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Work and Energy Loss; Electrostatics

  1. May 15, 2012 #1
    Griffith's Problem 2.40 (a) and (b)

    Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance ε, as a result of their mutual attraction.

    a) Use

    [tex]
    P= \frac{\epsilon_0}{2}E^2
    [/tex]

    to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.

    b) Use

    [tex]
    \frac{\epsilon_0}{2}E^2 = energy per unit volume
    [/tex]

    to express the energy lost by the field in this process.


    I solved the problems in my (they're both the same answer in the answer key) way getting

    [tex]
    \frac{(q^2)\epsilon}{2A\epsilon_0}
    [/tex]

    But Griffiths doesn't do that, he doesnt even delve into what E is, which I solved ambiguously using A. His answer was:

    [tex]
    \frac{\epsilon_0}{2}(E^2)A\epsilon
    [/tex]

    Is it alright that I expanded my E?
     
  2. jcsd
  3. May 16, 2012 #2

    rude man

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    Homework Helper
    Gold Member

    Well, not really. The poblem demands that the answer be in terms of A and E.

    However: was it specified whether the potential difference across the plates was kept constant, or was the voltage source used to charge the capacitor removed before the displacement took place? Makes a big difference ....
     
  4. May 16, 2012 #3
    It didn't specify. What I posted was as much as he wrote in for the particular question. But the point you brought about about A and E was overlooked and that makes sense.
     
  5. May 16, 2012 #4
    No, it makes a small difference (infinitesimally small, to be precise ;-P).
     
  6. May 17, 2012 #5

    rude man

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    Homework Helper
    Gold Member

    Well, if you don't consider 2:1 as big, then OK, it makes no big difference.
     
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