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Finding the Electric Field of a Parallel Plate Capacitor

  1. Jul 26, 2016 #1
    1. The problem statement, all variables and given/known data

    An electron is launched at a 45∘ angle and a speed of Vo = 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor towards the negative plate. The electron lands 4.0 cm away.

    2. Relevant equations

    F = ma = qE

    v = d/t

    s = vt -(1/2)at^2

    x = 0.04 m

    3. The attempt at a solution

    We can find t by using trigonometry:

    t = x/(cos45)

    yf - yi = 0 = Vo * sin(45) * t - (1/2)at^2

    Vo*sin(45) = (1/2)at

    using F = ma = qE,

    a = qE/m

    so,

    Vo*sin(45) = t * qE/(2m)

    and using t =x/cos(45)

    Solving for E,

    E = (2m * Vo * sin(45) * cos(45))/(qx)

    but cos(45)*sin(45) = 1/2

    Therefore,

    E = (Vo * m)/(q*x)

    At this point I plug in my constants:

    Vo = 5*10^6 m/s
    m = 9.11 * 10^-31
    q = 1.6 * 10^-19
    x = 0.04

    However I am getting a wrong answer. Could anybody point out where it is I've gone wrong? Thanks!
     
  2. jcsd
  3. Jul 26, 2016 #2

    BvU

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    Seems strange. Doesn't have the same dimensions left and right ....:rolleyes:
     
  4. Jul 26, 2016 #3

    Charles Link

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    I think your error is the equation t=x/cos(45). It needs a ## v_o ## in it.
     
  5. Jul 26, 2016 #4
    Of course -_- . Thanks! Silly mistake.

    For anyone else who wants to use this thread:

    My mistake was writing t=x/cos(45) instead of t=x/(cos(45)*Vo)

    The end formula would then be:

    E = (m* Vo^2)/(q*x)
     
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