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Homework Help: Is Potential Energy Work?

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to look for where the potential is zero on the same line where two different charges lie.
    I'm given the values of the two charges and the distance between them, I have to find where the potential is zero.

    2. Relevant equations

    3. The attempt at a solution
    I'm thinking about using the equation
    W = kQ1Q2 ( 1/d1 - 1/d2)
    I tried plugging in for Q1 and Q2 and the given distance into either d1 or d2. Where's my mistake? Or I'm supposed to use another equation? Also, is the W the same as the required "potential?"
  2. jcsd
  3. Feb 23, 2008 #2
    Potential energy(and potential!) and work are not the same thing, but they're related in these cases when dealing with electric fields(and also gravitational fields)

    More importantly you appear to have also confused potential and potential energy

    Do you know the equation for potential? You can write an equation for both charges, add them and set them equal to 0 to solve for the required position

    As far as the difference between PE and work in a conservative force field(like gravity or electric fields)if you change something's potential energy you have DONE work to it against its respective force field

    The potential function is technically potential energy per charge in an electric field, which is a little abstract, and it may not seem obvious why it's useful, but look at this problem. You found where the potential is 0, so you know F=-grad(potential) so you've actually found where the force cancels out between the two of them, without having to screw around with the force equation(and also from that equation you know wherever the potential is constant the force also equals 0
    Last edited: Feb 23, 2008
  4. Feb 24, 2008 #3
    Thank you for clearing that up. I'm using the V=kQ/d equation now. I'm actually working on the same problem as the one another user has asked about. But I got it now, thanks.
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