In Fig., each capacitor has C = 4.00 mF and Vab = +28.0 V. Calculate

(a) the charge on each capacitor

(b) the potential difference across each capacitor;

(c) the potential difference between points a and d.

To get the charge on each capacitor we need to find the total capacitance so,

For the first two capacitors ${\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_2$in series net capacitance is

$\frac{1}{C_{12}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}\Rightarrow C_{12}=\frac{C}{2}=\frac{4}{2}=2\mu \mathrm{F}$

The result of two in parallel , ${\mathrm{C}}_{12},\mathrm{C}$is given by

$C^{\mathrm{\prime }}=C_{12}+C=2+4=6.0\mu \mathrm{F}$

The total capacitance $C_t$equivalent to $C^i$, C is

$\frac{1}{C_t}=\frac{1}{C^{\mathrm{\prime }}}+\frac{1}{C}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\Rightarrow C_t=2.4\mu \mathrm{F}$

The total charge on the capacitor $C_t$is

$Q_t=V_{ab}{C}_t=2.4\times {10}^{-6}\times 28=6.7\times {10}^{-5}\mu \mathrm{C}$

Since the capacitor in series should carry the same charge

$Q^{\mathrm{\prime }}=Q_4=6.7\times {10}^{-5}\mathrm{C}$

The voltageacross $\begin{array}{l}{C}_{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{C}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}C\end{array}$is the same so get that

$V_{12}=V_3=V^{\mathrm{\prime }}\Rightarrow \frac{Q_3}{C_3}=\frac{Q^{\mathrm{\prime }}}{C^{\mathrm{\prime }}}\Rightarrow Q_3=\frac{6.7\times {10}^{-5}\times 4}{6}=4.47\times {10}^{-5}\mathrm{C}$

So the charge $Q_{12}$is

$Q_{12}=Q^{\mathrm{\prime }}-Q_3=(6.7-4.47)\times {10}^{-5}=2.23\times {10}^{-5}\mathrm{C}$

The charge $Q_1\&Q_2$is

$Q_1=Q_2=Q_{12}=2.23\times {10}^{-5}\mathrm{C}$

Hence the charge on $Q_1\&Q_2$ is same.

The voltage across any capacitor is

$V_i=\frac{Q_i}{C_i}\Rightarrow \left(1\right)$

The voltage across the first capacitor is

$V_1=\frac{Q_1}{C_1}=\frac{2.23\times {10}^{-5}}{4\times {10}^{-6}}=5.6\mathrm{V}$

The voltage across the second capacitor is

$V_2=\frac{Q_2}{C_2}=\frac{2.23\times {10}^{-5}}{4\times {10}^{-6}}=5.6\mathrm{V}$

The voltage across the third capacitor is

$V_3=\frac{Q_3}{C_3}=\frac{4.47\times {10}^{-5}}{4\times {10}^{-6}}=11.2\mathrm{V}$

The voltage across the fourth capacitor is

$V_4=\frac{Q_4}{C_4}=\frac{6.7\times {10}^{-5}}{4\times {10}^{-6}}=16.8\mathrm{V}$

Hence the potential difference between a and d is

$V_{ad}=V^{\mathrm{\prime }}=V_3=11.2\mathrm{V}$