\begin{problem*}{17.40} % capacitor circuits
A $C_1 = 7.7\U{$\mu$F}$ capacitor is charged by a $V = 125\U{V}$
battery (Fig. 17-29a) and then is disconnected from the battery.  When
this capacitor ($C_1$) is then connected (Fig. 17-29b) to a second
(initially uncharged) capacitor, $C_2$, the final voltage on each
capacitor is $V_2 = 15\U{V}$.  What is the value of $C_2$?
[\emph{Hint:} charge is conserved.]
\begin{center}
\begin{asy}
%    C1         C1
% +--||--+   +--||--+
% |      |   |      |
% +--i|--+   +--||--+
%    V          C2
%
%   (a)        (b)
import Circ;

real u = 1cm;
real dx = u;
real dy = 1.5u;

MultiTerminal bat = battery("$V$");
MultiTerminal c1 = capacitor("$C_1$", draw=false);
two_terminal_centerto(bat, c1, dy); c1.draw();
wire(bat.terminal[1], c1.terminal[1], rlsq, dx/2);
wire(bat.terminal[0], c1.terminal[0], rlsq, -dx/2);
label("(a)", bat.center + (0, -dy/2));

c1.shift(3*dx); c1.draw();
MultiTerminal c2 = capacitor("$C_2$", draw=false);
two_terminal_centerto(c1, c2, -dy); c2.draw();
wire(c2.terminal[1], c1.terminal[1], rlsq, dx/2);
wire(c2.terminal[0], c1.terminal[0], rlsq, -dx/2);
label("(b)", c2.center + (0, -dy/2));
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
\begin{center}
\begin{asy}
import Circ;

real u = 1cm;
real dx = u;
real dy = 1.5u;

MultiTerminal bat = battery("$V$");
MultiTerminal c1 = capacitor("$C_1$", draw=false);
two_terminal_centerto(bat, c1, dy); c1.draw();
wire(bat.terminal[1], c1.terminal[1], rlsq, dx/2);
wire(bat.terminal[0], c1.terminal[0], rlsq, -dx/2);
label("(a)", bat.center + (0, -dy/2));
label("$Q_{1a}$", c1.terminal[1], NE);
label("$-Q_{1a}$", c1.terminal[0], NW);

c1.shift(3*dx); c1.draw();
MultiTerminal c2 = capacitor("$C_2$", draw=false);
two_terminal_centerto(c1, c2, -dy); c2.draw();
wire(c2.terminal[1], c1.terminal[1], rlsq, dx/2);
wire(c2.terminal[0], c1.terminal[0], rlsq, -dx/2);
label("(b)", c2.center + (0, -dy/2));
label("$Q_{1b}$", c1.terminal[1], NE);
label("$-Q_{1b}$", c1.terminal[0], NW);
label("$Q_{2b}$", c2.terminal[1], SE);
label("$-Q_{2b}$", c2.terminal[0], SW);
\end{asy}
\end{center}

Because the voltage drop across $C_1$ in situation $a$ is the same as
the voltage drop across the battery ($V$), we have
$$
  Q_{1a} = C_1 V
$$
When we connect $C_2$ in situation $b$, this charge redistributes
between $C_1$ and $C_2$.  Because charge is conserved, we know
$$
  Q_{1a} = Q_{1b} + Q_{2b}
$$
We also know that the voltage drop across both capacitors in situation
$b$ must be equal ($\text{both} = V_2$), so
\begin{align*}
  Q_{1b} &= C_1 V_2 \\
  Q_{2b} &= C_2 V_2
\end{align*}
Plugging each of these formulas for charge ($Q_{1a}$, $Q_{1b}$, and
$Q_{2b}$) into the charge conservation formula yeilds
\begin{align*}
  C_1 V &= C_1 V_2 + C_2 V_2 \\
  C_1 (V - V_2) &= C_2 V_2 \\
  V_2 C_2 &= C_1 (V - V_2) \\
  C_2 &= C_1 \p({\frac{V}{V_2} - \frac{V_2}{V_2}}) \\
  C_2 &= C_1 \p({\frac{V}{V_2} - 1}) \\
  C_2 &= 7.7\U{$\mu$F} \cdot \p({\frac{125\U{V}}{15\U{V}} - 1})
       = \ans{56\U{$\mu$F}}
\end{align*}

\end{solution}
