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Force to open a capacitor

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A parallel plates capacitor has charge Q and the plates are at distance x. they are separated incremental distance dx. what is the force F to separate them.

    2. Relevant equations
    The capacity: ##C=\varepsilon\frac{A}{x}##
    The work done to charge a capacitor: ##W=\frac{1}{2}\frac{Q^2}{C}##

    3. The attempt at a solution
    The differential of the capacity:
    $$dC=-\frac{\varepsilon A}{x^2}dx$$
    The differential of the work:
    $$dW=-\frac{Q^2}{2C^2}dC=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}dx$$
    $$F\cdot dx=dW=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}dx$$
    $$F=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}$$
    It should be ##F=\frac{Q^2}{2\varepsilon A}##
    Is it reasonable that the force depends on the distance x, as in my result?
     
  2. jcsd
  3. Dec 10, 2014 #2

    ehild

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    It is correct for an ideal capacitor.
    The derivation could have been a simpler by using that the force between the plate is the negative gradient of the potential energy.
    potential energy of the capacitor with charge Q: ##U=-\frac{Q^2}{2C}=-\frac{Q^2x}{2 \varepsilon_0 A}##
    ##F=-\frac{dU}{dx}=\frac{Q^2}{2\varepsilon_0 A}##.
     
  4. Dec 10, 2014 #3
    Thanks, it's correct, but i was instructed this way by the book and i want to know why my way isn't correct.
    Not only x is in my result, it's totally different with the constant's placement
     
  5. Dec 10, 2014 #4

    ehild

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    It is correct, but substitute the expression for C. x will cancel.
     
  6. Dec 10, 2014 #5
    Thanks
     
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