Added Luis's English solution of 7.1.6

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