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Proton release inside a parallel capacitor Find velocity

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    The electric field strength is 47100 N/C inside a parallel-plate capacitor with 8.60 mm spacing. A proton is released from rest at the positive plate. What is the speed of the proton (in m/s) when it reaches the negative plate?

    2. Relevant equations
    Vf=√(-2U/m) I think?
    ΔU=-.5eEd



    3. The attempt at a solution

    I tried using that equation to find ΔU=-3.768*10^-15
    then plugged it in seem to have the wrong final answer though
     
    Last edited by a moderator: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    tiny-tim

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    welcome to pf!

    hi conov3! welcome to pf! :wink:
    why .5 ?
     
  4. Feb 5, 2012 #3
    Thank you! I figured since I am in a college level physics, I might as well try to get some help if I need it! ha

    Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)
    Then under that it says =-1/2eEd

    I am confused about it also and am really unsure.
     
  5. Feb 5, 2012 #4

    tiny-tim

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    i don't understand that … it doesn't seem to fit the question :confused:
     
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