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Work done in moving the plates of the capacitors

  1. Nov 23, 2013 #1
    A parallel plate capacitor of circular cross section r>>d, and separation d. It is charged to potential V then disconnected from the circuit . What will be the work done in moving the capacitor from d to d1?

    Answer:
    Here the initial energy is 1/2C[itex]V^{2}[/itex]
    Where C=Aε/d
    While moving the capacitor both the capacitance and potential change:
    Cf= Aε/d1
    Vf=∫E.dl (with the limits 0 to d1)
    Vf= E*d1
    We know that the E doesn't change ( Under the approx. as r>>d )
    Therefore initial potential and final potential can be related as V= E*d
    Vf=V*d1/d
    Work done = 1/2( Cf*[itex]Vf^{2}[/itex] - C*[itex]V^{2}[/itex] )
    = Aε/2 ([itex](V*d1/d)^{2}/d1[/itex] - C*[itex]V^{2}/d[/itex] )
    = [itex]AεV^{2}( d1/d -1)/2d[/itex]
    I want to know if this answer is correct.
     
    Last edited: Nov 23, 2013
  2. jcsd
  3. Nov 23, 2013 #2

    Meir Achuz

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    It's easier to use U=Q^2/2c because Q doesn't change for a disconnected capacitor, but your answer is correct.
     
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