Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric force between 2 parallel plates

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    An electron is projected with an initial velocity of 1.6x10[tex]^{6}[/tex] m/s. If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field?


    2. Relevant equations

    Electric force equation

    3. The attempt at a solution

    I am stuck trying to figure out the magnitude of the electric field, once I can figure this out I know how to solve the problem. Any pointers on how to find the magnitude of the electric field?
     
  2. jcsd
  3. Jan 24, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Use ordinary kinematic means to determine the velocity.

    You know the speed, hence how long for it to emerge.

    In that time you also know the deflection so you can determine the acceleration.

    That acceleration then yields the additional sideways component of velocity to calculate it's speed at that point right?
     
  4. Jan 24, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You don't need to know the magnitude of the electric field.
     
  5. Jan 24, 2009 #4
    I was planning to use Vf^2 = Vo^2 +2ad to find the final velocity, but I am missing the acceleration component. To find the acceleration I wanted to use F = ma (knowing the mass of the electron). But I would need the magnitude of the electric field to find the force from E = F / q.
     
  6. Jan 24, 2009 #5
    So how would I do this without using the electric field? I don't know what you mean about using the "deflection" to find the acceleration. How would I calculate that?
     
    Last edited: Jan 24, 2009
  7. Jan 24, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    The horizontal velocity to the end of the plate gives you time.

    Use the distance, acceleration, time relationship to determine acceleration.

    Then you can use your V2, acceleration and distance.
     
  8. Jan 24, 2009 #7
    Thanks for the advice, but this does not work. I get the answer wrong, the magnitude of the electric field is 364N/C, I am just not sure how to find it. When I use this electric field with the method I stated above I get the right answer. I just can't figure out how they got 364N/C :(
     
    Last edited: Jan 24, 2009
  9. Jan 24, 2009 #8

    LowlyPion

    User Avatar
    Homework Helper

    Once you determine the acceleration from the trajectory, then you can use f = ma to determine field intensity.

    They ask really though for just the speed.

    That equals (Vx2 +Vy2)1/2
     
  10. Jan 24, 2009 #9
    Yes! Thank you so much, I figured it out now =)
     
  11. Jan 24, 2009 #10

    LowlyPion

    User Avatar
    Homework Helper

    I wondered if that wasn't it.

    Glad you got it.

    Good luck.
     
  12. Jul 25, 2009 #11
    Please will someone explain how I am meant to calculate the time taken for the electron to reach the end of the plates. I'm just not getting it.
     
  13. Jul 25, 2009 #12
    you have the horizontal distance and the velocity in x-direction is constant, so t = v / s
     
  14. Jul 25, 2009 #13
    oh dear. that was rather dense of me. thank you! :)
     
  15. Jul 25, 2009 #14
    You're welcome ^^
     
  16. Jul 26, 2009 #15
    Sorry, i'm not getting the right answer..not sure where i'm going wrong..should it not be t=s/v because t=v/s yields an answer of 80,000,000
     
  17. Jul 26, 2009 #16

    Doc Al

    User Avatar

    Staff: Mentor

    To check if an equation makes sense, look at the units. t = v/s → (m/s)/s = m/s^2; these are units of acceleration, not time, so this equation makes no sense.

    Since v = s/t, t = s/v is correct. The units would be m/(m/s) = m(s/m) = s. Makes sense.
     
  18. Jul 26, 2009 #17
    oh sorry, it's my mistake
    it should be t = s/v
     
  19. Jul 26, 2009 #18
    thank you :)
     
  20. Jul 26, 2009 #19
    You're welcome
    sorry for the mistake earlier ^^
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?