Added Luis's English solution of 6.5.1

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+ <title>Savchenko Solutions</title>
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+ content="A website with solutions to physics problems from Savenko Textbook" />
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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>
+ $6.5.1$
+ a) The surface densities of charges in two planes are $\pm \sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act?
+ </p>
+ <p class="TxtSolutions">
+ b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes?
+ </p>
+ <p class="TxtSolutions">
+ c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.
+ </p>
+ <h4>Solution</h4>
+ <p class="TxtSolutions">
+ a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is
+ </p>
+ <center>
+ $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$
+ </center>
+ <p calss="TxtSolutions">
+ As $dq = \pm\sigma~dS$,
+ </p>
+ <center>
+ $dF = \frac{\sigma^2~dS}{2\varepsilon_0}$
+ </center>
+ <p class="TxtSolutions">
+ and moving $dS$ to another member of equation,
+ </p>
+ <center>
+ $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$
+ </center>
+ <p class="TxtSolutions">
+ b) Let's see following figure
+ </p>
+ <center>
+ <figure>
+ <img src="drawb.png"
+ loading="lazy" alt="6.5.1" width="300" />
+ <figcaption>
+ Electric field of plates
+ </figcaption>
+ </figure>
+ </center>
+ <p class="TxtSolutions">
+ According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is
+ </p>
+ <center>
+ $\vec{E} = \vec{E_1} + \vec{E_2}$
+ </center>
+ <center>
+ $E = \frac{\sigma}{\varepsilon_0}$
+ </center>
+ <center>
+ $\boxed{\sigma = \varepsilon_0 E} \;(1)$
+ </center>
+ <p class="TxtSolutions">
+ The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so,
+ </p>
+ <center>
+ $dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$
+ </center>
+ <p class=" TxtSolutions">
+ Hence,
+ </p>
+ <center>
+ $P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$
+ </center>
+ <p class="TxtSolutions">
+ Putting $(2)$ into $(1)$
+ </p>
+ <center>
+ $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$
+ </center>
+ <p class="TxtSolutions">
+ c) See following figure
+ </p>
+ <center>
+ <figure>
+ <img src="drawc.png"
+ loading="lazy" alt="6.5.1" width="300" />
+ <figcaption>
+ Electric field of plates
+ </figcaption>
+ </figure>
+ </center>
+ <p class="TxtSolutions">
+ Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so,
+ </p>
+ <center>
+ $\vec{E} = 2\vec{E_1} = 2\vec{E_2}$
+ </center>
+ <center>
+ $E = \frac{\sigma}{\varepsilon_0}$
+ </center>
+ <p class="TxtSolutions">
+ with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m
+ </p>
+ <center>
+ $\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$
+ </center>
+ <p class="TxtSolutions">
+ Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$
+ </p>
+ <center>
+ $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$
+ </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>
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