1 \begin{problem*}{20.40}
2 Two capacitors, $C_1 = 5.00\U{$\mu$F}$ and $C_2 = 12.0\U{$\mu$F}$, are
3 connected in series, and the resulting combination is connected to a
4 $\Delta V = 9.00\U{V}$ battery. Find
5 \Part{a} the equivalent capacitance of the combination,
6 \Part{b} the potential difference across each capacitor, and
7 \Part{c} the charge on each capacitor.
8 \end{problem*} % problem 20.40
12 The wire connecting the inner plates of $C_1$ and $C_2$ contains no
13 net charge, so we know that any charge on the inner plate of $C_1$
14 must have come from the inner plate of $C_2$. Because these charges
15 are equal and opposite, the total charge $Q$ on each capacitor
16 seperately is the same for both ($Q_1 = Q_2$). So using the
17 definition of capacitance for both cases we have
19 \Delta V_1 &= Q / C_1 \label{eqn.V1} \\
20 \Delta V_2 &= Q / C_2 \label{eqn.V2} \\
21 \Delta V &= \Delta V_1 + \Delta V_2
22 = Q \left( \frac{1}{C_1} + \frac{1}{C_2} \right)
23 = \frac{Q}{C_{eq}} \label{eqn.VQC}
27 C_{eq} = \left(\frac{1}{C_1}+\frac{1}{C_2}\right)^{-1}
28 = \left(\frac{1}{5.00\E{-6}\U{F}}+\frac{1}{12.0\E{-6}\U{F}}\right)^{-1}
29 = \ans{3.53\U{$\mu$F}}
33 Plugging back into equation \ref{eqn.VQC} we have
35 Q = \Delta V \cdot C_{eq}
36 = 3.53\U{$\mu$F}\cdot 9.00\U{V} = \ans{31.8\U{$\mu$C}}
40 And plugging into equations \ref{eqn.V1} and \ref{eqn.V2} we have
42 \Delta V_1 &= \frac{31.8\E{-6}\U{C}}{5.00\E{-6}\U{F}} = \ans{6.35\U{V}} \\
43 \Delta V_2 &= \frac{31.8\E{-6}\U{C}}{12.0\E{-6}\U{F}} = \ans{2.65\U{V}} \\