Added Luis's English solution of 6.2.11

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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
+ </div>
+ <p class="author">
+ Solutions of Savchenko Problems in Physics <br>
+ <i><b>knowledge must be free</b></i>
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+ <h3 id="back-link"><a href="../../#">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $6.2.11$ Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is $\rho$, and the second plate -$\rho$. Find the maximum electric field intensity.
+ </p>
+
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="6.2.11" width="350" />
+ <figcaption>
+ For problem 6.2.11
+ </figcaption>
+ </figure>
+ </center>
+
+ <h3>Solution</h3>
+ <p>
+ Absolute value of total charge of each plate is $\rho S h$, where $S$ is the surface of plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is:
+ </p>
+ <p>
+ $$E = \frac{\rho h}{\varepsilon_0}$$
+ </p>
+ <p>
+ Then, field generated by one of these biggest faces is the half of $E$. Let's consider now a point inside of one of these infinite plates. The resultant field over this point is the vectorial sum of the fields generated by lower negative plate, a plate with thickness $x$ (where $x$ is the distance from boundary between plates and the point) and a upper plate with thickness $h-x$:
+ </p>
+ <p>
+ $$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$
+ </p>
+ <p>
+ $$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$
+ </p>
+ <p>
+ Hence, maximum field intensity is achieved when $x=0$, i.e., point is in boundary between plates.
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$E_{max} = \frac{\rho h}{\varepsilon_0}$$
+ </p>
+
+ <p 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>
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