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Work to separate a plane capacitor

  1. Jun 15, 2017 #1
    1. The problem statement, all variables and given/known data
    A capacitor with C is charged by a battery to a voltage V and then disconnected. The distance between plates is slowly increased by an external force. What is the change of the electrostatic energy of capacitor during this process?

    2. Relevant equations
    I have two ways to solve the problem but I don't know which is correct.

    3. The attempt at a solution
    If the distance changed from d1 to d2, then
    [tex]
    C'=C\frac{d_{1}}{d_{2}}
    \\
    V'=V\frac{d_{2}}{d_{1}}
    \\
    \frac{1}{2}C'V'^{2}-\frac{1}{2}CV^{2}=\frac{1}{2}(C\frac{d_{1}}{d_{2}})(V\frac{d_{2}}{d_{1}})^{2}=\frac{1}{2}CV^{2}(\frac{d_{1}}{d_{2}}-1)
    [/tex]
    On the other hand, work can be defined as W=Q[V'-V], then
    [tex]
    W=Q(V'-V)=QV(\frac{d_{1}}{d_{2}}-1)=CV^{2}(\frac{d_{1}}{d_{2}}-1)
    [/tex]
    But it seems that there is a missing factor (1/2) in the second solution!
    Where did I go wrong?
     
  2. jcsd
  3. Jun 15, 2017 #2

    BvU

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    Here already :$$
    \frac{1}{2}C'V'^{2}-\frac{1}{2}CV^{2}=\frac{1}{2}(C\frac{d_{1}}{d_{2}})(V\frac{d_{2}}{d_{1}})^{2}=\frac{1}{2}CV^{2}(\frac{d_{1}}{d_{2}}-1)
    $$W changes from + to - halfway ?
    And try ##dW = Q dV## for the second approach
     
  4. Jun 15, 2017 #3
    Okay!! I got it!!
    Thanks a lot! haha!
     
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