Problem 20.56

20.56. a)
To find the equivalent capacitance, we first find the equivalent capacitance along each branch. Since the two branches are in parallel, then we just add the value for the capacitance along each branch and get the total equivalent capacitance.
$C_{eff} = \left(\frac{1}{3} + \frac{1}{6}\right)^{-1} + \left(\frac{1}{2} + \frac{1}{4}\right)^{-1}$
$C_{eff} = 3.33 \mu F$

c)
Its easier to do part c) first. The charge along a wire that doesn't branch is a constant. Therefore, the charge is the same across the $3 \mu F$ and $6 \mu F$ capacitors as well as across the $2 \mu F$ and the $4 \mu F$ capacitors. Let's call the branch with the $3 \mu F$ and $6 \mu F$ capacitors, branch A, and the branch with the $2 \mu F$ and the $4 \mu F$ capacitors, branch B.
We need to find the equivalent capacitance along both branches.
$C_{A} = \left(\frac{1}{3} + \frac{1}{6}\right)^{-1}$
$C_{A} = 2 \mu F$
so, along branch A,
Q_A = C_A V_A
$Q_{A} = (2 \mu F)(90 V)$
$Q_{A} = 180 \mu C ~=~ Q_{6} ~=~ Q_{3}$
Similarly,
$C_{B} = \left(\frac{1}{2} + \frac{1}{4}\right)^{-1}$
$C_{B} = \frac{8}{6} \mu F$
then, Q_B = C_B V_B
$Q_{B} = (1.33 \mu F)(90 V)$
$Q_{B} = 120 \mu C ~=~ Q_{2} ~=~ Q_{4}$

b)
Voltage drop is given by
$V = \frac{Q}{C}$
We know the charge and capacitance for each capacitor, so we have but to plug in the numbers.
$V_{3} = \frac{Q_{3}}{C_{3}}$
$V_{3} = \frac{180 \mu C}{3 \mu F} ~=~ 60 V$
$V_{6} = \frac{Q_{6}}{C_{6}}$
$V_{6} = \frac{180 \mu C}{6 \mu F} ~=~ 30 V$
$V_{2} = \frac{Q_{2}}{C_{2}}$
$V_{2} = \frac{120 \mu C}{2 \mu F} ~=~ 60 V$
$V_{4} = \frac{Q_{4}}{C_{4}}$
$V_{4} = \frac{120 \mu C}{4 \mu F} ~=~ 30 V$

d)
The energy stored by this configuration is the same as the energy stored by one capacitor of value C_eff. So,
$U_{total} = \frac{1}{2} C_{eff} V^{2}$
$U_{total} = \frac{1}{2} (3.3 \mu F) (90 V)^{2}$
$U_{total} = 1.34 \times 10^{-2} J$