work done to move a capacitor in parrallel a distance x

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work done to move a capacitor in parrallel a distance x

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1364587708013:dw|
I was just tagging myself.

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Other answers:

herre i mean to move it away from the other by applying a force \[F_A\]
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work done = change in internal energy
1) Energy in a capacitor \(U={1\over2}{Q^2\over C}\)
F=qE
2) what changes when you separate the plates? capacitance: \[C={\epsilon_0kA\over d}\]
so, \[U={1\over2}{Q^2d\over\epsilon_0kA}\]
W = final energy - initial energy
remember that Q does not change in this case, but C and V do
@Jonask you are increasing the separation between the plates, right?
yes i was hoping for \[dW=Fdx\]
i dont see x in you expressions above
https://www.edx.org/courses/MITx/8.02x/2013_Spring/courseware/Week_3/sequence_3_1/
i would like us to do it intergration way
https://www.youtube.com/watch?feature=player_embedded&v=bV0zosBkKC0#at=116
you do not necessarily have to use the integration way. total work done is independent of path followed; depends on the initial and final states only!

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