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Find the velocity of the electrons at the end of the plates

  1. Sep 18, 2014 #1
    Hey guys, I'm having some trouble figuring this part out, so I've done most of this problem just can't figure out this part...so here is the question. (Sorry in advance as I don't know how to use LaTeX on here).
    What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.

    The diagram: http://puu.sh/bERqv/67e68b7551.png [Broken]

    I'm having a hard time figuring out delta y2, I figured out y1, as the electron is enters the electric field, using equations x = v0t and y = 1/2at^2 and using equation F = qE, I solved for a, and then solved for y1 = (qEx^2)/(2v0^2m)

    As I mentioned above, I'm not sure exactly how to solve for y2, but I know the equations for constant velocity x = x0+vxt and same goes for y, so I eliminate t, and then I'm totally lost as to what to solve for/ vx and vy and finding the expression for the end of the plates.

    Thanks
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 18, 2014 #2

    Simon Bridge

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    Oh you mean for a CRT.

    Hint: look at the triangle.
    How do you Δy2 is the opposite side of a right-angled triangle. The adjacent side has length L. How would you normally find the opposite side knowing the adjacent and the angle?
     
    Last edited by a moderator: May 6, 2017
  4. Sep 18, 2014 #3
    Using tan theta ratio, but how would you know the angle?
     
  5. Sep 18, 2014 #4

    Simon Bridge

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    From the angle that the electron leaves the plates - what is the velocity vector of the electron?
     
  6. Sep 18, 2014 #5
    As the electron enters the field, it follows a curved path.
     
  7. Sep 18, 2014 #6
    I'm still unsure how exactly to get an expression for vx and vy of the electrons at the end of the plates? How would I go on to combine uniform motion and constant acceleration for between the plates?

    (x-x0)/vx = (y-y0)/vy

    vx, vy = (x-x0)*(vy)/(y-y0), and the same thing for vy but I have no idea how to relate this to the parallel plate...
     
    Last edited: Sep 18, 2014
  8. Sep 18, 2014 #7

    Simon Bridge

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    How did you do this for ballistics?
     
  9. Sep 19, 2014 #8
    I'm sorry, I don't remember, is it possible you can just show me how to do it, I have some sense as in t = 0 so I'll have x = vxt and y = vyt, but I have no idea at all how to go from there, I've tried many things and spent many hours...x=v0t, y = 1/2at^2, I tried something with this as well got no where. I just can't relate the terms as I did with the first part with this one.

    Or if anything I could use a few more hints.
     
    Last edited: Sep 19, 2014
  10. Sep 19, 2014 #9

    Simon Bridge

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    Do each component separately.
    At t=0, you know the speed in the y direction and the speed in the x direction.
    You also know the acceleration in the y direction and the acceleration in the x direction.
    You know some kinematic equations.

    If you are having trouble remembering how to do this, just look it up in your notes or online.
    Keyword "ballistics".
     
  11. Sep 19, 2014 #10
    I got it...all I needed was tan theta = vy/vx and was simple from there.
     
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