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How much charge will flow from the battery?

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A 3500 pF air-gap capacitor is connected to a 32 V battery. If a piece of mica fills the space between the plates, how much charge will flow from the battery?

    2. Relevant equations

    C= K*C_0
    V= V_0/ K
    Q= C*V
    K of mica is 7

    3. The attempt at a solution

    solved for the _0's and plugged into Q=CV
    i used (C/K) * (V*K) = (3500 *10^-12F)/7 * 32V*7 = 1.12*10^-7C

    its wrong but i dunno what else to do. help.
  2. jcsd
  3. Oct 4, 2008 #2
    Re: Dielectrics

    well you've cancelled out the mica with your equation.
    You know that capacitance depends linearly on the k value of the dielectric (assuming area and distance between parallel plates remains constant).
    So effectively the capacitance of with the mica inserted is 7 * 3500E-12.
    Q = CV.
    Do they want the answer in C or e?
  4. Oct 5, 2008 #3
    Re: Dielectrics

    ok so it would be

    Q = (7*3500*10^-12) * 32V

    they want it in C. that gives 7.84*10^-7C which was my first answer but it is wrong.
  5. Oct 5, 2008 #4
    Re: Dielectrics

    can someone help?
  6. Oct 5, 2008 #5
    Re: Dielectrics

    The question is "how much charge flows from the battery?" (when you insert the mica).
    Your (above) calculation gives the charge after inserting the mica. This includes the charge that was there already.
    You just have to replace 7 by 6.

    delta Q = kCo*V-Co*V = (k-1)Co*V

    Is this the right answer?
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