91Ó°ÊÓ

The potential difference is V = 10 V

${\mathrm{C}}_1={\mathrm{C}}_2={\mathrm{C}}_3={\mathrm{C}}_4={\mathrm{C}}_5=10\mathrm{μ¹ó}\left(\frac{{10}^{-6}\mathrm{F}}{1\mathrm{μ¹ó}}\right)=1.0×{10}^{-5}\mathrm{F}$

Find the charge on capacitor1by using the concept of capacitance. To find the charge on capacitor2, find the equivalent capacitance and potential difference across capacitors2.

Formulae are as follows:

q = CV

For parallel combination,${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}\underset{\mathbf{j}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{∑}}}{\mathbf{C}}_{\mathbf{j}}$

For series combination,$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}\mathbf{=}\underset{\mathbf{j}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{∑}}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{j}}}$

Where C is capacitance, V is the potential difference, and q is the charge on the capacitor.

It is known that,

q = CV

The potential difference across the capacitor ${\mathrm{C}}_1$is,

${\mathrm{V}}_1=10\mathrm{V}$

So the charge on the capacitor 1 is,

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1{\mathrm{V}}_1 \\ {\mathrm{q}}_1=1.0×{10}^{-5}\mathrm{F}×10\mathrm{V} \\ =1.0×{10}^{-4}\mathrm{C} \end{gathered}$$

Hence, the charge on the capacitor 1 is $1.0×{10}^{-4}\mathrm{C}$

For finding the charge${\mathrm{q}}_2$, first, find the equivalent capacitance.

Consider the three-capacitor combination consisting of${\mathrm{C}}_2$and its two closest neighbors, each of capacitance C. Using the formula for parallel and series combination of the capacitor, write the equivalent capacitance of this combination as

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_2\mathrm{C}}{\mathrm{C}+{\mathrm{C}}_2}$

By substituting the values,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}_{×}\mathrm{C}}{\mathrm{C}+\mathrm{C}} \\ {\mathrm{C}}_{\mathrm{eq}}=\mathrm{C}+\frac{{\mathrm{C}}^2}{2\mathrm{C}} \\ =\frac{3{\mathrm{C}}^2}{2\mathrm{C}} \\ =1.5\mathrm{C} \end{gathered}$$

The voltage drop in this combination is,

$\mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+{\mathrm{C}}_{\mathrm{eq}}}$

By outing the value of ${\mathrm{C}}_{\mathrm{eq}}$,

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{CV}}{\mathrm{C}+1.5\mathrm{C}} \\ =\frac{{\mathrm{CV}}_1}{2.5\mathrm{C}} \\ =0.4{\mathrm{V}}_1 \end{gathered}$$

This voltage difference is divided into two equal parts between ${\mathrm{C}}_2$and the capacitor connected in series with it. So, the total voltage across the capacitor 2 is,

$$\begin{gathered} {\mathrm{V}}_2=\frac{\mathrm{V}}{2} \\ =\frac{0.4{\mathrm{V}}_1}{2} \\ =0.2{\mathrm{V}}_1 \end{gathered}$$

Thus, the total charge on the capacitor 2 will be,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_2 \\ {\mathrm{q}}_2=1.0×{10}^{-6}\mathrm{F}×0.2{\mathrm{V}}_1 \end{gathered}$$

By substituting the value of${\mathrm{V}}_1$,

$$\begin{gathered} {\mathrm{q}}_2=1.0×{10}^{-5}\mathrm{F}×0.2×10\mathrm{V} \\ =2×{10}^{-5}\mathrm{C} \end{gathered}$$

Hence, the charge on the capacitor 2 is$2×{10}^{-5}\mathrm{C}$

Therefore, we can find the charge on both capacitors by using the concept of capacitance.