91Ó°ÊÓ

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

$\frac{1}{C_{12}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}⇒C_{12}=\frac{C}{2}=\frac{4}{2}=2\mathrm{μ}\mathrm{F}$

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

$C^{\mathrm{′}}=C_{12}+C=2+4=6.0\mathrm{μ}\mathrm{F}$

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

$\frac{1}{C_t}=\frac{1}{C^{\mathrm{′}}}+\frac{1}{C}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}⇒C_t=2.4\mathrm{μ}\mathrm{F}$

The total charge on the capacitor $C_t$is

$Q_t=V_{ab}{C}_t=2.4×{10}^{−6}×28=6.7×{10}^{−5}\mathrm{μ}\mathrm{C}$

Since the capacitor in series should carry the same charge

$Q^{\mathrm{′}}=Q_4=6.7×{10}^{−5}\mathrm{C}$

The voltageacross $\begin{array}{l}{C}_{12}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots C_3â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots C\end{array}$is the same so get that

$V_{12}=V_3=V^{\mathrm{′}}⇒\frac{Q_3}{C_3}=\frac{Q^{\mathrm{′}}}{C^{\mathrm{′}}}⇒Q_3=\frac{6.7×{10}^{−5}×4}{6}=4.47×{10}^{−5}\mathrm{C}$

So the charge $Q_{12}$is

$Q_{12}=Q^{\mathrm{′}}−Q_3=(6.7−4.47)×{10}^{−5}=2.23×{10}^{−5}\mathrm{C}$

The charge $Q_1\&Q_2$is

$Q_1=Q_2=Q_{12}=2.23×{10}^{−5}\mathrm{C}$

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

The voltage across any capacitor is

$V_i=\frac{Q_i}{C_i}⇒\left(1\right)$

The voltage across the first capacitor is

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

The voltage across the second capacitor is

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

The voltage across the third capacitor is

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

The voltage across the fourth capacitor is

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

Hence the potential difference between a and d is

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