91影视

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

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:

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

For series combination
$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{鈭�}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}} \\ \mathrm{q}=\mathrm{CV} \end{gathered}$$

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,

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

So, find the capacitance of the capacitor 1 by,

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

By substituting the values,

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

Similarly, find the capacitance of the capacitor 2 by,

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

By substituting the values,

e

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

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

For series combination,$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{鈭�}}\frac{1}{{\mathrm{C}}_{\mathrm{j}}}$

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

Let鈥檚 find the charge on the positive plate of each capacitor.

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

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

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

So, the new equivalent capacitance is,

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

Now, the new charge on the positive plate is,

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

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,

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

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

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.