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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    <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>

Let's subdivide the cylindrical capacitor into a series of capacitors with cross-sectional area $S$ and thickness $dx$

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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