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Parallel Plates, Capacitor and Dielectrics

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    The parallel Plates of a capacitor have an area of 2.00 x 10-1m2 a separation distance of 1.00 X 10-2 m and are connected to a 3000 V power supply.
    The capacitor is then connected from the supply, and a dielectric is inserted between the plates. We find that the potential difference decreases to 1000 V while the charge on each plate remains constant.

    Find the following:
    a. original capacitance
    b. magnitude of charge
    c. capacitance, C after a dielectric is inserted
    d. the dielectric constant k of the dielectric
    e. permittivity of the dielectric

    2. Relevant equations
    a. C=ε0 A/D
    b. Q = CΔV ---im not sure

    sadly the handouts given to me ends on that chapter on computing magnitude of electric field.
    i wanted to know about this badly.

    3. The attempt at a solution
    a. C=(8.85X10-12 C2(Nm2))(2.00X10-2 m2)/(1.00X10-2 m) = 1.77X10-11
    F=1.77 μF

    b. Q=CΔV = (1.77X10-11 F)(3000V) = 5.31X10-8 C

    this is all i know please help, also please explain things tome or did i messed this up? i know there is google but i really have a lot of things to do. please help.
     
  2. jcsd
  3. Dec 3, 2015 #2

    SammyS

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    Hello XYTRIX. Welcome to PF !

    You should easily find the answer to (c) . You know the charge, (It hasn't changed.) and you know the new potential difference.
     
  4. Dec 3, 2015 #3
    i'm very bad at this so my answers at a and b are both right?
     
  5. Dec 3, 2015 #4

    SammyS

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    They look right to me.
     
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