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Algebra Clarification Requested - The Fate of an Electron in a Uniform Electric Field

  1. Jan 20, 2005 #1
    The electron, having been held at height h0, is now released from rest. Calculate its speed v when it reaches the top plate.

    I found the initial energy to be:

    -q_e*E*h_0

    and the final energy to be:

    1/2*m*v^2+-q_e*E*h_1


    And I need to solve for v, so I set the two equal to each other as per the conservation of energy thereom and solve for v, correct?

    -q_e*E*h_0 = 1/2*m*v^2+-q_e*E*h_1

    My problem is isolating v and cancelling/clearing up variables. My algebra is lagging a bit, admittedly. Thanks for any help.
     
  2. jcsd
  3. Jan 20, 2005 #2

    dextercioby

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    Pay attention with the signs...The skeleton of your equation is
    [tex] A=B+v^{2}C [/tex]

    Isolate v^{2} and then take root of second order...

    Daniel.
     
  4. Jan 20, 2005 #3
    Thanks. I only get one shot at an answer (it's an online-based assignment), so I would appreciate any response saying if this is correct or not, and what I did wrong if it is incorrect:

    [tex] sqrt((2(h_0-h_1))/(m))=v[/tex]
     
  5. Jan 20, 2005 #4
    Can anyone else help me out please?

    EDIT: Nevermind, I got it.
     
    Last edited: Jan 20, 2005
  6. Jan 21, 2005 #5

    dextercioby

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    Excellent,that was the idea,to get it,without too much help... :smile: Anyways,i was asleep... :tongue2:

    Daniel.

    P.S.I didn't dream of your exercise... :tongue2:
     
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