\end{problem*}
\begin{solution}
+\Part{a}
+Using Gauss' law, we can find the electric field from an infinite
+plate with charge density $\sigma$.
+\begin{align}
+ \Phi_E &= \oint_S \vect{E}\cdot\vect{dA} = 2EA \\
+ &= \frac{q_\text{in}}{\kappa\varepsilon_0}
+ = \frac{\sigma A}{\kappa\varepsilon_0} \\
+ E &= \frac{\sigma}{2\kappa\varepsilon_0} \;.
+\end{align}
+
+The capacitance of a ciruit element is defined by $C\equiv q/V$, where
+the potential difference $V$ is measured with $+q$ on one side of the
+element and $-q$ on the other. The electric field between the plates
+will be
+\begin{equation}
+ E = 2\frac{\sigma}{2\kappa\varepsilon_0}
+ = \frac{q}{\kappa\varepsilon_0 A} \;,
+\end{equation}
+and the potential difference is
+\begin{equation}
+ V = Ed = \frac{qd}{\kappa\varepsilon_0 A} \;.
+\end{equation}
+
+Now that we have a formula for $V(q)$, we can plug into the capacitor
+equation to find the capacitance.
+\begin{equation}
+ C \equiv \frac{q}{V} = \frac{\kappa\varepsilon_0 A}{d}
+ = \frac{1.00 \cdot 8.85\E{-12}\U{C$^2$/N$\cdot$m$^2$}
+ \cdot 2.30\E{-4}\U{m$^2$}}
+ {1.50\E{-3}\U{m}}
+ = \ans{1.36\E{-12}\U{F}} = \ans{1.36\U{pF}} \;.
+\end{equation}
+As you'd expect, $C$ increases with increasing $A$, and decreases with
+increasing $d$.
+
+\Part{b}
+Just plugging into the capacitance equation,
+\begin{equation}
+ q = CV = 1.36\E{-12}\U{F} \cdot 12.0\U{V}
+ = \ans{1.63\E{-11}\U{C}} = \ans{16.3\U{pC}} \;.
+\end{equation}
+
+\Part{c}
+Plugging into my earlier equation for electric field,
+\begin{equation}
+ E = \frac{V}{d} = \frac{12.0\U{V}}{1.50\E{-3}\U{m}}
+ = \ans{8.00\U{kV/m}} \;.
+\end{equation}
\end{solution}
\end{problem*}
\begin{solution}
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+
+TwoTerminal C1 = capacitor((0,0), ang=0, "$C_1$", "2.50\U{$\mu$F}");
+TwoTerminal C2 = capacitor(C1.end + (u,0), ang=0, "$C_2$", "6.25\U{$\mu$F}");
+TwoTerminal B = battery("$V$", "6.00\U{V}", draw=false);
+B.centerto(C1.beg, C2.end, offset=2u);
+B.draw();
+wire(B.beg, C1.beg, rlsq);
+wire(B.end, C2.end, rlsq);
+wire(C1.end, C2.beg);
+
+label("\Part{a}", B.mid + (0, -5u));
+
+B.shift((4u, 0));
+B.draw();
+centerto(B, C1, offset=-2u);
+C1.draw();
+centerto(B, C2, offset=-4u);
+C2.draw();
+wire(B.beg, C1.beg, rlsq, dist=-u/2);
+wire(B.beg, C2.beg, rlsq, dist=-u/2);
+wire(B.end, C1.end, rlsq, dist=u/2);
+wire(B.end, C2.end, rlsq, dist=u/2);
+
+label("\Part{b}", B.mid + (0, -5u));
+\end{asy}
+\end{center}
+
+\Part{a}
+The net capacitance of the two capacitors in series is
+\begin{equation}
+ C = \p({\frac{1}{C_1} + \frac{1}{C_2}})^{-1}
+ = 1.79\U{$\mu$F} \;.
+\end{equation}
+The equivalent capacitor has the total battery voltage across itself,
+so it carries a charge of
+\begin{equation}
+ Q = CV = 1.79\U{$\mu$F} \cdot 6.00\U{V} = \ans{10.7\U{$\mu$C}} \;.
+\end{equation}
+Because $C_1$ and $C_2$ are in series, they each have the same $Q$ as
+the equivalent capacitor.
+
+\Part{b}
+When the capacitors are in parallel, they each have the total battery
+voltage, so
+\begin{align}
+ Q_1 &= C_1 V = 2.50\U{$\mu$F} \cdot 6.00\U{V} = \ans{15.0\U{$\mu$C}} \\
+ Q_2 &= C_2 V = 6.25\U{$\mu$F} \cdot 6.00\U{V} = \ans{37.5\U{$\mu$C}} \;.
+\end{align}
\end{solution}
Find \Part{a} the equivalent capacitance of the capacitors in
Figure~P26.18, \Part{b} the charge on each capacitor, and \Part{c} the
potential difference across each capacitor.
