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| + | <a href="../../" style="text-decoration: none;"> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
| + | </div> | ||
| + | </a> | ||
| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <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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| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
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| <a href="../../" style="text-decoration: none;"> | |||
| <div id="logo"> | |||
| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
| </div> | |||
| </a> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <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> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||