Added 7.1.12, 7.1.23, 7.3.9 & 11.5.11

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+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $7.1.23^*.$ Find the period of small vibrations of a body of mass $m$, whose charge is $q$, inside a smooth sphere of radius $R$, if the charge $Q$ is fixed at the top point of the sphere.
+</p>
+<center>
+ <figure>
+ <img src="7.1.23.png"
+ loading="lazy" width="200" />
+ <figcaption>
+ For problem $7.1.23^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="7.1.23_1.png"
+ loading="lazy" width="200" />
+ <figcaption>
+ Forces acting on the system of two charges
+ </figcaption>
+ </figure>
+</center>
+<p>
+
+Considering the fact that oscillations are small, $\varphi \ll 1$, we could use the approximations for small angle $\varphi$
+$$\sin\varphi \approx \varphi; \quad\cos\varphi \approx 1\quad(1)$$
+From where, we could approximate that the distance between charges remains approximately the same
+$$r\approx2R$$
+Thus, the сoulomb's law for two charges
+$$F_c = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}$$
+Let's write Newton's second law on a tangential trajectory, in the direction of motion of a charge of mass $m$
+$$ma = mg \sin\varphi + F_c \sin\frac{\varphi}{2}$$
+Considering the approximation $(1)$ and the expression
+$$ma = -mg\frac{x}{R} - \frac{x}{2R} \frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}$$
+$$a = -x\left(\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}\right)$$
+Solving the equation of harmonic oscillations $(\ddot{x}+\omega^2x(t)=0)$, we could obtain the value for angular velocity
+$$\omega=\sqrt{\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}}$$
+From here the period of oscillation
+$$\boxed{T=\frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}}}}$$
+Since the expression under the square root cannot be less than zero, the ratio at which there will be no oscillations
+$$\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3} > 0 \Leftrightarrow \boxed{\frac{qQ}{32\pi\varepsilon_0R^2}>-mg}$$
+ </p>
+
+ <h4>Answer</h4>
+ <div class="scroll-wrapper">
+ <p>
+ $$T=2\pi\left(\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}\right)^{-1/2}\text{ with }\frac{qQ}{32\pi\varepsilon_0R^2}>-mg$$
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