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Calculate Potential Energy of Electron at a Parallel Plate Capacitor

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data

    An electron with a speed of 5.0x10^6 m/s is injected into a parallel plate apparatus, in a vacuum, through a hole in the positive plate. The electron collides with the negative plate at 1.0x10^6 m/s. What is the potential difference between the plates?
    Vi= 5.0x10^6

    Me= 9.11x10^-31
    2. Relevant equations
    Ke = 1/2 mv^2
    Ee =q∆V
    3. The attempt at a solution
    I believe I should calculate the change in KE of the electron.
    Since masses are the same, I can cancel them out and only use velocities.
    (1.0*10^6)^2 - (5.0x10^6)^2 = -2.4x10^13

    -2.4x10^13/(1.60x10^-19) = -1.5x10^32

    That is obviously an incredibly high potential difference. The answer in the textbook is 68V.

    What's the proper ay to solve this?
  2. jcsd
  3. Nov 8, 2014 #2


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    Staff: Mentor

    Work with units (a very good idea in general), then the error will be obvious.
  4. Nov 9, 2014 #3

    rude man

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    Gold Member

    What are E and e? Does not look right. q∆V has units of energy.
    This is 2 x ∆KE. Why 2?
    How about final KE = initial KE - loss in potential energy.
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