In Fig. 25-51,$\mathbf{V}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}\mathbf{,}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{\mu F}$and ${\mathbf{C}}_{\mathbf{3}}\mathbf{=}{\mathbf{C}}_{\mathbf{4}}\mathbf{=}\mathbf{15}\mathbf{\mu F}$.What is the charge on capacitor 4?

The given values are:

1. V = 9V
2. ${\mathrm{C}}_1={\mathrm{C}}_2=30\mathrm{\mu F}$
   
3. ${\mathrm{C}}_2={\mathrm{C}}_4=15\mathrm{\mu F}$
   

The potential difference on each capacitor is the same in parallel combination but the charge on the capacitor may be different. In a series combination, the charges on the capacitors are equal; the potential difference may be different. Using this concept, we can find the charge on the capacitor.

Formulae:

The equivalent capacitance of a parallel connection of capacitors,

${\mathrm{C}}_{\mathrm{equivalent}}=\underset{}{\sum {\mathrm{C}}_{\mathrm{i}}}$ …(i)

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

The capacitors ${\mathrm{C}}_3$and ${\mathrm{C}}_4$are in parallel combination, so the equivalent capacitance can be given using equation (i) as follows:

$$\begin{gathered} {\mathrm{C}}_{34}={\mathrm{C}}_3+{\mathrm{C}}_4 \\ =15\mathrm{\mu F}+15\mathrm{\mu F} \\ =30\mathrm{\mu F} \end{gathered}$$

Now,$C_1,C_2$ and $C_{34}$are in series and all are numerically equal to 30$\mathrm{\mu F}$ .

So, each capacitor will have one-third of the battery voltage across it. Therefore, the potential across capacitor 4 is$V_4=3V$

So, the charge on capacitor can be given using equation (ii) as:

$$\begin{gathered} {\mathrm{q}}_4=15\mathrm{\mu F}\times 3\mathrm{V} \\ =45\mathrm{\mu C} \end{gathered}$$

Hence, the value of the charge is$45\mathrm{\mu C}$ .