anonymous
  • anonymous
work done to move a capacitor in parrallel a distance x
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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yrelhan4
  • yrelhan4
*
anonymous
  • anonymous
|dw:1364587708013:dw|
yrelhan4
  • yrelhan4
I was just tagging myself.

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