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Jonask

  • 2 years ago

work done to move a capacitor in parrallel a distance x

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  1. yrelhan4
    • 2 years ago
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    *

  2. Jonask
    • 2 years ago
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    |dw:1364587708013:dw|

  3. yrelhan4
    • 2 years ago
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    I was just tagging myself.

  4. Jonask
    • 2 years ago
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    herre i mean to move it away from the other by applying a force \[F_A\]

  5. Jonask
    • 2 years ago
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    *

  6. electrokid
    • 2 years ago
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    work done = change in internal energy

  7. electrokid
    • 2 years ago
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    1) Energy in a capacitor \(U={1\over2}{Q^2\over C}\)

  8. Jonask
    • 2 years ago
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    F=qE

  9. electrokid
    • 2 years ago
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    2) what changes when you separate the plates? capacitance: \[C={\epsilon_0kA\over d}\]

  10. electrokid
    • 2 years ago
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    so, \[U={1\over2}{Q^2d\over\epsilon_0kA}\]

  11. electrokid
    • 2 years ago
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    W = final energy - initial energy

  12. electrokid
    • 2 years ago
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    remember that Q does not change in this case, but C and V do

  13. electrokid
    • 2 years ago
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    @Jonask you are increasing the separation between the plates, right?

  14. Jonask
    • 2 years ago
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    yes i was hoping for \[dW=Fdx\]

  15. Jonask
    • 2 years ago
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    i dont see x in you expressions above

  16. Jonask
    • 2 years ago
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    https://www.edx.org/courses/MITx/8.02x/2013_Spring/courseware/Week_3/sequence_3_1/

  17. Jonask
    • 2 years ago
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    i would like us to do it intergration way

  18. Jonask
    • 2 years ago
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    https://www.youtube.com/watch?feature=player_embedded&v=bV0zosBkKC0#at=116

  19. electrokid
    • 2 years ago
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    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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