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Potential energy of a capacitor

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data

    What happens to the potential energy of a capacitor with length d when said length d is doubled? Use the formula for the potential energy of a capacitor [U = 1/(2C) * Q]. After that, refer to the work done by moving the capacitor the distance d. These two approaches should yield the same result!

    2. Relevant equations

    U = 1/(2C) * Q, and, I guess, the traditional formula for work, W = F * d.

    3. The attempt at a solution

    I know how to solve the first part, because C is halved when d is doubled, so the potential energy is doubled. But I have absolutely no clue as to how to even begin working on the second part. Any tips?
     
    Last edited: Oct 10, 2013
  2. jcsd
  3. Oct 10, 2013 #2

    LCKurtz

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    What does the "length of a capacitor" mean? How does moving a capacitor change its potential energy which apparently depends on Q and C only? I'm not a EE, but I don't understand this problem.
     
  4. Oct 10, 2013 #3
    Sorry, d is the length of the space between the capacitor.
     
  5. Oct 10, 2013 #4

    LCKurtz

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    So are you really asking about moving one plate of the capacitor a distance d, thus changing its potential energy? If so, what formulas do you have relating C, d, Q, and F? I'm asking because I don't know them and I think you need them to work the problem.
     
  6. Oct 10, 2013 #5

    BruceW

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    yeah, that's the right place to start for the second part. What is the force in this case? p.s. sorry for jumping in.
     
  7. Oct 10, 2013 #6
    Well, we have F according to Coulomb's law, and U = 1/(2C) * Q, and then there is C = (εr0*A/d, where A is the area of the capacitor and d is the aforementioned length.

    The trick, I think, I think, is to figure out exactly what F is. If we find out F, the rest should be easy.
     
  8. Oct 10, 2013 #7

    BruceW

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    yep. And true, it is electrostatics, so it is technically Coulomb's law for each charge. But you have many charges. Do you remember the equation for the electric field inside a parallel-plate capacitor?
     
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