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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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 $V_0$.
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $7.1.12^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <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
+$$U=\frac{U_0}{2\ln\frac{R_2}{R_1}}$$
+ </p>
+ <h4>Answer</h4>
+ <p>
+ $$V=(V_0/2)/\ln(R_2/R_1)$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Lutfulloyev Shukurullo<br>
+ </p>
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