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Electric Circuit Potential

  1. Nov 27, 2006 #1
    During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.0 x 10^8 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

    I have no idea how to do this problem. Haven't done electric circuits in a year. Any tips or suggestions or formulas to start me off? I was thinking since the number given is change in Potential...can we use a potential formula like V = kq/r? Im not sure what applies here. Any tips at all would be really appreciated, thank you.
     
  2. jcsd
  3. Nov 27, 2006 #2

    berkeman

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    Staff: Mentor

    Hint -- I believe that electron movement was probably part of a lightning strike....

    Also, do you know what units are used with electrical potential?
     
  4. Nov 27, 2006 #3
    Hint -- I believe that electron movement was probably part of a lightning strike....

    I don't get that hint, I'm probably really dumb since it's probably sarcasm haha. Does that mean it's equal to the potential difference given?


    electrical potential = voltage?
     
  5. Nov 27, 2006 #4

    berkeman

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    Not dumb at all. The cloud is described at being at the higher positive potential, and that means that the electric field is pointing from the clouds to the earth. F=qE, so the force on a positive charge would be from the clouds down, but the physical situation is that there is a force on electrons in the ground, and they all arc over at once to generate the lightning bolt. So yes, at least from my perspective, the electrons *lost* the potential energy associated with the initial voltage difference. Does that make sense?
     
  6. Nov 27, 2006 #5
    It makes a little sense...let me see if I can understand...you're saying the electrons that go from the clouds to the ground, lose the given potential difference? Since they are afterall going from higher to lower. But now that it's the other way...is it simply gaining this change?
     
  7. Nov 27, 2006 #6

    berkeman

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    No, the electrons are starting on the ground and zinging up to the cloud. The - charge of the electrons is attracted by the + charge (lack of electrons) in the cloud. Remember that electron flow is always in the opposite direction of the imaginary "positive current", so electrons are accelerated in the opposite direction of the electric field vector.
     
  8. Nov 27, 2006 #7
    AHI understand that now...but what just came to mind is... Work = qV. I used this formula and got the right answer for this problem.
     
  9. Nov 27, 2006 #8

    berkeman

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    Yes, one of the units for energy is electron volts. Quiz question -- 1J = how many eV?
     
  10. Nov 27, 2006 #9
    6.25 x 10^18 eV? :-)
     
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