1 \begin{problem*}{26.67}
2 Capacitors $C_1=6.00\U{$\mu$F}$ and $C_2=2.00\U{$\mu$F}$ are charged
3 as a parallel combination across a $250\U{V}$ battery. The capacitors
4 are disconnected from the battery and from each other. Then they are
5 connected positive plate to negative plate and negative plate to
6 positive plate. Calculate the resulting charge on each capacitor.
10 After charging, the charges on the capacitors are
12 Q_1 &= C_1 V = 1.50\U{mC} \\
13 Q_2 &= C_2 V = 500\U{$\mu$C} \;.
16 After disconnecting the battery, flipping $C_2$, and reconnecting, the
17 total charge on one side is $Q_t=Q_2+(-Q_2)=1.00\U{mC}$. This charge
18 is divided into $Q_1'$ and $Q_2'$ such that
21 V_1' = \frac{Q_1'}{C_1} &= V_2' = \frac{Q_2'}{C_2} \\
22 Q_2' &= Q_1'\frac{C_2}{C_1} \\
23 Q_1' + Q_1'\frac{C_2}{C_1} &= Q_t \\
24 Q_1' &= \frac{Q_t}{1+\frac{C_2}{C_1}} = \ans{750.0\U{$\mu$C}} \\
25 Q_2' &= Q_1'\frac{C_2}{C_1}
26 = \frac{Q_t}{\frac{C_1}{C_2} + 1} = \ans{250.0\U{$\mu$C}}