Question

Each capacitor in figure (31-E21) has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.

Answer 1

Capacitors b and c are in parallel; their equivalent capacitance is 20 µF.
Thus, the net capacitance of the circuit is given by

$$\begin{gathered} \frac{1}{C_{\mathrm{net}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{10} \\ \Rightarrow \frac{1}{C_{\mathrm{net}}}=\frac{2+1+2}{20}=\frac{5}{20} \\ \Rightarrow C_{\mathrm{net}}=4\mathrm{\mu F} \end{gathered}$$

The total charge of the battery is given by

$Q=C_{\mathrm{net}}V=(4\mathrm{\mu F})\times (100\mathrm{V})=4\times {10}^{-4}\mathrm{C}$

For a and d,
$$\begin{gathered} q=4\times {10}^{-4}\mathrm{C}\mathrm{and}\mathrm{C}={10}^{-5}\mathrm{F} \\ \therefore E=\frac{q^2}{2C}=\frac{4\times {10}^{-4}}{2\times {10}^{-5}} \\ \Rightarrow E=8\times {10}^{-3}\mathrm{J}=8\mathrm{mJ} \end{gathered}$$

For b and c,
$$\begin{gathered} q=4\times {10}^{-4}\mathrm{C}\mathrm{and}{C}_{eq}=2C=2\times {10}^{-5}\mathrm{F} \\ \therefore V=\frac{q}{C_{eq}}\frac{4\times {10}^{-4}}{2\times {10}^{-5}}=20\mathrm{V} \\ \Rightarrow E=\frac{1}{2}C{V}^2 \\ \Rightarrow E=\frac{1}{2}\times {10}^{-5}\times 400 \\ \Rightarrow E=2\times {10}^{-3}\mathrm{J}=2\mathrm{mJ} \end{gathered}$$