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en/6.1.13.md
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| + | ### Statement | ||
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| + | $6.1.13.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $6.1.13.$ Four identical particles, each of mass $m$ and charge $-q$, revolve in a circular orbit, located at the corners of a square of side $\ell$. A charge $q$ is located at the center of this square. Determine the angular velocity of the particles in orbit. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | <div align="center"> | ||
| + | <img src="https://savchenkosolutions.com/6/6.1.13/drawing.png" alt="Forces acting on the charge"> | ||
| + | </div> | ||
| + | |||
| + | Let us find the magnitude of the net radial force $F$ acting on one of the $-q$ charges and directed towards the center of the orbit. | ||
| + | This force is the superposition of the attractive force from the central charge $q$ and the repulsive forces from the other three $-q$ charges. | ||
| + | |||
| + | The radius of the orbit (distance to the center) is $r = \frac{\ell}{\sqrt{2}}$. | ||
| + | Using Coulomb's law, we project all forces onto the radial axis: | ||
| + | $$F = F_{\text{center}} - 2F_{\text{adj}} \cdot \cos 45^\circ - F_{\text{opp}} \quad (1)$$ | ||
| + | |||
| + | Where: | ||
| + | * The 1st term is the interaction with the central charge $q$. | ||
| + | * The 2nd term is the interaction with the two adjacent $-q$ charges (at distance $\ell$). | ||
| + | * The 3rd term is the interaction with the opposite $-q$ charge (at distance $\ell\sqrt{2}$). | ||
| + | |||
| + | Substituting the expressions for Coulomb's law: | ||
| + | $$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\ell/\sqrt{2})^2} - 2 \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\ell\sqrt{2})^2} \quad (2)$$ | ||
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| + | $$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( 2 - \sqrt{2} - \frac{1}{2} \right) = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( \frac{3 - 2\sqrt{2}}{2} \right) \quad (3)$$ | ||
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| + | <i>Note: Since $3 > 2\sqrt{2}$, we have $F > 0$. The net force is indeed directed towards the center, which makes circular motion possible.</i> | ||
| + | |||
| + | According to Newton's second law for rotational motion (where $a_{\text{cp}}$ is the centripetal acceleration): | ||
| + | $$\omega^2 r = a_{\text{cp}} = \frac{F}{m} \quad (4)$$ | ||
| + | |||
| + | From this, we express the angular velocity $\omega$: | ||
| + | $$\omega = \sqrt{\frac{F}{mr}} = \sqrt{ \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( \frac{3 - 2\sqrt{2}}{2} \right) \cdot \frac{1}{m \ell / \sqrt{2}} } \quad (5)$$ | ||
| + | |||
| + | After mathematical simplifications (multiplying by $\sqrt{2}$), we obtain the final expression: | ||
| + | $$\omega = \sqrt{ \frac{3\sqrt{2} - 4}{2} \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{m\ell^3} } \quad (6)$$ | ||
| + | |||
| + | #### Answer | ||
| + | $$\omega = \sqrt{ \frac{3\sqrt{2} - 4}{2} \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{m\ell^3} }$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.1.13.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.1.13.$ Four identical particles, each of mass $m$ and charge $-q$, revolve in a circular orbit, located at the corners of a square of side $\ell$. A charge $q$ is located at the center of this square. Determine the angular velocity of the particles in orbit. | |||
| ### Solution | |||
| <div align="center"> | |||
| <img src="https://savchenkosolutions.com/6/6.1.13/drawing.png" alt="Forces acting on the charge"> | |||
| </div> | |||
| Let us find the magnitude of the net radial force $F$ acting on one of the $-q$ charges and directed towards the center of the orbit. | |||
| This force is the superposition of the attractive force from the central charge $q$ and the repulsive forces from the other three $-q$ charges. | |||
| The radius of the orbit (distance to the center) is $r = \frac{\ell}{\sqrt{2}}$. | |||
| Using Coulomb's law, we project all forces onto the radial axis: | |||
| $$F = F_{\text{center}} - 2F_{\text{adj}} \cdot \cos 45^\circ - F_{\text{opp}} \quad (1)$$ | |||
| Where: | |||
| * The 1st term is the interaction with the central charge $q$. | |||
| * The 2nd term is the interaction with the two adjacent $-q$ charges (at distance $\ell$). | |||
| * The 3rd term is the interaction with the opposite $-q$ charge (at distance $\ell\sqrt{2}$). | |||
| Substituting the expressions for Coulomb's law: | |||
| $$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\ell/\sqrt{2})^2} - 2 \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\ell\sqrt{2})^2} \quad (2)$$ | |||
| $$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( 2 - \sqrt{2} - \frac{1}{2} \right) = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( \frac{3 - 2\sqrt{2}}{2} \right) \quad (3)$$ | |||
| <i>Note: Since $3 > 2\sqrt{2}$, we have $F > 0$. The net force is indeed directed towards the center, which makes circular motion possible.</i> | |||
| According to Newton's second law for rotational motion (where $a_{\text{cp}}$ is the centripetal acceleration): | |||
| $$\omega^2 r = a_{\text{cp}} = \frac{F}{m} \quad (4)$$ | |||
| From this, we express the angular velocity $\omega$: | |||
| $$\omega = \sqrt{\frac{F}{mr}} = \sqrt{ \frac{1}{4\pi\varepsilon_0} \frac{q^2}{\ell^2} \left( \frac{3 - 2\sqrt{2}}{2} \right) \cdot \frac{1}{m \ell / \sqrt{2}} } \quad (5)$$ | |||
| After mathematical simplifications (multiplying by $\sqrt{2}$), we obtain the final expression: | |||
| $$\omega = \sqrt{ \frac{3\sqrt{2} - 4}{2} \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{m\ell^3} } \quad (6)$$ | |||
| #### Answer | |||
| $$\omega = \sqrt{ \frac{3\sqrt{2} - 4}{2} \cdot \frac{1}{4\pi\varepsilon_0} \frac{q^2}{m\ell^3} }$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||