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Spring and Capacitor

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=80f528f2f9b90f379aac09ad5190c4f8.png

    2. Relevant equations

    Force of attraction between the plates = (1/2)(QV)/d

    3. The attempt at a solution

    Initial charge on the capacitor Q1=CV
    Final charge on the capacitor Q2=2CV

    When switch is open force between the plates F1 = (1/2)(Q1V)/d1 = (1/2)(CV2)/d1

    When switch is closed force between the plates F2 = (1/2)(Q2V)/d2 = (1/2)(2CV2)/d2

    d2 = (3/2)d1

    F1/F2 = 3/4 or F2 = (4/3)F1

    The spring force also becomes 4/3 times of the initial force i.e (4/3)F0

    Is it correct ?

    Many Thanks
     

    Attached Files:

  2. jcsd
  3. Aug 24, 2015 #2

    mfb

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    Be careful here, the capacitance changes.
    The potential changes as well, which influences the calculation of F2.
     
  4. Aug 24, 2015 #3
    :sorry:

    F1/F2 = 8/9 or F2 = (9/8)F1

    The spring force also becomes 9/8 times of the initial force i.e (9/8)F0

    Is it correct now?
     
  5. Aug 24, 2015 #4

    mfb

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    Can you show your steps? I get a different result.
     
  6. Aug 24, 2015 #5
    Sorry once again .

    It should be (16/9)F0 .

    If it is wrong , i will surely show you the steps :smile: .
     
  7. Aug 24, 2015 #6

    mfb

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    That agrees with the answer I got.
     
  8. Aug 24, 2015 #7
    Thanks mfb .
     
  9. Aug 24, 2015 #8
    The spring stretches in both the situations . More in the latter case , as attractive force between the plates gets stronger.

    Right ???
     
  10. Aug 24, 2015 #9

    mfb

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    Right. Which also means the system has to be moved a bit, otherwise the solution doesn't make sense.
     
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