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work done to move a capacitor in parrallel a distance x

Physics
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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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