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Parallel Plates Question

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data

    An electron traveling 1.3*10^2m/s enters the region between two parallel plates at the distances shown. When it collides with the plates, it has a speed of 1.8*10^3 m/s. What is the electric field between the plates? What is the electrostatic force on the electron? And how far along the plate does the electron hit?

    ------- +
    |3.0cm (3cm between electron and top plate (positive)
    |
    electron
    |
    |
    |10cm distance between electron and negative (bottom plate)
    ------------- negative
    2. Relevant equations



    3. The attempt at a solution

    I tried using V= ΔE/q but than realized that i probably had to use E = ΔV/ΔD

    I really dont know how to go about this. You dont have to give me all the answers, just some steps please
     
  2. jcsd
  3. Jan 7, 2013 #2

    gneill

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    Staff: Mentor

    Why don't you start with the change in kinetic energy? You're given the initial and final speeds. How does the change in KE relate to the change in electric potential over the given distance?
     
  4. Jan 7, 2013 #3
    I tried change in energy divided by charge it didnt work for potential difference
     
  5. Jan 7, 2013 #4

    gneill

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    Sure it will. But keep in mind it's the potential difference through which the electron falls; the change in potential between the initial position of the electron (distance 3.0 cm from the + plate) to the final position of the electron (at the + plate); it's not the full potential difference between the two plates.
     
  6. Jan 7, 2013 #5
    ok so i did V= .5 (9.11*10^-31) * ((1800^2) - (130^2)) / 1.6*10^-19

    = 9.175762813*10^-6 - its on the sheet but is that right? i must have done something wrong before

    Than for electric field i would do 9.17...*10^-6 / (10-3cm)?

    Fe= Eq

    than how about how far along the plate how do i find that?
     
  7. Jan 7, 2013 #6

    gneill

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    Looks okay.
    Make that 3x10-2 m, not cm. The field is the change in potential divided by the distance, and here the distance is 3 cm but to make the units match write 3x10-2 m
    It's a projectile problem. Use the force and known mass to find the acceleration in the y-direction.
     
    Last edited: Jan 8, 2013
  8. Jan 8, 2013 #7
    so (10*10^-2 - 3*10^-2) ?

    and i forget how to do projectile can you guide me in the right direction?
     
  9. Jan 8, 2013 #8

    gneill

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    No, just 3x10-2. The electron is only falling through 3 cm. It just so happens that the field between the plates is (essentially) uniform, so knowing the field strength anywhere will give it to you for the whole region.
    Start with the separate equations of motion for the x and y components of the trajectory. Find out how long (time) it takes for the electron to strike the plate (y-direction motion).
     
  10. Jan 8, 2013 #9
    Is it not falling through 10cm? because it starts at 3cm below the top plate, than drops 10cm to the bottom plate?

    And like kinematics equations? d=1/2at-vit^2? and such?
     
  11. Jan 8, 2013 #10

    gneill

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    Which direction is the electric force acting on the electron? (Which plate will attract the negatively charged electron? Which repels it?)
    Yup.
     
  12. Jan 8, 2013 #11
    Isnt it positive to negative, so down?

    Ah alright
     
  13. Jan 8, 2013 #12

    gneill

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    Can you clarify that? Isn't what positive to negative?

    According to your diagram the top plate is positive, the bottom plate is negative. There are positive charges on the top plate, negative charges on the bottom plate. Will the negative electron be attracted to the negative charges on the bottom, or to the positive charges on the top?
     
  14. Jan 8, 2013 #13
    OH positive to negative i think i was thinking of magnetism lol even than i dont even remember if thats correct. It will be attracted to the top plate since electron is negative i got it.
     
  15. Jan 9, 2013 #14
    I dont think i got the potential difference right, because i tried all 9.2*10^-6 / 3*10^-2 and 10^-2 and 7^-2 i tried all the denominators and i never got an answer on the sheet
     
  16. Jan 10, 2013 #15

    CAF123

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    Gold Member

    Do they give you the distance between the plates in the problem statement? I don't see how you can tell what distance the electron has travelled without knowing this distance. If I understand the problem statement correctly, they give you the relative distances of the electron to the plates at either end but not the separation. Or maybe it's the case that it enters at 3cm below top plate in which case you would know the distance.
     
  17. Jan 10, 2013 #16

    gneill

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    Well, 3.0 cm is 0.03 m, or 3 x 10-2 m.

    Why, what are the choices on the sheet? Is the problem statement exactly as you've written it?
     
  18. Jan 10, 2013 #17
    Yes it is the distance between plates is given the length of plates is not.

    The distance is 3cm from the top plate to the electron ( just coming in) an 10cm from the electron to the bottom plate.

    Would itbe bettet if i took a picture of it?
     
  19. Jan 10, 2013 #18

    gneill

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    This is my understanding of the setup:

    attachment.php?attachmentid=54607&stc=1&d=1357838722.gif
     

    Attached Files:

  20. Jan 10, 2013 #19
    Yes thats right except the curvature isnt drawn on mine but that diagram is accurate

    The 4ans given are
    7.1 x10^-5
    6.1x10^-5
    1.2x10^-6
    1x10^-6
    All N/C
     
  21. Jan 10, 2013 #20

    gneill

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    It's possible that the given answers are incorrect (they may correspond to a previous revision of the problem where the given values were different).

    What value did you calculate for E?
     
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