In Fig. 25-55,$\mathit{V}\mathbf{=}\mathbf{12}\mathit{V}\mathbf{,}\mathbf{}{\mathit{C}}_{\mathbf{1}}\mathbf{=}{\mathit{C}}_{\mathbf{5}}\mathbf{=}{\mathit{C}}_{\mathbf{6}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}\mathbf{}\mathbf{and}\mathbf{}{\mathit{C}}_{\mathbf{2}}\mathbf{=}{\mathit{C}}_{\mathbf{3}}\mathbf{=}{\mathit{C}}_{\mathbf{4}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}$. What are (a) the net charge stored on the capacitors and (b) the charge on capacitor 4?

The given values are:

1. Potential difference,$V=12V$
2. Capacitance,$C_1=C_5=C_6=6.0\mu F$
3. Capacitance,$C_2=C_3=C_4=4.0\mu F$

We find the equivalent capacitance of the capacitor to find the net charge on the system. Using the relation between charge and capacitance we can find the charge on the capacitor.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\sum _{}\frac{1}{C_j}$ …(i).

The equivalent capacitance of a parallel connection of capacitors,

$\frac{1}{C_{equivalent}}=\sum _{}{C}_j$ …(ii)

The charge stored between the plates of the capacitor, $q=CV$ …(iii)

The capacitor $C_2$and $C_3$and $C_4$are parallel.

Their equivalent capacitance is given by using the given data in equation (ii) as follows:

$$\begin{gathered} C\text{'}=C_{234} \\ =4.0\mathrm{\mu F}+4.0\mathrm{\mu F}+4.0\mathrm{\mu F} \\ =12\mathrm{\mu F} \end{gathered}$$

The equivalent capacitance of the capacitors 5 and 6 which are in parallel is given by using equation (ii) as follows:

$$\begin{gathered} C\text{'}\text{'}=C_{56} \\ =6.0\mathrm{\mu F}+6.0\mathrm{\mu F} \\ =12\mathrm{\mu F} \end{gathered}$$

Now, the capacitors$C_1,C\text{'},C\text{'}\text{'}$are in series and their equivalent capacitance is given by using equation (i) as follows:

$$\begin{gathered} \frac{1}{C_{system}}=\frac{1}{C_1}+\frac{1}{C\text{'}}+\frac{1}{C\text{'}\text{'}} \\ \frac{1}{C_{system}}=\frac{C_{1}C\text{'}C\text{'}\text{'}}{C\text{'}C\text{'}\text{'}+C_{1}C\text{'}\text{'}+C_{1}C\text{'}} \\ =\frac{\left(6\right)\left(12\right)\left(12\right)}{12\times 12+6\times 12+6\times 12}\mu F \\ =3.0\mu F \end{gathered}$$

Thus, the value of the charge can be given using equation (iii) as follows:

$$\begin{gathered} q_{system}=3.0\mu F\times 12V \\ =36\mu C \end{gathered}$$

Hence, the value of the charge of the system is $36\mu C$.

Since, charge of the system is equal to charge on capacitor 1, thus $q_{sys}=q_1$

Then, the voltage across $C_1$is calculated using the given data in equation (iii) as,

$$\begin{gathered} V_1=\frac{q_1}{C_1} \\ =\frac{36\mu C}{6\mu F} \\ =6.0V \end{gathered}$$

Then the voltage across series pair of $C\text{'}\text{'}\mathrm{and}C\text{'}$is given as:

$V_{bat}=V_1+V$, where $V$is voltage across $C\text{'}\text{'}\mathrm{and}C\text{'}$

$V=\frac{q}{C_{eq}}$, where q is the charge on equivalent capacitance of $C\text{'}\text{'}\mathrm{and}C\text{'}\mathrm{and}{C}_{eq}$is the equivalent capacitance of $C\text{'}\text{'}\mathrm{and}C\text{'}$, is given by using equation (ii) in equation (iii) as:

$$\begin{gathered} V=36\mu C\frac{C\text{'}\text{'}+C\text{'}}{C\text{'}\text{'}C\text{'}} \\ =\frac{\left(36\mathrm{\mu C}\right)\left(12\mathrm{\mu F}+12\mathrm{\mu F}\right)}{\left(12\mathrm{\mu F}+12\mathrm{\mu F}\right)} \\ =0.6\mathrm{V} \end{gathered}$$

Now, the potential difference across $V_1$is given as:

$$\begin{gathered} V_1=V_{bat}-V \\ =12V-6.0V \\ =6.0V \end{gathered}$$

Since, $C\text{'}\text{'}=C\text{'}$, therefore, the voltage value is given by,

$$\begin{gathered} V\text{'}=V\text{'}\text{'} \\ =\frac{6.0V}{2} \\ =6.0V \end{gathered}$$

which is also the voltage across $V_4$.

Thus, the charge on the capacitor $C_4$ is given using equation (iii) as:

$$\begin{gathered} q_4=\left(4.0\mathrm{\mu F}\right)\left(3.0\mathrm{V}\right) \\ =12\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is $12\mathrm{\mu C}$.