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Capacitor Problem

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img4.imageshack.us/img4/3144/capaeo.jpg [Broken]




    2. Relevant equations
    Q=CV



    3. The attempt at a solution
    a) i found Ceq=4.00 μF
    Qtotal =CV=(4.00 μF) (210 V)=8.40×10−4C. How i can find Vcd=?
    b)i have solved this.
    c)i have no idea about this.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 27, 2009 #2
    (a) The switch S is open. The top two capacitors are in series: Ctop = 1/(1/3+1/6) = 2 μF, and the
    bottom two are also in series: Cbot = 1/(1/6 + 1/3) = 2 μF. The top and bottom capacitors are in
    parallel: Ceq = Ctop + Cbot = 4 μF. The total charge on the capacitors is: Qtot = CeqVab = 840 μC.
    Recall that capacitors in parallel split charge. From symmetry, the top capacitors have half the
    charge and the bottom capacitors have half the charge. Capacitors in series have the same charge.
    Thus, both top capacitors each have 420 μC and both bottom capacitors each have 420 μC. That
    is, all four capacitors have exactly the same charge Q = 420 μC. The potential drop from a to d is
    thus: Vad = Q/C = (420 μC)/(3 μF) = 140 V, and the potential drop from a to c is: Vac = Q/C =
    (420 μC)/(6 μF) = 70 V. Thus, the potential difference between c and d is: Vcd = Vad −Vac = 70 V.



    (b) The switch S is now closed. Now the two left capacitors are in parallel: Cleft = 3 + 6 = 9 μF,
    and the two right capacitors are in parallel: Cright = 6 + 3 = 9 μF. The left and right sets are in
    series: Ceq = 1/(1/9 + 1/9) = 4.5 μF. Recall that for capacitors in series, the potential drops add.
    From symmetry, the potential drop across the left capacitors is the same as the potential drop across
    the right capacitors: Vleft = Vright = Vab/2 = 105 V. The two right capacitors must have the same
    potential drop because they are parallel, and the same goes for the two left capacitors. That is, the
    potential drop across each and every capacitor in this configuration is: V = Vad = Vdb = Vac =
    Vcb = 105 V.



    (c) After the switch is closed, the charge on the top left capacitor is: QTL = CTLVad = 315 μC;
    the charge on the top right capacitor is: QTR = CTRVdb = 630 μC; the charge on the bottom
    left capacitor is: QBL = CBLVac = 630 μC; and the charge on the bottom right capacitor is:
    QBR = CBRVcb = 315 μC. The right side of the top left plate is the negative side, and has charge
    −315 μC, and the left side of the top right plate is the positive side, and has charge +630 μC. Thus,
    the total charge on the two plates closest to point d is +630 − 315 = 315 μC. Before the switch
    closed, there was zero net charge on the two plates closest to d: 420− 420 = 0 μC. So 315 μC must
    have flowed through the switch from bottom to top.




    I hope this would help you . c ya
     
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