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  <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>