\begin{problem*}{40}
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.]
\end{problem*}

\empaddtoprelude{
  input makecirc; % circuit drawing functions
  initlatex("");
  pair A, B, Cl, Cr, Dl, Dr, El, Er;
  numeric a;
  a := 3cm;
  A := (-3a/4,0);
  B := (3a/4, 0);
  def drawA = 
    % add elements
    centreof.b(A+(1,0), A-(1,0), bat);
    battery.B(c.b, phi.b, "V", "");
    centreof.c(B.B.n, B.B.p, cap);
    capacitor.C(c.c+(0,a), normal, phi.c, "C_1", "");
    % add wiring along the bottom
    wire(B.B.n, A+(a/2,0), nsq);
    wire(B.B.p, A-(a/2,0), nsq);
    % add wiring along the sides and top
    wire(A+(a/2,0), C.C.l, udsq);
    wire(A-(a/2,0), C.C.r, udsq);
    puttext.bot("($a$)", A-(0,24pt));
    Cl = C.C.r;
    Cr = C.C.l;
  enddef;
  def drawB = 
    % add elements
    centreof.d(B+(1,0), B-(1,0), cap);
    capacitor.D(c.d, normal, phi.d, "C_2", "");
    centreof.e(C.D.l, C.D.r, cap);
    capacitor.E(c.e+(0,a), normal, phi.e, "C_1", "");
    % add wiring along the bottom
    wire(C.D.l, B+(a/2,0), nsq);
    wire(C.D.r, B-(a/2,0), nsq);
    % add wiring along the sides and top
    wire(B+(a/2,0), C.E.l, udsq);
    wire(B-(a/2,0), C.E.r, udsq);
    puttext.bot("($b$)", B-(0,24pt));
    Dl = C.D.r;
    Dr = C.D.l;
    El = C.E.r;
    Er = C.E.l;
  enddef;
}

\begin{nosolution}
\begin{center}
\begin{empfile}[5p]
\begin{emp}(0cm, 0cm)
  drawA;
  drawB;
\end{emp}
\end{empfile}
\end{center}
\end{nosolution}

\begin{solution}
\begin{center}
\begin{empfile}[5]
\begin{emp}(0cm, 0cm)
  drawA;
  drawB;
  puttext.ulft("$Q_{1a}$", Cl);
  puttext.urt("$-Q_{1a}$", Cr);
  puttext.ulft("$Q_{2b}$", Dl);
  puttext.urt("$-Q_{2b}$", Dr);
  puttext.ulft("$Q_{1b}$", El);
  puttext.urt("$-Q_{1b}$", Er);
  labeloffset := 9pt;
  puttext.top("$V$", (Cl+Cr)/2);
  puttext.top("$V_2$", (Dl+Dr)/2);
  puttext.top("$V_2$", (El+Er)/2);
\end{emp}
\end{empfile}
\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}
