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Work Done/Electric Potential

  1. Jun 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Two conducting spherical shells are concentric, with radii of 0.70 m and 1.60 m. The electric potential of the inner shell, with respect to the outer shell, is +1050 V. An electron is transported by an external force from the inner shell to the outer shell. The work done by the external force is closest to:

    Solution: 1.68 * 10^16 J


    2. Relevant equations
    [​IMG]


    3. The attempt at a solution

    Well, I thought I would have to encorporate one of the above formulas, however, I couldnt seem to use any of them.....then, I decided I would use

    qV = U....and plugging this in it works...

    ie (1.6*10^-19C)*(1050V) = 1.68*10^-16 J

    HOWEVER- I hate simply plugging in values and would like to actually understand...would anyone be able to explain where this formula comes from/how it is derived...or point me in the right direction as to understanding...cheers i appreciate your time!



    cheers
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 30, 2007 #2
    im an idiot.....

    seen as Potential Energy Difference = Work Done

    And we know Potential = Energy per unit charge, it makes sense that the potential difference of 1050V multiplied by the charge concerned (ie electron) will give the Change in Potential Energy and thus the work done.

    Is this corect thinking?

    Thanks again if you read this
     
  4. Jul 10, 2007 #3

    GmL

    User Avatar

    Yes, that's correct. You're looking for the potential difference betwen the two shells which happens to be +1050 V, so PE = qV is the correct formula to use.
     
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