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Parallel Capacitor

  1. Apr 8, 2009 #1
    An electron is launched at a 45 angle and a speed of 5.0*10^6 m/sfrom the positive plate of the parallel-plate capacitor shown in the figure (Intro 1 figure) . The electron lands 4.0 cm away.

    1. The problem statement, all variables and given/known data
    Find the electric field inside the parrallel and the smallest possible spacing between the plates


    2. Relevant equations
    E = m*a/Q


    3. The attempt at a solution
    I have found the electric field inside already and it equals 3.5 *10^3 N/C. However I don't know how to find the smallest possible spacing between the plates

    Thanks for helping
     
  2. jcsd
  3. Apr 8, 2009 #2

    rl.bhat

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    How did you get the acceleration?
    Treat the motion of the electron as a projectile motion and find the mximum height. That will be spacing between the plates.
     
  4. Apr 8, 2009 #3
    Okay, so I think I screwed up the first part too
    I got wrong answer
    Can anyone help???????
     
  5. Apr 8, 2009 #4

    rl.bhat

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    Find the horizontal component of the velocity of the electron. Range is given. From that find the time of flight. Using the vertical component of the velocity and time, find the acceleration.
     
  6. Apr 8, 2009 #5
    So I did it and I found out that a=-3.125 *10^14, but when I plug it into equation to find E= am/Q I found the result which is as much as half of the correct answer.

    Any helps???????
     
  7. Apr 8, 2009 #6

    rl.bhat

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    To calculate acceleration you have to take half the time of flight.
     
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