## Solution to 1996 Problem 62

 Let $C_1 = 1 \;\mu\mathrm{F}$ and $C_2 = 2 \;\mu\mathrm{F}$. Let $Q_2$ denote the charge on the positive side of the the $1 \;\mu\mathrm{F}$ capacitor and let $Q_2$ denote the charge on the positive side of the $2 \;\mu\mathrm{F}$ capacitor. Let $V_1$ denote the voltage across the $1 \;\mu\mathrm{F}$ capacitor and let $V_2$ denote the voltage across the $2 \;\mu\mathrm{F}$ capacitor. Let $R$ denote the total resistance of the wire (so R include the resistance from both of the two disconnected sections of the wire).The current must be the same everywhere in the circuit. Therefore, the rate at which positive charges leave the $1 \;\mu\mathrm{F}$ capacitor must equal the rate at which positive charges leave the $2 \;\mu\mathrm{F}$ capacitor. That is\begin{align*}\frac{d Q_1}{dt} = \frac{d Q_2}{dt}\end{align*}This implies that\begin{align*}\frac{d (Q_1 - Q_2)}{dt} = 0 \Rightarrow Q_1 - Q_2 = \mbox{const}\end{align*}Initially, $Q_1 = 5 \mbox{ V} \cdot C_1 = 5 \;\mu\mathrm{C}$ and $Q_2 = 5 \mbox{ V} \cdot C_2 = 10 \;\mu\mathrm{C}$. So, $Q_1 - Q_2 = - 5 \;\mu\mathrm{C}$.The current through the resistor must be zero when the circuit reaches a steady state, so the magnitudes of the voltages across the capacitors must be equal, and their sign must be opposite (because of Kirchhoff's voltage law). That is,\begin{align*}\frac{Q_1}{C_1} = -\frac{Q_2}{C_2} \Rightarrow 2 Q_1 = -Q_2\end{align*}The solution to the system of equations $2 Q_1 = -Q_2$ and $Q_1 - Q_2= -5 \;\mu\mathrm{C}$ is $Q_1 = - 5/3 \;\mu\mathrm{C}$ and $Q_2 = 10/3 \;\mu\mathrm{C}$.\begin{align*}V_2 = \frac{Q_2}{C_2} = \frac{10/3 \;\mu\mathrm{C}}{2 \;\mu\mathrm{F}} = \boxed{\frac{5}{3} \mbox{ V}}\end{alig...Therefore answer (C) is correct.