Added 6.4.7

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		+	<header style="text-align:center;">
		+	<a href="../../" style="text-decoration: none;">
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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<h3 id="back-link"><a href="../../#6.4">$\leftarrow$Back</a></h3>
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		+	<h3> Statement </h3>
		+	<p>
		+	$6.4.7^*.$ Find the capacity of a cylindrical capacitor formed by two coaxial cylinders of radius $R_1$ and $R_2$. Cylinder length $l \gg R_1$, $R_2$.
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="6.4.7_1.png"
		+	loading="lazy" width="140" />
		+	<figcaption>
		+	For problem $6.4.7^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+
		+	Let's subdivide the cylindrical capacitor into a series of capacitors with cross-sectional area $S$ and thickness $dx$
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="6.4.7_2.png"
		+	loading="lazy" width="320" />
		+	<figcaption>
		+	Series of capacitors
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	then the total capacitance $C_0$ of the capacitor could be found as
		+	$$\frac{1}{C_0} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots +\frac{1}{C_{n-1}}+\frac{1}{C_n} = \sum_i\frac{1}{C_i}=\int\frac{1}{dC}\quad(1)$$
		+	The capacitance of the small capacitor with the cross-sectional area $S$ and thickness $dx$ placed on the distance $x$ from the central axis of the cylinder
		+	$$dC = \frac{\varepsilon_0 S}{dx} = \varepsilon_0\frac{2\pi x l}{dx}\quad(2)$$
		+	After substituting $(2)$ into $(1)$
		+	$$\boxed{\frac{1}{C_0} = \frac{1}{2\pi\varepsilon_0l}\int_{R_1}^{R_2} \frac{dx}{x}}$$
		+	After integration of the following expression
		+	$$\frac{1}{C_0} = \frac{1}{2\pi\varepsilon_0l}\cdot\ln{\frac{R_2}{R_1}} $$
		+	After the transformations, we obtain the capacity of a cylindrical capacitor:
		+	$$\boxed{C_0 = \frac{2\pi\varepsilon_0l}{\ln{\frac{R_2}{R_1}}}} $$
		+
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$C_0 = \frac{2\pi\varepsilon_0l}{\ln{\frac{R_2}{R_1}}}$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Aliaksandr Melnichenka<br>
		+	</p>
		+
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			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="Find the capacity of a cylindrical capacitor formed by two coaxial cylinders of radius R_1 and R_2. Cylinder length l \gg R_1, R_2.">
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			</p>
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			<h3 id="back-link"><a href="../../#6.4">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$6.4.7^*.$ Find the capacity of a cylindrical capacitor formed by two coaxial cylinders of radius $R_1$ and $R_2$. Cylinder length $l \gg R_1$, $R_2$.
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="6.4.7_1.png"
			loading="lazy" width="140" />
			<figcaption>
			For problem $6.4.7^*$
			</figcaption>
			</figure>
			</center>
			<p>

			Let's subdivide the cylindrical capacitor into a series of capacitors with cross-sectional area $S$ and thickness $dx$
			</p>
			<center>
			<figure>
			<img src="6.4.7_2.png"
			loading="lazy" width="320" />
			<figcaption>
			Series of capacitors
			</figcaption>
			</figure>
			</center>
			<p>
			then the total capacitance $C_0$ of the capacitor could be found as
			$$\frac{1}{C_0} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots +\frac{1}{C_{n-1}}+\frac{1}{C_n} = \sum_i\frac{1}{C_i}=\int\frac{1}{dC}\quad(1)$$
			The capacitance of the small capacitor with the cross-sectional area $S$ and thickness $dx$ placed on the distance $x$ from the central axis of the cylinder
			$$dC = \frac{\varepsilon_0 S}{dx} = \varepsilon_0\frac{2\pi x l}{dx}\quad(2)$$
			After substituting $(2)$ into $(1)$
			$$\boxed{\frac{1}{C_0} = \frac{1}{2\pi\varepsilon_0l}\int_{R_1}^{R_2} \frac{dx}{x}}$$
			After integration of the following expression
			$$\frac{1}{C_0} = \frac{1}{2\pi\varepsilon_0l}\cdot\ln{\frac{R_2}{R_1}} $$
			After the transformations, we obtain the capacity of a cylindrical capacitor:
			$$\boxed{C_0 = \frac{2\pi\varepsilon_0l}{\ln{\frac{R_2}{R_1}}}} $$

			</p>

			<h4>Answer</h4>
			<p>
			$$C_0 = \frac{2\pi\varepsilon_0l}{\ln{\frac{R_2}{R_1}}}$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Aliaksandr Melnichenka<br>
			</p>



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