Translated 3.1.1-3.1.17

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+ <meta name="description" content="A bead with charge q can move without friction along a stretched string of length 2 L with charges Q attached to its ends. Find the increment of potential energy when the bead moves x along the string from its center. Find that for small displacements the dependence of the increment of potential energy on x is quadratic. Find how much the bead of mass m will be displaced if it is given a small velocity v at the equilibrium position.">
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+ <title>A bead with charge q can move without friction along a stretched string of length 2 L with charges Q attached to its ends. Find the increment of potential energy when the bead moves x along the string from its center. Find that for small displacements the dependence of the increment of potential energy on x is quadratic. Find how much the bead of mass m will be displaced if it is given a small velocity v at the equilibrium position.</title>
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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="../../#3.1">$\leftarrow$Back</a></h3>
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+ <h3> Statement </h3>
+ <p>
+ $3.1.10^*.$ A bead with charge $q$ can move without friction along a stretched string of length $2 L$ with charges $Q$ attached to its ends. Find the increment of potential energy when the bead moves $x$ along the string from its center. Find that for small displacements the dependence of the increment of potential energy on $x$ is quadratic. Find how much the bead of mass $m$ will be displaced if it is given a small velocity $v$ at the equilibrium position.
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+ For problem $3.1.10^*$
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+ <h3>Solution</h3>
+ <p>
+ Initial energy of the electric field
+$$ U_1=\frac{2kQq}{L} $$
+Energy of the system after the bead is displaced
+$$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$
+Change in energy of the system
+$$ \Delta U=U_2-U_1=kQq\left(\frac{1}{L-x}+\frac{1}{L+x}-\frac{2}{L}\right) $$
+Let us consider the transformation for small values ​​of $(1+\xi)^\alpha=1+\alpha\xi$, where $\xi\ll1$
+$$ \boxed{\Delta U \approx\frac{Qqx^2}{2\pi\varepsilon_0L^3}} $$
+From the principle of conservation of energy it follows that the kinetic energy will be equal to the change in electrical energy $\Delta U$
+$$ E_k=\Delta U\Leftrightarrow \frac{m\upsilon^2}{2}=\frac{Qq\Delta x^2}{2\pi\varepsilon_0L^3} $$
+Where do we find the change in length $\Delta x$
+$$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\Delta U=\frac{Qqx^2}{2\pi\varepsilon_0L^3};\quad\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}$$
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