringTest

oh goodie, I can :) ok, what is the exact info you are given? just what's in the pic of the CRT you showed me?

11. JenniferSmart1

yes

12. TuringTest

so we are supposed to get the initial x-velocity from the voltage generated by the AC, which you say is 500V. I don't know where you get that number, but if it is correct answer @experimentX 's question

13. experimentX

yep .. it's called accelerating potential.|dw:1362064638627:dw|

14. JenniferSmart1

I'm not worried about that anymore. do we know that the V_acc is exactly 500V or is there some uncertainty. should i mention that there is some uncertainty

15. experimentX

this is how you accelerate electrons http://en.wikipedia.org/wiki/File:WaterCooledXrayTube.svg No it doesn't matter for theory. Just for error analysis ..

16. JenniferSmart1

for my analysis I mean

17. JenniferSmart1

would you like to see my graph?

18. experimentX

19. experimentX

hmm ... looks like you could that data to calculate errors.

20. experimentX

I"v never been good with error analysis ... but let's see what I can do.

21. JenniferSmart1

no need to I guess....the xcel program did that already

22. experimentX

he who?? I know couple of error analysis method like Standard ERror

23. JenniferSmart1

never mind @experimentX I'm running out of time....I'm gonna make changes and submit what I have.

24. experimentX

hmm ... what methods are asked to use?

25. JenniferSmart1

he said to just describe the method but don't go into too much detail. I described it already and I just need to tweek a couple of things in my report to make it look better. The equations are fine I guess, I'll just use the final displacement equation without too just some derivation steps. Thanks for you help though. I'm not gonna learn anything new in the next 15 mins

26. experimentX

okay .. I'll send you the details of derivation if you want ... within next 15 hrs. (optional: if you want)

27. JenniferSmart1

how do I insert a picture in latex?

28. experimentX

wow ... i don't know ... maybe these guys know http://tex.stackexchange.com/

29. JenniferSmart1

thanks everyone...I learned my lesson, but it's not like I was playing video games, I've been busy...maybe I should quit my job

30. experimentX

wow!! ... looks like I am the only free loader at home here.

31. JenniferSmart1

wait, you don't work?

32. experimentX

Not since i joined grad school.

33. JenniferSmart1

I see, anyhow gotta go....thanks though @TuringTest and @experimentX ....I'll work on my time commitments

34. experimentX

well ... didn't do much except troll around. Maybe next time .. :P

35. experimentX

|dw:1362138326802:dw||dw:1362138537113:dw|

36. experimentX

|dw:1362138832951:dw|

37. experimentX

So until now we have expression for $$v_x$$ and $$a$$ ( and 'a' is perpendicular to the original path) |dw:1362139377040:dw|

38. experimentX

|dw:1362139616525:dw||dw:1362139773642:dw|

39. experimentX

Your this expression is incorrect $\Delta y_1=v_0\frac{L}{v_0}+\frac 12 a(\frac{L}{v_0})^2$ $$v_0$$ is zero by default ... no y-component.

40. experimentX

|dw:1362142311512:dw||dw:1362142342862:dw||dw:1362142369122:dw||dw:1362142375743:dw|

41. experimentX

$\Delta y = \Delta y_1 + \Delta y_2 = \frac{EL^2 }{4V_{acc} } + \frac{EDL}{2V_{acc}}$

42. experimentX

in my equations read d as L ... rest are same.