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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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			<h3> Statement </h3>
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			$3.1.10^*.$ 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
		@@ -72,7 +72,7 @@
			$$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$
			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 $\xi\ll1$
		+	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 $\Delta x$
			$$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$

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			<h4>Answer</h4>
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			$$\Delta U=\frac{Qqx^2}{2\pi\varepsilon_0L^3};\quad\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}$$
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	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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.		$3.1.10^*.$ 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 $3.1.10^*$		For problem $3.1.10^*$
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	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	Initial energy of the electric field		Initial energy of the electric field
	$$ U_1=\frac{2kQq}{L} $$		$$ U_1=\frac{2kQq}{L} $$
	Energy of the system after the bead is displaced		Energy of the system after the bead is displaced
	@@ -72,7 +72,7 @@
	$$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$		$$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$
	Change in energy of the system		Change in energy of the system
	$$ \Delta U=U_2-U_1=kQq\left(\frac{1}{L-x}+\frac{1}{L+x}-\frac{2}{L}\right) $$		$$ \Delta U=U_2-U_1=kQq\left(\frac{1}{L-x}+\frac{1}{L+x}-\frac{2}{L}\right) $$
	Let us consider the transformation for small values of $(1+\xi)^\alpha=1+\alpha\xi$, where $\xi\ll1$		Let us consider the transformation for small values of $(1+\xi )^\alpha =1+\alpha\xi$, where $\xi\ll1$
	$$ \boxed{\Delta U \approx\frac{Qqx^2}{2\pi\varepsilon_0L^3}} $$		$$ \boxed{\Delta U \approx\frac{Qqx^2}{2\pi\varepsilon_0L^3}} $$
	From the principle of conservation of energy it follows that the kinetic energy will be equal to the change in electrical energy $\Delta U$		From the principle of conservation of energy it follows that the kinetic energy will be equal to the change in electrical energy $\Delta U$
	$$ E_k=\Delta U\Leftrightarrow \frac{m\upsilon^2}{2}=\frac{Qq\Delta x^2}{2\pi\varepsilon_0L^3} $$		$$ E_k=\Delta U\Leftrightarrow \frac{m\upsilon^2}{2}=\frac{Qq\Delta x^2}{2\pi\varepsilon_0L^3} $$
	Where do we find the change in length $\Delta x$		Where do we find the change in length $\Delta x$
	$$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$		$$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$

	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$\Delta U=\frac{Qqx^2}{2\pi\varepsilon_0L^3};\quad\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}$$		$$\Delta U=\frac{Qqx^2}{2\pi\varepsilon_0L^3};\quad\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}$$
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	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Dzikan Mikita<br>		Dzikan Mikita<br>
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