Question

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?

Short answer

The charge on capacitor 4 is 45$\mathrm{\mu C}$ .

Step-by-step solution

The given data

The given values are:

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

Understanding the concept of the equivalent capacitance and charge

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)

Calculation of charge on capacitor 4

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$45\mathrm{\mu C}$ .