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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| $7.1.6$ | | $7.1.6$ |
| Particles with mass $m$ and charge $q$ enter in a capacitor of length $l$ under an angle $\alpha$ respect to plate's plane and exit forming an angle $\beta$. Find the initial kinetic energy of particles, if the field intensity inside the capacitor is $E$. | | Particles with mass $m$ and charge $q$ enter in a capacitor of length $l$ under an angle $\alpha$ respect to plate's plane and exit forming an angle $\beta$. Find the initial kinetic energy of particles, if the field intensity inside the capacitor is $E$. |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="7.1.6" width="300" /> | | loading="lazy" alt="7.1.6" width="300" /> |
| <figcaption> | | <figcaption> |
| For problem 7.1.6 | | For problem 7.1.6 |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Initial kinetic energy is defined as | | Initial kinetic energy is defined as |
| $$K_0 = \frac{mv_0^2}{2} \;(1)$$ | | $$K_0 = \frac{mv_0^2}{2} \;(1)$$ |
| If we apply Newton's Second Law, acceleration can be defined, | | If we apply Newton's Second Law, acceleration can be defined, |
| $$a_y = \frac{qE}{m} \;(2)$$ | | $$a_y = \frac{qE}{m} \;(2)$$ |
| As there are no forces in horizontal direction, the component of velocity in that direction is unaffected | | As there are no forces in horizontal direction, the component of velocity in that direction is unaffected |
| $$v_0\cos{\alpha}=v_1\cos{\beta} \;(3)$$ | | $$v_0\cos{\alpha}=v_1\cos{\beta} \;(3)$$ |
| For the horizontal path until exit's point | | For the horizontal path until exit's point |
| $$l = v_0\cos{\alpha}~t_v \;(4)$$ | | $$l = v_0\cos{\alpha}~t_v \;(4)$$ |
| The time expended by particles during the path through the capacitor can be found from acceleration definition | | The time expended by particles during the path through the capacitor can be found from acceleration definition |
| $$\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$$ | | $$\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$$ |
| $$a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$$ | | $$a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$$ |
| $$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \;(5)$$ | | $$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \;(5)$$ |
| Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically | | Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically |
| $$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \;(6)$$ | | $$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \;(6)$$ |
| Putting $(6)$ into $(4)$ and separating $v_0$ | | Putting $(6)$ into $(4)$ and separating $v_0$ |
| $$v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$$ | | $$v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$$ |
| and according to $(2)$, | | and according to $(2)$, |
| $$v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \;(7)$$ | | $$v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \;(7)$$ |
| Finally, substituting $(7)$ into $(1)$ | | Finally, substituting $(7)$ into $(1)$ |
| $$\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$$ | | $$\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$$ |
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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;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
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