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Circuit with a parallel plate capacitor

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    1.) Consider a parallel plate capacitor with plates of area 1.36 m2 whose plates are 1.24 cm apart. The gap between the plates is filled with air (assume that κair is unity) and the positive plate has a charge of 10.5 nC on it while the negative plate has a negative charge of equal magnitude on it.

    The capacitor has its plate-spacing reduced to 1.12 mm, and the plates are connected to a 1.51 V battery. So that the charge on the positive side of the capacitor is now 16.2nC.

    The capacitor is disconnected from the battery and the plates are returned to their initial spacing. So that the voltage across the capacitor is now 16.7V.

    The space between the plates is now filled with a fluid with a dielectric constant of 5.07.
    So that the voltage across the capacitor is now 3.297V.

    The area of overlap between the capacitor plates is increased to 2.01 m2.

    What is the charge on the positive plate of the capacitor?


    2. Relevant equations



    3. The attempt at a solution

    V = Q/C
    C = AE/d

    Q = CV = AEV/d
    d = 0.00112m
    V = 1.51V
    A = 1.36
    Q = 16.2nC

    V = Q/C = Q/ (AE/d) = Qd/AE
    d = 0.0124m
    Q = 16.2nC
    A = 1.36m2
    V = 16.7V

    V = V(0)/k
    V(0) = 16.7V
    k = 5.07
    V = 3.297V

    Q = CV = AEV/d
    A = 2.01m2
    V = 3.297V
    d = 0.0124m
    Q = 4.73nC
    BUT Q IS WRONG? WHY IS IT WRONG?
     
  2. jcsd
  3. Sep 26, 2008 #2

    LowlyPion

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    Re: Circuits

    Where does charge come from after the battery is disconnected?
     
  4. Sep 26, 2008 #3
    Re: Circuits

    i guess there is no new charge so is the charge still 16.2nC?
     
  5. Sep 26, 2008 #4
    Re: Circuits

    if it is...why is V = Q/C = 11.2V? If C = AE/d = 2.01(8.85*10^-12)/0.0124
    According to the answer key it is supposed to be 2.23V using the formula initial VC = final VC.
     
  6. Sep 26, 2008 #5

    LowlyPion

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    Re: Circuits

    Yes. That's what it looks like to me.
     
  7. Sep 26, 2008 #6

    LowlyPion

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    Re: Circuits

    Wasn't the question just how much charge?

    The voltage will drop when the area increases if the charge is the same.
     
  8. Sep 30, 2008 #7
    Re: Circuits

    oh okay. thank you so much!
     
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