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Homework Help: Calculating the magnitude of an electric field

  1. Dec 8, 2018 #1
    1. The problem statement, all variables and given/known data
    An electron enters the lower left side of a parallel plate capacitor and exits precisely at the upper right side (just clearing the upper plate). The initial velocity of the electron is 7*10^6m/s parallel to the plates. The capacitor is 2cm long and its plates are separated by 0.15m. Assume that the electric field is uniform at every point between the plates and find its magnitude.
    THE ANSWER IS = 2.09*10^5N/C
    Don't mind my weak attempt at diagram.

    2. Relevant equations
    U(y-direction) = U(x-direction)*tanΦ
    F=ma=qE
    x=U(x-direction)t
    y=U(y-direction)+½at2



    3. The attempt at a solution
    tanθ=0.15/0.02=7.5
    U(x-direction)= 7*106m/s
    U(y-direction)= U(x-direction)*tanΦ= 7.6*106*7.5=5.25*107m/s
    x=2cm=0.02m, y=0.15m
    x=U(x-direction)*t
    0.02 = 7*106*t
    t = 2.86*10-9s
    y = U(y-direction)+½at
    0.15 = 5.25*107*2.86*10-9 + ½(2.86*10-9)2a
    0.15=0.15015 + 4.0898*10-18a
    a= 3.668*1013ms-2
    F = ma = qE
    = 9.11*10-31 * 3.668*1013 = qE
    ⇒ 1.6*10-19 * E = 3.3415*10-17
    E=208.85NC-1
     

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    Last edited by a moderator: Dec 10, 2018 at 11:09 AM
  2. jcsd
  3. Dec 8, 2018 #2

    neilparker62

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    I would guess you don't need any angle since the initial direction is parallel to plate and that component of velocity will not change. You just use it to get Δt which can then be used to determine acceleration perpendicular to plate.
     
  4. Dec 8, 2018 #3

    gneill

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    Staff: Mentor

    The problem statement says:
    (my emphasis on the direction). Why have you assigned a y-component to the initial velocity?

    Your diagram showing a diagonal trajectory does not make sense. The electron will be accelerating in the y-direction, remaining at constant velocity in the x-direction (no forces acting that way). So the path will be parabolic in form.
     
  5. Dec 9, 2018 #4
    But the electron entered at the lower left side and exited at the upper right side of the capacitor.. That's from the question. So, I thought it's path should be diagonal.
     
  6. Dec 9, 2018 #5
    I think I did that.. from my solution but are you insinuating I use the same velocity?
     
  7. Dec 9, 2018 #6
    Its path is a curve, but INITIALLY the velocity is horizontal and the U in those equations is INITIAL velocity.

    The idea is the same as throwing a ball horizontally off a cliff. Initial velocity is all horizontal. In a ball off a cliff, gravity acts on one component but not the other, so the equations you use for one component are accelerated and the equations for the other component use a = 0.
     
  8. Dec 9, 2018 #7

    gneill

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    No. It'll be parabolic, like a thrown stone in a uniform gravitational field.
    upload_2018-12-9_10-27-33.png
     
  9. Dec 9, 2018 #8
    TYSM
     
  10. Dec 9, 2018 #9
    I'm sincerely grateful. I get it now
     
  11. Dec 9, 2018 #10

    gneill

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    Staff: Mentor

    Great! Cheers!
     
  12. Dec 10, 2018 at 9:07 AM #11

    SammyS

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    @Haddon ,

    By the way: . . . . :welcome: .
     
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