Fig.25-39 represents two air-filled cylindrical capacitors connected in series across a battery with potential V = 10 VCapacitor 1 has an inner plate radius of 5.0mm an outer plate radius of 1.5 cmand a length of 5.0 cm.Capacitor 2 has an inner plate radius of 2.5mman outer plate radius of 1.0 cmand a length of 9.0 cm. The outer plate of capacitor 2 is a conducting organic membrane that can be stretched, and the capacitor can be inflated to increase the plate separation. If the outer plate radius is increased to 2.5 cmby inflation, (a) how many electrons move through point P and (b) do they move toward or away from the battery?

$$\begin{gathered} \mathrm{V}=10\mathrm{V} \\ {\mathrm{a}}_1=5.0\mathrm{mm} \\ {\mathrm{b}}_1=1.5\mathrm{cm} \\ {\mathrm{L}}_1=5\mathrm{cm} \\ {\mathrm{a}}_2=2.5\mathrm{mm} \\ {\mathrm{b}}_2=1\mathrm{cm} \\ {\mathrm{b}}_2\text{'}=2.5\mathrm{cm} \end{gathered}$$

By using the formula of capacitance for cylindrical capacitors, find the capacitance of both the capacitors. By using the series combination of capacitors, find the equivalent capacitance of the circuit. Then find the charge on the positive plate. Similarly, after modifying capacitor 2, find the new charge on the positive plate, and then the charge is transferred from the battery. Get the total number of the electron by dividing the charge by the charge of an electron.

Formulae are as follows:

$\mathrm{C}=\frac{2{\mathrm{\tau \tau \epsilon }}_0\mathrm{L}}{\ln \left({\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\right)}$

For series combination
$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}} \\ \mathrm{q}=\mathrm{CV} \end{gathered}$$

Where C is capacitance, V is the potential difference, q is the charge, L is length, and a and b are the radii.

According to the equation 25 - 14, the capacitance for the cylindrical capacitor is given by,

$\mathrm{C}=\frac{2{\mathrm{\tau \tau \epsilon }}_0\mathrm{L}}{\ln \left({\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\right)}$

So, find the capacitance of the capacitor 1 by,

${\mathrm{C}}_1=\frac{2{\mathrm{\tau \tau \epsilon }}_0{\mathrm{L}}_1}{\ln \left({\displaystyle \frac{{\mathrm{b}}_1}{{\mathrm{a}}_1}}\right)}$

By substituting the values,

$$\begin{gathered} {\mathrm{C}}_1=\frac{2\mathrm{\tau \tau }\times \left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)\times 5\times {10}^{-2}\mathrm{m}}{\ln \left({\displaystyle \frac{1.5\times {10}^{-2}\mathrm{m}}{5\times {10}^{-3}\mathrm{m}}}\right)} \\ =\frac{2.78\times {10}^{-12}}{\ln \left(3\right)} \\ =2.53\times {10}^{-12}\mathrm{F}\left(\frac{1\mathrm{pF}}{{10}^{-12}\mathrm{F}}\right) \\ =2.53\mathrm{pF} \end{gathered}$$

Similarly, find the capacitance of the capacitor 2 by,

${\mathrm{C}}_2=\frac{2{\mathrm{\tau \tau \epsilon }}_0{\mathrm{L}}_2}{\ln \left({\displaystyle \frac{{\mathrm{b}}_2}{{\mathrm{a}}_2}}\right)}$

By substituting the values,

e

$$\begin{gathered} {\mathrm{C}}_2=\frac{5\times {10}^{12}\mathrm{F}}{\ln \left(4\right)} \\ =3.61\times {10}^{-12}\mathrm{F}\left(\frac{1\mathrm{pF}}{{10}^{-12}\mathrm{F}}\right) \\ =3.61\mathrm{pF} \end{gathered}$$

Since ${\mathrm{C}}_1\mathrm{and}{\mathrm{C}}_2$are connected in series, their equivalent capacitance can be calculated.

For series combination,$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{12}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2} \\ {\mathrm{C}}_{12}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2} \\ {\mathrm{C}}_{12}=\frac{2.53\mathrm{pF}\times 3.61\mathrm{pF}}{2.53\mathrm{pF}\times 3.61\mathrm{pF}} \\ {\mathrm{C}}_{12}=1.49\mathrm{pF} \end{gathered}$$

Let’s find the charge on the positive plate of each capacitor.

$$\begin{gathered} \mathrm{q}={\mathrm{C}}_{12}\mathrm{V} \\ =1.49\mathrm{pF}\times 10\mathrm{V} \\ =14.9\mathrm{pC} \end{gathered}$$

Now, the capacitor 2 is modified as described in the problem. So, the new capacitance for capacitor 2 is,

role="math" localid="1661751513348" $$\begin{gathered} {\mathrm{C}}_2\text{'}=\frac{2{\mathrm{\tau \tau \epsilon }}_0{\mathrm{L}}_2}{\ln \left({\mathrm{b}}_2\text{'}/{\mathrm{a}}_2\right)} \\ \mathrm{By}\mathrm{substituting}\mathrm{the}\mathrm{values}, \\ {\mathrm{C}}_2\text{'}=\frac{2\mathrm{\tau \tau }(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m})\times 9\times {10}^{-12}\mathrm{m}}{\ln \left({\displaystyle \frac{2.5\times {10}^{-2}\mathrm{m}}{2.5\times {10}^{-3}\mathrm{m}}}\right)} \\ =\frac{5\times -{10}^{-12}\mathrm{F}}{\ln \left(10\right)} \\ =2.17\mathrm{pF} \end{gathered}$$

So, the new equivalent capacitance is,

$$\begin{gathered} {\mathrm{C}}_{12}\text{'}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2\text{'}}{{\mathrm{C}}_1+{\mathrm{C}}_2\text{'}} \\ =\frac{2.53\mathrm{pF}\times 2.17\mathrm{pF}}{2.53\mathrm{pF}+1.17\mathrm{pF}} \\ =1.17\mathrm{pF} \end{gathered}$$

Now, the new charge on the positive plate is,

$$\begin{gathered} \mathrm{q}\text{'}={\mathrm{C}}_{12}\text{'}\mathrm{V} \\ =1.17\mathrm{pF}\times 10\mathrm{V} \\ =11.7\mathrm{pC} \end{gathered}$$

So, the charge transfer from thebattery is,

$$\begin{gathered} \mathrm{Q}=\mathrm{q}-\mathrm{q}\text{'} \\ =14.9\mathrm{pC}-11.7\mathrm{pC} \\ =3.2\mathrm{pC} \end{gathered}$$

Get the total number of electrons that pass through point P by dividing this charge by the charge of an electron,

$$\begin{gathered} \mathrm{N}=\frac{3.2\times {10}^{-12}\mathrm{C}}{1.6\times {10}^{-19}\mathrm{C}} \\ =2\times {10}^7 \end{gathered}$$

Hence, the number of an electron moving through point P is $2\times {10}^7$.

The plate closest to point P is a positive plate and it has suffered a decrease in the positive charges. So from the figure, the electrons move rightward to the battery i.e. they move away from the battery.

Hence, the electrons moveaway fromthe battery.

Therefore, by using the formula for the capacitance of the cylindrical capacitor, find the number of electrons moving through point P and their direction.