1 \begin{problem*}{20.41}
2 Four capacitors are connected as shown in Figure P20.41.
3 \Part{a} Find the equivalent capacitance between points $a$ and $b$.
4 \Part{b} Calculate the charge on each capacitor, taking
5 $\Delta V_{ab} = 15.0\U{V}$
6 \end{problem*} % problem 20.41
10 First consider the top two capacitors, $C_1 = 15.0\U{$\mu$F}$ and
11 $C_2 = 3.00\U{$\mu$F}$.
12 They are in series, so the effective capacitance of the top line is
15 C_t = \left(\frac{1}{C_1} + \frac{1}{C_2}\right)^{-1}
19 We can find the effective capacitance of the box, because $C_t$ is
20 in parallel with $C_3 = 6.00\U{$\mu$F}$.
22 C_b = C_t + C_3 = 8.50\U{$\mu$C}
25 We can find the total equivalent capacitance, because $C_b$ is in
26 series with $C_4 = 20.0\U{$\mu$F}$.
28 C_{eq} = \left(\frac{1}{C_b} + \frac{1}{C_4}\right)^{-1}
29 = \ans{ 5.96\U{$\mu$F}}
33 Working backwards to find the charges, using $Q = CV$, we have
35 Q_4 = Q_b = C_{eq} V_{ab} = \ans{89.5\U{$\mu$C}}
37 So the voltage across the box is
39 V_b = \frac{Q_b}{C_b} = 10.5\U{V}
43 Q_3 = C_3 V_b = \ans{63.2\U{$\mu$C}}
47 Q_1 = Q_2 = C_t V_b = \ans{26.3\U{$\mu$C}}