Three capacitors having capacitances of 8.4, 8.4 and $\mathbf{4}\mathbf{.}\mathbf{2}\mathit{\mu }\mathit{F}$are connected in series across a 36V potential difference. (a) What is the charge on the $\mathbf{4}\mathbf{.}\mathbf{2}\mathit{\mu }\mathit{F}$capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

$\begin{array}{r}{C}_1=8.4\mu F\\ C_2=8.4\mu F\\ C_3=4.2\mu F\\ V_{ab}=36\mathrm{V}\end{array}$

The total charge flow through the system is given by

$Q_t=C_t{V}_{ab}$ (1)

The total capacitance is given by

$\frac{1}{C_t}=\sum \frac{1}{C_i}=\frac{1}{8.4}+\frac{1}{8.4}+\frac{1}{4.2}=\frac{10}{21}=2.1\mu F$

Substituting in (1) yields

$Q=36\times 2.1\times {10}^{-6}=75.6\times {10}^{-6}\mathrm{C}$

$Q_3=75.6\times {10}^{-6}\mathrm{C}$

The total charge stored in capacitors is given by

$U=\frac{1}{2}{C}_t{V}^2=0.5\times 2.1\times {10}^{-6}\times {36}^2=1.36\times {10}^{-3}\mathrm{J}$

The charge in the system will be redistributed on each capacitor but the total charge is the same.

$Q_1+Q_2+Q_3=3\times 75.6=226.8\mu \mathrm{C}$

$V_1=V_2=V_3=\frac{Q_1}{C_1}=\frac{Q_2}{C_2}=\frac{Q_3}{C_3}$ (2)

Substituting in (1) yields

$\begin{array}{r}\frac{Q_1}{8.4}=\frac{Q_2}{8.4}=\frac{Q_3}{4.2}\\ Q_1=Q_2=2Q_3\end{array}$

And for $c_3$

$Q_3=Q_1/2=45.4\mu C$

$V=\frac{Q_1}{C_1}=\frac{0.72\times {10}^{-6}}{8.4\times {10}^{-8}}=10.8\mathrm{V}$

The new energy stored in the system is given by

$U=\frac{1}{2}{C}_P{V}^2$ (4)

The capacitance is given by

$C_t=C_1+C_2+C_3=8.4+8.4+4.2=21.0\mu F$

$U_P=0.5\times 21\times {10}^{-6}\times (10.8{)}^2=1.22\times {10}^{-3}\mathrm{J}$