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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    <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.

    <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>
        $$\Delta U=\frac{Qqx^2}{2\pi\varepsilon_0L^3};\quad\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}$$
    </p>

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