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Physics 12 Electric Energy Question

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data

    A proton starts at rest and travels through a potential difference of 8000V. What's it's final speed.
    2. Relevant equations
    ΔEe+ΔEk=0
    ΔEk=-ΔEe
    Ee=qV

    3. The attempt at a solution
    I know how to find the correct magnitude, but I am getting the wrong sign.

    Ek=-ΔEe
    1/2mv^2=-qΔV
    v=root [-(2qΔV)/m]

    Note mass of proton=1.67 x 10^-27 kg

    v=root[-2(1.6.x10^-19)(8000))/1.67 x 10^-27]

    I get an error because obviously there isnt a root for a negative value of [-2(1.6.x10^-19)(8000))/1.67 x 10^-27].
    I want to know where I am going wrong in terms of positive/negative.
     
  2. jcsd
  3. Sep 5, 2015 #2

    haruspex

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    The question doesn't state which way the potential difference is, it only gives you the magnitude.
    But you are given that it starts at rest and only moves because of the field, so which way will it move, to a higher potential or to a lower one?
     
  4. Sep 5, 2015 #3
    Won't it be lower because it's losing potential electric energy and getting kinetic energy?
     
  5. Sep 5, 2015 #4

    haruspex

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    An electron has a negative charge. If it moves to a lower potential does it gain or lose PE?
     
  6. Sep 5, 2015 #5
    Lose
     
  7. Sep 5, 2015 #6

    haruspex

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    Try a different question. If a positively charged particle moves to a lower electrical potential, does it gain or lose PE?
     
  8. Sep 5, 2015 #7
    Oh, does the elctron gain due to the negative charge? If so then the positive particle would lose PE.
     
  9. Sep 5, 2015 #8

    haruspex

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    An electron moving to a lower electrical potential would gain PE: ##q \Delta V > 0##. So which way is this electron moving?
     
  10. Sep 5, 2015 #9
    The proton would be moving against the electric field I believe.

    However, this question is talking about a proton.
     
  11. Sep 5, 2015 #10

    haruspex

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    Oh, I'm sorry - don't know why I read it as electron.
    So, restart.
    The proton will move to a lower potential, so ##\Delta V < 0##, ##\Delta PE = q \Delta V < 0##. Now, is ##\Delta KE## equal to ##\Delta PE## or to ##-\Delta PE##?
     
  12. Sep 5, 2015 #11
    KE will be equal to -PE
     
  13. Sep 5, 2015 #12

    haruspex

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    Right. So does that get rid of your minus sign?
     
  14. Sep 5, 2015 #13
    Yes, but how come when solving algebraically there is a minus sign. I want to understand it algebraically.
     
  15. Sep 5, 2015 #14

    haruspex

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    I thought I explained it algebraically.
    The proton will move to a lower potential, so ##\Delta V < 0##, ##\Delta PE = q \Delta V < 0##, ##\Delta KE = -\Delta PE= -q \Delta V > 0##.
    If it were an electron, both the sign of the charge and the sign of the potential difference would change, so the change in KE would still be positive.
     
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