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Electron in a parallel plate capacitor - finding plate's magnitude

  1. Mar 24, 2008 #1
  2. jcsd
  3. Mar 24, 2008 #2
    anyone? thanks.
     
  4. Mar 24, 2008 #3
    I believe the posters in the previous thread hinted at the strategy quite nicely, but I suppose I'll detail it a bit more. First note the equations for kinematics. Let's define the vertical direction as our y-direction, and the horizontal direction as our x-direction. What is happening to the electron in the x-direction in terms of force (and subsequently accelertion)? Consider the same for the y-direction. You should be able to use these kinematic equations to find the acceleration of the electron. Then consider Newton's Second Law, and how the acceleration fits into this equation. This will allow you to find the net force, which is an electric force (hint hint). How can you write down the electric force in terms of the charge of the particle and the electric field?
     
    Last edited: Mar 24, 2008
  5. Feb 18, 2009 #4
    I have the acceleration, but nothing else makes sense after that. Care to explain?
     
  6. Feb 19, 2009 #5
    Oh wow, this is quite an old thread. I guess the problem was to find the magnitude of the electric field. If you already have the acceleration of the electron, you're all set. What's the equation for Newton's Second Law? What's the equation for electric force? How do these equations relate to one another in this particular problem?
     
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