\begin{problem*}{20.47}
\Part{a} A $C = 3.00\U{$\mu$F}$ capacitor is connected to a $\Delta
V_a = 12.0\U{V}$ battery.  How much energy $U_a$ is stored in the
capacitor?
\Part{b} If the capacitor had been connected to a $\Delta V_b =
6.00\U{V}$ battery, how much energy would have been stored?
\end{problem*} % problem 20.47

\begin{solution}
Simply plugging into the formula for energy stored in a capacitor we have
\begin{align}
 U_a &= \frac{1}{2} C (\Delta V)^2
      = \frac{1}{2} (3.00\E{-6}\U{F}) \cdot (12.0\U{V})^2
      = \ans{ 216 \U{$\mu$J}} \\
 U_b &= \frac{1}{2} (3.00\E{-6}\U{F}) \cdot (6.0\U{V})^2
      = \ans{ 54 \U{$\mu$J}}
\end{align}
\end{solution}
