<h3 id="back-link"><a href="/#7.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $7.1.12^*.$ Determine what the accelerating potential difference $V$ should be in order for the electrons to follow the path shown in the figure. Radii of cylindrical capacitor plates $R_1$ and $R_2$. Potential difference between the plates $V_0$.

    <h3>Solution</h3>
    <p>
        Gauss theorem

$$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi \varepsilon_0 r}$$
Let's write down the small change of the electrostatic field and then integrate it
$$dU=E \,dr\Rightarrow \int dU=\int E \,dr$$
Integrate from $R_1$ to $R_2$
$$U_0=\frac{q}{2\pi \varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi \varepsilon_0 r}\ln\frac{R_2}{R_1}$$
Law of conservation of energy
$$\frac{mv^2}{2}=eU$$
$$e\frac{q}{2\pi \varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$
From where we get
$$\frac{q}{2\pi \varepsilon_0}=\frac{mv^2}{e}$$
Substituting the previously obtained values
$$\frac{U_0}{\ln\frac{R_2}{R_1}}=\frac{2eU}{e}$$
From where
$$U=\frac{U_0}{2\ln\frac{R_2}{R_1}}$$

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