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Work on equipotential surface

  1. May 2, 2007 #1
    The work in joules required to carry a 6.0 C charge from a 5.0 V equipotential surface to a 6.0V equipotential surface and back again to the 5.0V surface is:

    A) 0
    B) 1.2 X 10^-5
    C) 3.0 X 10^-5
    D) 6.0 X 10^-5
    E) 6.0X10^-6

    I was thinkin the work is 0, but then again that seems too easy
  2. jcsd
  3. May 2, 2007 #2


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    Gold Member

    Do you have a reason for thinking it's 0 ?
  4. May 2, 2007 #3
    Is it 0 because of [tex]W=F*d=qEd[/tex]?? The work done would be equal to 6C(1V)=6J when you bring it from 5V to 6V (6-5=1) and the work done would be equal to 6C(-1V)=-6J when it is going down the voltage from 6V to 5V (5-6=-1). Therefore when you add 6J and -6J you get 0? Is my solution correct?
  5. May 2, 2007 #4


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    Gold Member

    I can't argue with that.
  6. May 2, 2007 #5
    thats what I was thinking, I just didn't know because mult choice questions are so tricky!

  7. May 3, 2007 #6
    Zero..Work is of equal manitude but of opposite signs..they cancel to make net work 00000
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