+\begin{center}
% +-6uF-+
% +-+ +-+
% | +-2uF-+ |
% 8uF 8uF
% | 9V |
% +---|i----+
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+
+TwoTerminal B = battery((0,0), ang=180, "$V$", "9.00\U{V}");
+TwoTerminal C1 = capacitor("$C_1$", "8.00\U{$\mu$F}", draw=false);
+TwoTerminal C2 = capacitor("$C_2$", "6.00\U{$\mu$F}", draw=false);
+TwoTerminal C3 = capacitor("$C_3$", "2.00\U{$\mu$F}", draw=false);
+TwoTerminal C4 = capacitor("$C_4$", "8.00\U{$\mu$F}", draw=false);
+
+centerto(B, C3, offset=3u, reverse=true); C3.draw();
+centerto(C3, C2, offset=2u); C2.draw();
+pair a = (B.end.x, (C2.beg.y + C3.beg.y)/2);
+pair b = (B.beg.x, a.y);
+C1.centerto(B.end, a, offset=2u); C1.draw();
+C4.centerto(B.beg, b, offset=-2u); C4.draw();
+
+a = (a.x - u, a.y); // shift to split x difference between C1.end and B.end
+b = (b.x + u, b.y);
+
+wire(B.end, C1.beg, rlsq);
+wire(C1.end, a, udsq);
+wire(a, C2.beg, udsq);
+wire(a, C3.beg, udsq);
+dot(a);
+wire(C2.end, b, rlsq);
+wire(C3.beg, b, rlsq);
+wire(b, C4.end, rlsq);
+dot(b);
+wire(C4.beg, B.beg, udsq);
+\end{asy}
+\end{center}
\end{problem*}
\begin{solution}
+\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}
\end{solution}
\end{problem*}
\begin{solution}
+Just plug into the capacitor energy formula,
+\begin{equation}
+ U = \frac{1}{2}CV^2
+ = \frac{1}{2}QV
+ = \frac{1}{2} \cdot (54.0\E{-6}\U{C}) \cdot (12.0\U{V})
+ = \ans{324\U{$\mu$J}} \;.
+\end{equation}
\end{solution}
\end{problem*}
\begin{solution}
+Using the capacitor energy formula,
+\begin{align}
+ U &= \frac{1}{2}CV^2 \\
+ V &= \sqrt{\frac{2U}{C}}
+ = \sqrt{\frac{2 \cdot 300\U{J}}{30.0\E{-6}\U{F}}}
+ = \ans{4.47\U{kV}} \;.
+\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+\Part{a}
+The capacitance of a parallel plate capacitor is
+\begin{equation}
+ C = \frac{\kappa\varepsilon_0 A}{d} \;.
+\end{equation}
+For other geometries, the constants change, but the capacitance is
+always proportional to $\kappa$, the dielectric constant of the gap.
+After removing the mica sheet, the capacitor will have a new
+capacitance of
+\begin{equation}
+ C' = \frac{\kappa'}{\kappa} C = \frac{1.00}{5.00} \cdot 2.00\U{nF}
+ = 400\U{pF} \;,
+\end{equation}
+where $\kappa' \approx 1.00$ is the dielectric constant for air (which
+replaces the mica).
+
+During the withdrawing process, the capacitance changes, and the
+voltage between the plates changes, but because the capacitor is
+isolated, the charge does not change. We can use the initial voltage
+to find that charge.
+\begin{equation}
+ Q = C \Delta V_i = (2.00\E{-9}\U{F}) \cdot (100\U{V}) = 200\U{nC}
+\end{equation}
+
+Conserving energy during the withdrawing process,
+\begin{align}
+ U_i + W &= U_f \\
+ \frac{Q^2}{2C} + W &= \frac{Q^2}{2C'} \\
+ W &= \frac{Q^2}{2} \cdot \p({\frac{1}{C'} - \frac{1}{C}})
+ = \frac{Q^2}{2} \cdot \p({\frac{\kappa}{\kappa' C} - \frac{1}{C}})
+ = \frac{Q^2}{2\kappa' C} \cdot (\kappa - \kappa')
+ = \frac{C\Delta V_i^2}{2\kappa'} \cdot (\kappa - \kappa') \\
+ &= \frac{(2.00\E{-9}\U{F}) \cdot (100\U{V})^2}{2 \cdot 1.00}
+ \cdot(5.00 - 1.00)
+ = \ans{40.0\U{$\mu$J}} \;.
+\end{align}
\end{solution}
\end{problem*}
\begin{solution}
+\begin{center}
+\begin{asy}
+import Circ;
+
+real u = 1cm;
+real s = 6u;
+
+TwoTerminal B = battery("$V$");
+TwoTerminal C1 = capacitor("$C_1$", draw=false);
+TwoTerminal C2 = capacitor("$C_2$", draw=false);
+centerto(B, C1, offset=2u); C1.draw();
+centerto(C1, C2, offset=2u); C2.draw();
+wire(B.end, C1.end, rlsq, dist=u/2);
+wire(B.end, C2.end, rlsq, dist=u/2);
+wire(B.beg, C1.beg, rlsq, dist=-u/2);
+wire(B.beg, C2.beg, rlsq, dist=-u/2);
+label("$+Q_1$", C1.end, align=dir(70));
+label("$-Q_1$", C1.beg, align=dir(110));
+label("$+Q_2$", C2.end, align=dir(70));
+label("$-Q_2$", C2.beg, align=dir(110));
+
+pair c = C1.mid + (s/2, 0);
+draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrow);
+
+C1.centerto(B.mid, C2.mid, -s); C1.draw();
+centerto(C1, C2, offset=u, reverse=true); C2.draw();
+label("$+Q_1$", C1.end, align=NW);
+label("$-Q_1$", C1.beg, align=SW);
+label("$+Q_2$", C2.end, align=SE);
+label("$-Q_2$", C2.beg, align=NE);
+
+c = C2.mid + (s/2, 0);
+draw((c-(u,0)) -- (c+(u,0)), kirchhoff_pen, Arrow);
+
+C1.shift(s+u); C1.draw();
+C2.shift(s+u); C2.draw();
+wire(C1.end, C2.beg);
+wire(C2.end, C1.beg);
+label("$+Q_1'$", C1.end, align=NW);
+label("$-Q_1'$", C1.beg, align=SW);
+label("$-Q_2'$", C2.end, align=SE);
+label("$+Q_2'$", C2.beg, align=NE);
+\end{asy}
+\end{center}
+
After charging, the charges on the capacitors are
\begin{align}
Q_1 &= C_1 V = 1.50\U{mC} \\
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