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Capacitance problem

  1. Jun 10, 2007 #1
    1. The problem statement, all variables and given/known data
    The potential difference V = 100 V is applied to the capacitor arrangement shown in the figure. Here C1 = 10 microF, C2 = 5 microF, and C3 = 4microF. If capacitor C3 undergoes electrical breakdown (i.e. becomes equivalent to a conducting wire), what is the increase in (a) the charge on capacitor 1, and (b) the potential difference across capacitor 1?

    3. The attempt at a solution
    I wasn't sure how to solve it so I started doing different things:
    I first found the equivalent capacitance, and with that I found the total charge to be Q_tot = 3.16 * 10^-4 C.
    Then, I thought since the equivalent of C1 and C2, and C3 are in series, Q1 + Q2 = Q3, so Q_tot = 2 * Q3.
    Then, from Q3 and C3, I find the potential difference across of C3 to be V3 = 39.5V, which in turn makes V1 = V2 = 60.5V.
    From V1 and V2, I found the initial charges Q1 = 605 microC, and Q2 = 302.5 microC.
    I'm not sure where to go from here...

    I tried also to find the final potential difference across C1:
    C12 = C1 + C2 = Q_tot / V_new
    V_new = 21V, but it's supposed to be 79V... what am I doing wrong here?

    Since Q = CV, if I plug in V = 79V, then Q1 = 790 microC, which is the answer for part (a), right? So should I try to solve part (b) first?

    Any help would be greatly appreciated!

    P.S. sorry for not using tex...
  2. jcsd
  3. Jun 10, 2007 #2
    you've made a mistake

    Q1 + Q2 = Q3, so Q_tot = 2 * Q3.

    this is the faulty assumption

    it should be
    Q1 + Q2 = Q3 = Q_tot
  4. Jun 10, 2007 #3
    But if Q1 + Q2 = Q3, and Q_tot = Q1 + Q2 + Q3, then Q_tot = Q3 + Q3 = 2*Q3 ... right?
  5. Jun 10, 2007 #4
    you just restated what i already said was wrong

    Q_tot = Q1 + Q2 = Q3
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