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Separating capacitor plates: (+) or (-) work?

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a parallel-plate capacitor with plates of area A and with separation d. Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.

    Hint i used
    If the plate separation were changed while the voltage was kept constant, the stored energy would change. The force between the plates would be the quantitiy that would be multiplied by the change in the plate separation to obtain the change in energy. In other words,

    F= -dU/dd

    Im not sure if im fully understanding this hint. Is the formula they've provided simply Work = force x distance? Besides im not actually supposed to move the plates, so what would i have to input for the the dd (change in distance) part?



    2. Relevant equations



    3. The attempt at a solution

    U = (ϵ0AV2)/(2d)

    The formula they told me to work out for energy stored in a capacitor.

    So im also feeling quite confused about whether work should be positive or negative.

    Because the two plates are oppositely charged, it means that they experience a force of attraction - therefore moving the plates further apart should mean positive work being done right?
    However in the hint for this question im working on, the work is deemed to be negative.

    From my experiences with work being done, i've always felt that doing positive work means that some external force has to act upon the object to cause that particular change in position. Therefore for negative work, it is like releasing an object and letting it move by itself without you having to put any extra effort in.

    If there is any easy way that people can determine whether work is positive or negative in the context of electric charge then please do tell.
     
    Last edited: Aug 21, 2011
  2. jcsd
  3. Aug 21, 2011 #2

    Doc Al

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    Staff: Mentor

    Are you given the potential energy function and then asked to find the force? That hint tells you how to find the force from the potential energy function--you're finding the derivative of U with respect to the distance.

    Right.
    Where does the hint say anything about work being positive?
     
  4. Aug 21, 2011 #3
    No i had to work out the function for U myself. It was an earlier part of the hint so i just chucked the equation in.

    Ok well starting from the equation F = -dU/dd, i thought that in general, dU would mean change in energy (like u2-u1) and dd means change in distance (d2-d1). Therefore the equation would become U2-U1 = - Force x (d2-d1) which is similar to the work = force x distance equation.

    Therefore i thought that the F = -dU/dd equation had somehow been derived from the concept of work being done. That then caused me to ponder why the negative sign was in there.
     
  5. Aug 21, 2011 #4

    Doc Al

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    Staff: Mentor

    Yes, that equation is related to the work being done. The negative sign is because you need to apply a force opposite to the force of attraction to calculate the work done. F is the force between the plates, so -F is the force you must exert.

    Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pen"
     
    Last edited by a moderator: Apr 26, 2017
  6. Aug 21, 2011 #5
    Ohhh ok, that makes sense and is simple too - i just thought too much and confused myself i think.

    Ok so back to the actual question where im supposed to find F(V). I differentiated U with respect to d and i got

    dU/dd = (-ϵ0AV2/2) x loged

    so does F(V) = dU/dd?
     
  7. Aug 21, 2011 #6

    Doc Al

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    Staff: Mentor

    What's the derivative of 1/x with respect to x?

    Don't forget the minus sign.
     
  8. Aug 21, 2011 #7
    woops yeah i was thinking of integration.

    instead of loged its supposed to be -1/d2

    so the two minus signs cancel each other out



    Got it correct! thnx heaps for clearing things up a little =]
     
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