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| <h2>Solutions of Savchenko Physics Textbook</h2> |
| <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> |
| <p class="author"> |
| Aliaksandr Melnichenka <br/> |
| Aliaksandr Melnichenka <br/> |
| October 2023 |
| October 2023 |
| </p> |
| </p> |
| </header> |
| </header> |
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| <main> |
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
| <article> |
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| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
| <h3> Statement </h3> |
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| <p> |
| <h4>Statement</h4> |
| $7.1.6$ |
| <p> |
| 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 $E$. |
| $7.1.6$ |
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| 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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| <figure> |
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| <img src="statement.png" |
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| loading="lazy" alt="7.1.6" width="300" /> |
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| <figcaption> |
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| For problem 7.1.6 |
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| </figcaption> |
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| </figure> |
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| </center> |
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| <h4>Solution</h4> |
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| <p class="TxtSolutions"> |
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| Initial kinetic energy is defined as |
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| </p> |
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| <center> |
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| $K_0 = \frac{mv_0^2}{2} \;(1)$ |
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| </center> |
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| <p class="TxtSolutions"> |
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| If we apply Newton's Second Law, acceleration can be defined, |
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| </p> |
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| <center> |
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| $a_y = \frac{qE}{m} \;(2)$ |
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| </center> |
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| <p class="TxtSolutions"> |
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| As there are no forces in horizontal direction, the component of velocity in that direction is unaffected |
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| </p> |
| </p> |
| <center> |
| <center> |
| $v_0\cos{\alpha}=v_1\cos{\beta} \;(3)$ |
| <figure> |
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| <img src="statement.png" |
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| loading="lazy" alt="7.1.6" width="300" /> |
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| <figcaption> |
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| For problem 7.1.6 |
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| </figcaption> |
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| </figure> |
| </center> |
| </center> |
| <p class="TxtSolutions"> |
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| <h3>Solution</h3> |
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| <p> |
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| Initial kinetic energy is defined as |
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| $$K_0 = \frac{mv_0^2}{2} \;(1)$$ |
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| If we apply Newton's Second Law, acceleration can be defined, |
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| $$a_y = \frac{qE}{m} \;(2)$$ |
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| As there are no forces in horizontal direction, the component of velocity in that direction is unaffected |
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| $$v_0\cos{\alpha}=v_1\cos{\beta} \;(3)$$ |
| For the horizontal path until exit's point |
| For the horizontal path until exit's point |
| </p> |
| $$l = v_0\cos{\alpha}~t_v \;(4)$$ |
| <center> |
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| $l = v_0\cos{\alpha}~t_v \;(4)$ |
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| </center> |
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| <p class="TxtSolutions"> |
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| 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 |
| </p> |
| $$\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$$ |
| <center> |
| $$a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$$ |
| $\vec{a_y} = \frac{\Delta \vec{v_y}}{\Delta t} = \frac{\vec{v_{1y}}-\vec{v_{0y}}}{t_v}$ |
| $$t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \;(5)$$ |
| </center> |
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| <center> |
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| $a_y = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{t_v}$ |
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| </center> |
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| <center> |
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| $t_v = \frac{v_1\sin{\beta}+v_0\sin{\alpha}}{a_y} \;(5)$ |
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| </center> |
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| <p clas="TxtSolutions"> |
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| Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically |
| Substituting $v_1$ from $(3)$ into $(5)$ and developing algebraically |
| </p> |
| $$t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \;(6)$$ |
| <center> |
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| $t_v = v_0\cos{\alpha}\frac{\tan{\alpha}+\tan{\beta}}{a_y} \;(6)$ |
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| </center> |
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| <p class="TxtSolutions"> |
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| Putting $(6)$ into $(4)$ and separating $v_0$ |
| Putting $(6)$ into $(4)$ and separating $v_0$ |
| </p> |
| $$v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$$ |
| <center> |
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| $v_0^2 = \frac{a_yl}{\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}$ |
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| </center> |
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| <p class="TxtSolutions"> |
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| and according to $(2)$, |
| and according to $(2)$, |
| </p> |
| $$v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \;(7)$$ |
| <center> |
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| $v_0^2 = \frac{qEl}{m\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})} \;(7)$ |
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| </center> |
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| <p class="=TxtSolutions"> |
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| Finally, substituting $(7)$ into $(1)$ |
| Finally, substituting $(7)$ into $(1)$ |
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| $$\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$$ |
| </p> |
| </p> |
| <center> |
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| $\boxed{K_0 = \frac{qEl}{2\cos^2{\alpha}(\tan{\alpha}+\tan{\beta})}}$ |
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| </center> |
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| <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> |
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| BSc. Luis Daniel Fernández Quintana<br> |
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| Physics Department (FCNE)<br> |
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| Universidad de Oriente, Cuba<br> |
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| </p> |
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| </article> |
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| BSc. Luis Daniel Fernández Quintana<br> |
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| Physics Department (FCNE)<br> |
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| Universidad de Oriente, Cuba<br> |
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