91Ó°ÊÓ

$$\begin{gathered} {\mathrm{C}}_1=4\mathrm{μ¹ó} \\ {\mathrm{C}}_2=8\mathrm{μ¹ó} \\ {\mathrm{C}}_3=12\mathrm{μ¹ó} \\ \mathrm{V}=12\mathrm{V} \end{gathered}$$

By using the concept of parallel and series combination of the capacitor, find the equivalent capacitance. Using the equivalent capacitance in the relation between capacitance and charge, find the charge. Find the number of electrons that travel through each point by dividing the charge by the charge on an electron.

Formulae are as follows:

$$\begin{gathered} \mathrm{q}=\mathrm{CV} \\ \mathrm{N}=\frac{\mathrm{q}}{\mathrm{e}} \end{gathered}$$

Where C is capacitance, V is the potential difference, q is the charge, and N is the no. of electrons.

First, calculate the equivalent capacitance.

${\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_2$are in parallel combination so its equivalent capacitance is,

This ${\mathrm{C}}_{12}$is in series with ${\mathrm{C}}_3$,

So the equivalent capacitance is,

$$\begin{gathered} {\mathrm{C}}_{123}=\frac{{\mathrm{C}}_3{\mathrm{C}}_{12}}{{\mathrm{C}}_3+{\mathrm{C}}_{12}} \\ =\frac{12\mathrm{μ¹ó}×12\mathrm{μ¹ó}}{12\mathrm{μ¹ó}+12\mathrm{μ¹ó}} \\ =6\mathrm{μ¹ó} \end{gathered}$$

Now, the total charge passing through the point is,

$$\begin{gathered} \mathrm{q}={\mathrm{C}}_{123}\mathrm{V} \\ =6\mathrm{μ¹ó}×12\mathrm{V} \\ =72\mathrm{μ°ä} \end{gathered}$$

The charge on an electron is$\mathrm{e}=1.6×{10}^{-19}\mathrm{C}$

So, the number of electronsthat travel through the pointis,

$$\begin{gathered} {\mathrm{N}}_{\mathrm{a}}=\frac{\mathrm{q}}{\mathrm{e}} \\ =\frac{72×{10}^{-6}\mathrm{C}}{1.6×{10}^{-19}} \\ =4.5×{10}^{14} \end{gathered}$$

Hence, the number of electrons that travel through the point is${\mathrm{N}}_{\mathrm{a}}=4.5×{10}^{14}$

The equivalent capacitance of the combination of capacitors 1 and 2 is${\mathrm{C}}_{12}=12\mathrm{μ¹ó}$

And the charge on that capacitor is the total charge due to the battery i.e.$\mathrm{q}=72\mathrm{μ°ä}$

So, the voltage across that pair is,

$$\begin{gathered} {\mathrm{V}}_{12}=\frac{\mathrm{q}}{{\mathrm{C}}_{12}} \\ =\frac{72\mathrm{μ°ä}}{12\mathrm{μ¹ó}} \\ =6\mathrm{V} \end{gathered}$$

Now, find the charge by${\mathrm{C}}_1$using the equation 25 - 1,

$$\begin{gathered} {\mathrm{q}}_1={\mathrm{C}}_1{\mathrm{V}}_{12} \\ =4\mathrm{μ¹ó}×6\mathrm{V} \\ =24\mathrm{μ°ä} \end{gathered}$$

So, the number of electrons that travel through the point b is,

$$\begin{gathered} {\mathrm{N}}_{\mathrm{b}}=\frac{{\mathrm{q}}_1}{\mathrm{e}} \\ =\frac{24×{10}^{-6}\mathrm{C}}{1.6×{10}^{-19}\mathrm{C}} \\ =1.5×{10}^{14} \end{gathered}$$

Hence, the number of electrons that travel through the point b is${\mathrm{N}}_{\mathrm{b}}=1.5×{10}^{14}$

Similarly the charge on${\mathrm{C}}_2$is,

$$\begin{gathered} {\mathrm{q}}_2={\mathrm{C}}_2{\mathrm{V}}_{12} \\ =8\mathrm{μ¹ó}×6\mathrm{V} \\ =48\mathrm{μ°ä} \end{gathered}$$

So, the number of electronsthat travel through point c is,

$$\begin{gathered} {\mathrm{N}}_{\mathrm{c}}=\frac{{\mathrm{q}}_2}{\mathrm{e}} \\ {\mathrm{N}}_{\mathrm{c}}=\frac{48×{10}^{-6}\mathrm{C}}{1.6×{10}^{-19}\mathrm{C}} \\ =3×{10}^{14} \end{gathered}$$

Hence, the number of electrons that travel through the point c is${\mathrm{N}}_{\mathrm{c}}=3×{10}^{14}$

To find the number of electronsthat travel through the point, use the concept of conservation of charge.

From the conservation of charge, the charge through the pointis the same as the through point d.

So, the total number of electrons through point d is,

${\mathrm{N}}_{\mathrm{d}}=4.5×{10}^{14}$

Hence, the number of electrons that travel through the point d is${\mathrm{N}}_{\mathrm{d}}=4.5×{10}^{14}$

As it is known that the electrons travel in the direction opposite to the current. Since the current is from the positive terminal of the battery to the negative terminal of the battery, so the electrons travel up through the point b

Hence, the electrons travel up through the point b

Similarly, through point, the electrons travel up.

Hence, the electrons travel up through the point c

Therefore, by using the relation between capacitance and charge, find the number of electrons at various points and also the direction of motion of that electron.