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Parallel plate capacitor question

  1. Feb 13, 2017 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate capacitor is connected to a 100 V battery until it is fully charged. The distance between the plates is d and the space between the plates is filled with air (k=1.0). Then, the battery is disconnected. If the distance between the plates is increased to 3d, the voltage of the capacitor will be ?
    2. Relevant equations
    C=ɛ0*A/d
    C=Q/V
    U=QV/2
    E=V/d

    3. The attempt at a solution
    So i know that capacitance is inversely prop. to distance. so if distance is tripled then capacitance will be 1/3 C but i cannot relate that to voltage. I know the correct answer is 33V but i cannot get my head about it. I know that if i multiply 100*1/3 i will get 33V but how does multiplying capacitance (F)by volts will give me volts.
     
  2. jcsd
  3. Feb 13, 2017 #2

    cnh1995

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    Which quantity remains the same in both the cases?
     
  4. Feb 13, 2017 #3
    Area i believe?
     
  5. Feb 13, 2017 #4

    cnh1995

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    ..and??
     
  6. Feb 13, 2017 #5
    Area and k (dielectric)
     
  7. Feb 13, 2017 #6

    cnh1995

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    Ok. What about charge?
     
  8. Feb 13, 2017 #7
    Charge is not related?
     
  9. Feb 13, 2017 #8

    cnh1995

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    Charge has nowhere to go I believe. So wouldn't it remain consatnt?
    V=Q/C.
     
  10. Feb 13, 2017 #9
    I GOT IT. Thank you.

    So as distance becomes tripled. capacitance becomes 1/3 and since capacitance is inversely proportional to voltage. so C goes 1/3 voltage is multiplied by 3 so 100*3=300V
     
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