\begin{problem*}{21.46}
A $C=10.0\U{$\mu$F}$ capacitor is sharged by a $\epsilon= 10.0\U{V}$
battery through a resistance $R$.  The capacitor reaches a potential
difference of $V_C(t_f)=4.00\U{V}$ at the instant $t_f = 3.00\U{s}$
after the charging begins.  Find $R$.
\end{problem*} % problem 12.46

\begin{solution}
Applying Kirchhoff's loop rule,
\begin{align}
 \epsilon - V_C - V_R &= \epsilon - \frac{q}{C} - IR = 0 \\
 R \frac{dq}{dt} &= \epsilon - \frac{q}{C} \\
 \frac{dq}{dt} &= \frac{C\epsilon - q}{RC} \\
 \frac{dq}{C\epsilon - q} &= \frac{dt}{RC} \\
 \int \frac{dq}{C\epsilon -q} &= \int \frac{dt}{RC} \\
 -\ln (C\epsilon-q) &= t/RC + A \\
 C\epsilon - q &= Be^{-t/RC} \\
 q &= C\epsilon - Be^{-t/RC} \\
 q &= C\epsilon(1-e^{-t/RC})
\end{align} 
Where $A$ is a constant of integration,
 $B = e^{-A}$ is another way of writing that constant,
 $C\epsilon = Q$ (because as $t \rightarrow \infty$, $q \rightarrow Q$),
 and $B = Q = C\epsilon$ (because at $t=0$, $q=0$).

Note: The book derives this on pages 709-710 if you want more details.

Now applying this to our particular problem,
\begin{align}
 V_C(t_f) &= \frac{q(t_f)}{C}
   = \frac{C\epsilon}{C}(1-e^{-t_f/RC}) \\
 \frac{V_C(t_f)}{\epsilon} &= 1 - e^{-t_f/RC} \\
 e^{-t_f/RC} &= 1 - \frac{V_C(t_f)}{\epsilon} \\
 \frac{-t_f}{RC} &= \ln\left(1 - \frac{V_C(t_f)}{\epsilon}\right) \\
 R &= \frac{-t_f}{C \ln(1 - V_C(t_f)/\epsilon)}
    = \frac{-3.00\U{s}}{10.0\E{-6}\U{F}\cdot\ln(1-4.00\U{V}/10.0\U{V})}
    = \ans{587\U{k$\Omega$}}
\end{align}
\end{solution}
