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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | | <meta name="date" content="2023-10" scheme="YYYY-MM"> |
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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> | | <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 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> |
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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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| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| 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 |
| $$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$ | | $$ U_2=\frac{kQq}{L-x}+\frac{kQq}{L+x} $$ |
| 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 $\Delta x$ |
| $$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$ | | $$ \boxed{\Delta x=\upsilon\sqrt{\frac{m\pi\varepsilon_0L^3}{Qq}}} $$ |
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| <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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