+\Part{a}
+Bundling down equivalent capacitances,
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+
+TwoTerminal B = battery((0,0), ang=180, "$V$");
+TwoTerminal C1 = capacitor("$C_1$", draw=false);
+TwoTerminal C23 = capacitor("$C_{23}$", draw=false);
+TwoTerminal C4 = capacitor("$C_4$", draw=false);
+centerto(B, C23, offset=2u, reverse=true); C23.draw();
+pair a = (B.end.x, C23.beg.y);
+pair b = (B.beg.x, a.y);
+C1.centerto(B.end, a, offset=u); C1.draw();
+C4.centerto(B.beg, b, offset=-u); C4.draw();
+wire(B.end, C1.beg, rlsq);
+wire(C1.end, C23.beg, udsq);
+wire(C23.end, C4.end, rlsq);
+wire(C4.beg, B.beg, udsq);
+
+B.shift((6u, 0));
+B.draw();
+TwoTerminal C1234 = capacitor("$C_{1234}$", draw=false);
+centerto(B, C1234, offset=2u, reverse=true); C1234.draw();
+wire(B.end, C1234.beg, rlsq, dist=-u/2);
+wire(B.beg, C1234.end, rlsq, dist=+u/2);
+
+pair c = ((C4.mid.x + B.end.x - u/2)/2, C4.mid.y);
+draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrows);
+\end{asy}
+\end{center}
+$C_{23}$ comes from $C_2$ and $C_3$ in parallel, so the equivalent
+capacitance is
+\begin{equation}
+ C_{23} = C_2 + C_3 = (6.00 + 2.00)\U{$\mu$F} = 8.00\U{$\mu$F} \;.
+\end{equation}
+$C_{1234}$ comes from $C_1$, $C_{23}$, and $C_4$ in series, so the
+equivalent capacitance is
+\begin{equation}
+ C_{1234} = \p({\frac{1}{C_1} + \frac{1}{C_{23}} + \frac{1}{C_4}})^{-1}
+ = \frac{8.00\U{$\mu$F}}{3} = \ans{2.67\U{$\mu$F}} \;.
+\end{equation}
+
+\Part{b}
+The equivalent capacitor has the total battery voltage across itself,
+so it carries a charge of
+\begin{equation}
+ Q_{1234} = C_{1234} V = (2.67\U{$\mu$F}) \cdot (9.00\U{V})
+ = 24.0\U{$\mu$C} \;.
+\end{equation}
+Because $C_1$, $C_{23}$, and $C_3$ are in series, they each have the
+same $Q$ as the equivalent capacitor.
+\begin{equation}
+ Q_1 = Q_{23} = Q_3 = Q_{1234} = \ans{24.0\U{$\mu$C}} \;.
+\end{equation}
+The voltage across $C_{23}$ is therefore
+\begin{equation}
+ V_{23} = \frac{Q_{23}}{C_{23}} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
+ = 3.00\U{V} \;.
+\end{equation}
+$C_2$ and $C_3$ are in parallel, so they each have the total $V_{23}$
+voltage.
+\begin{align}
+ Q_2 &= C_2 V_{23} = 6.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{18.0\U{$\mu$C}} \\
+ Q_3 &= C_3 V_{23} = 2.00\U{$\mu$F} \cdot 3.00\U{V} = \ans{6.00\U{$\mu$C}} \;.
+\end{align}
+Note that $Q_2+Q_3=Q_{23}$, which must be true because charge is conserved.
+
+\Part{c}
+During \Part{b}, we already found $\ans{V_{23}=V_2=V_3=3.00\U{V}}$.
+Using the charges on $C_1$ and $C_4$, we find
+\begin{align}
+ V_1 &= \frac{Q_1}{C_1} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
+ = \ans{3.00\U{V}} \\
+ V_4 &= \frac{Q_4}{C_4} = \frac{24.0\U{$\mu$C}}{8.00\U{$\mu$F}}
+ = \ans{3.00\U{V}} \;.
+\end{align}