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Homework Help: Work required to move capacitor plates

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine the work required to move the plates of an ideal parallel plate capacitor from a seperation distance of D to an integer multiple of D, (assume only one moves for simplicity). The each plate has an area A.

    2. Relevant equations
    F = q^2 /( 2 e_0 A) field strength between the plates
    [tex]\Delta[/tex] V = U/q
    [tex]\Delta[/tex]U = -W
    [tex]\Delta[/tex]V= Qd/e_0 A



    3. The attempt at a solution

    moving the terms above around I came up with U = Qdq/e_0A
    I know I need an integral here, probably evaluated between D and multiple of D(its a general solution thats needed)
    Am I on the right track?
     
  2. jcsd
  3. May 25, 2010 #2
    e6909617aaece890a9d11784ece37423.png

    Though it's a little cheap and probably not what your professor had in mind, you can use the equation for energy density(J/m^2) stored in an electric field to solve this problem. Do final energy stored minus initial energy stored to calculate what energy was added (or done) to the system.
     
  4. May 25, 2010 #3
    I appreciate the reply. I thought about that, but I use that in another part of the problem, not that I can't use it again since its different variables, but Im fairly certain I'll be graded on the diversity of methods used.

    Anyone know if that integral approach would work or am I grasping at fog?
     
  5. May 25, 2010 #4
    Why not use the appropriate equation for energy stored in a capacitor?
     
  6. May 25, 2010 #5
    well since you know the force between the two plates

    Work is

    W=Integral (Fdr)
     
  7. May 26, 2010 #6
    Thats what I thought but I encounter a problem, when I compare the integral result to one obtained by use of potential energy, Im getting a different result depending on different terms. Considering that the work should be the same, Im stuck. Is there anyway Q^2= Vbat?
     
  8. May 26, 2010 #7
    In the question has the capacitor been charged and disconnected from the battery?If so Q remains constant and V changes if the plate separation is changed.If the capacitor remains connected to the battery V remains constant and Q changes.
     
  9. May 26, 2010 #8
    Also remember the potential energy of a system is (CV^2)/2
     
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