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Homework Help: Calculating distance of an electron moving between 2 plates

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Electrons are accelerated from REST by a potential difference of 1.0 x 10^3 volts
    They pass midway between two flat parallel plates 0.15m long and 0.030m apart. The electric field between the plates has a magnitude of 2.0 x 10^3 N/C. Determine how far an electron will be deflected (x) from it's original horizontal path by the time that it reaches the other end of the plates

    I can't use most of the formulas because there is constant acceleration involved after it reaches the 0.35 m mark, thus I can't get a x nor a y and the acceleration is just in the X axis,
    I'm really stuck!

    can anybody please help me?

    2. Relevant equations
    e = 1.602 x 10^-19
    m = 9.11 x 10^-11
    k = 9.0 x 10^9 N M^2/C^2

    ε = V/r
    ε = Fe/q
    Vf^2 = Vi^2 + 2ad
    d = v2t - 1/2at^2
    d = v1t - 1/2at^2

    3. The attempt at a solution
    First find the acceleration
    F = ma
    qε = ma
    (1.602x10^-19)(2000) = 9.11 x 10^-31a
    a = 3.52 x 10^14

    next find the distance between starting position and the end of the plates
    ε = V/r
    2000 = 1000/r
    r = 0.50

    after that find the distance between starting position and the start of the plates
    0.50 - 0.15 = 0.35m

    next you find the final velocity (the second before the electron reach's the plates)
    Vf^2 = Vi^2 + 2(3.52x10^24)(0.35)
    vf = 1.6 x 10^7

    Attached Files:

    Last edited: Jan 15, 2012
  2. jcsd
  3. Jan 15, 2012 #2
    The first thing to do is calculate the velocity of electrons accelerated through 1000V
    use eV = 0.5mv^2
    These electrons then travel into the space betwen the parallel plates where they experience a vertical force.
    Hope this helps
  4. Jan 15, 2012 #3
    Well it only kinda helped, I am still unsure if that is the answer.

    eV = 1/2mv^2
    1.602 x 10^-16 = 4.555 x x10^-31V^2

    v = 1.9 x 10^7

    t = 0.15/1.9 x 10^7
    t = 7.0 x 10^-9

    d = V1t + 1/2at^2
    d = 0 + 1/2(3.52 x 10^14)(7.0 x 10^-9)
    d = 0.008624

    The answer makes logical sense, because we know that the space between the plates is 0.030m

    and it's only deflected a bit so it would make sense that the vertical deflection would be 0.008624m

    can anyone confirm this?
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