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Electric Fields and Electric Potential

  1. Jul 20, 2017 #1
    1. The problem statement, all variables and given/known data
    An Electron enters a parallel plate apparatus that is 8cm long and 4cm wide. The electron has a horizontal speed of 6*10^7 m/s. The potential difference between the plates is 6*10^2 V. Calculate the electrons velocity as it leaves the plates. (ans. 6*10^7 m/s [E 3.3° N])

    2. Relevant equations
    ΔE=qΔV
    V=kq/r
    E=kq/r2
    θ = Tan-1 = (Vfy/Vfx)
    ay=qE/m
    ΔT=L/vi
    Vf = (Vfy2 + Vfx2)1/2

    3. The attempt at a solution
    I FOUND OUT HOW TO DO

    Electric Field Magnitude --> E

    E = -
    ΔV/ΔD
    E = -600V/0.04m
    E = -1500 V/m

    Vertical Acceleration:
    ay = (qE)/m
    ay = ((-1.6022*10-19C)(-1500V/m))/0.08m
    ay=2.64098901*1014m/s2

    Time to reach end of apparatus:
    vi = L/Δt
    Δt = L/vi
    Δt = 0.08m/6*107m/s
    Δt = 4/3*10-9s

    Final Vertical Velocity:
    Vfy = AyΔt
    Vfy = (2.64098901*1014 m/s2)(4/3*10-9s)
    Vfy = 352131.868m/s

    Velocity Vectors added:
    Vf = (V2fx+V2fy)0.5
    Vf = 60001033.2982m/s

    Direction of speed:
    θ = Tan-1(Vfy/Vfx)
    θ = 3.34°

    Therefore, final velocity of electron as it exits apparatus is 60001033.2982m/s [E 3.34° N]
     
    Last edited: Jul 20, 2017
  2. jcsd
  3. Jul 20, 2017 #2
    This is a step in the right direction. With this, what is the force on the electron? Once you have the force, what is the acceleration?
     
  4. Jul 22, 2017 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Since you're computing to 9-12 significant digits you'll have to correct for relativistic kinematics!
     
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