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<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>
− $6.5.2$
− Two electroconductor pistons of cross section $S$ form a plane capacitor in a tube of an isolating material which contains air at atmospheric pressure $P_0$. How changes the distance between pistons if charges of different signs $\pm Q$ are applied to them? The system is good heat conductor and there is no friction.
− </p>
− <center>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $6.5.2$
+ Two electroconductor pistons of cross section $S$ form a plane capacitor in a tube of an isolating material which contains air at atmospheric pressure $P_0$. How changes the distance between pistons if charges of different signs $\pm Q$ are applied to them? The system is good heat conductor and there is no friction.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="6.5.2" width="200" />
+ <figcaption>
+ For problem 6.5.2
+ </figcaption>
+ </figure>
+ </center>
+
+ <h3>Solution</h3>
+ <p>
+ Let's consider the following figure
+ </p>
+ <center>
<figure>
− <img src="statement.png"
− loading="lazy" alt="6.5.2" width="200" />
− <figcaption>
− For problem 6.5.2
− </figcaption>
− </figure>
− </center>
− <h4>Solution</h4>
− <p class="TxtSolutions">
− Let's consider the following figure
− </p>
− <center>
− <figure>
<img src="draw.png"
loading="lazy" alt="6.5.2" width="450" />
<figcaption>
Forces over pistons
</figcaption>
</figure>
− </center>
− <p class="TxtSolutions">
− Applying Newton Second Law over one of the piston, considering the pistons are massless,
− </p>
− <center>
− $F_p = F_{p0} + F_e$
− </center>
− <p class="TxtSolutions">
− where $F_p$, $F_{p0}$ and $F_e$ are forces due to enclosed air between pistons, due to atmospheric pressure and due to electric field, respectively.
− </p>
− <center>
− $P = P_0 + \frac{F_e}{S} \;(1)$
− </center>
− <p class="TxtSolutions">
− The electric force $F_e$ is defined from the electric field generated by a charged plate that is $E = \frac{\sigma}{2\varepsilon_0}=\frac{Q}{2\varepsilon_0 S}$, so
− </p>
− <center>
− $F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$
− </center>
− <p class="TxtSolutions">
− Putting $(2)$ into $(1)$
− </p>
− <center>
− $P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$
− </center>
− <p class="TxtSolutions">
− Taking account that the system is good heat conductor, we can suppose that thermal equilibrium is instantaneously established or that the container acts as a thermos, keeping the temperature unchanged.Then, applying Boyle-Mariotte law
− </p>
− <center>
− $P_0 S L = P S l$
− </center>
− <center>
− $\frac{L}{l} = \frac{P}{P_0} \;(4)$
− </center>
− <p class="TxtSolutions">
− Substituting $(3)$ into $(4)$
− </p>
− <center>
− $\boxed{\frac{L}{l} = 1 + \frac{Q^2}{2P_0\varepsilon_0 S^2}}$
</center>
− <p class="TxtSolutions" style="text-align: center; font-style: italic; font-size: 14;">
− BSc. Luis Daniel Fernández Quintana<br>
− Physics Department (FCNE)<br>
− Universidad de Oriente, Cuba<br>
+ <p>
+ Applying Newton Second Law over one of the piston, considering the pistons are massless,
+ $$F_p = F_{p0} + F_e$$
+ where $F_p$, $F_{p0}$ and $F_e$ are forces due to enclosed air between pistons, due to atmospheric pressure and due to electric field, respectively.
+ $$P = P_0 + \frac{F_e}{S} \;(1)$$
+ The electric force $F_e$ is defined from the electric field generated by a charged plate that is $E = \frac{\sigma}{2\varepsilon_0}=\frac{Q}{2\varepsilon_0 S}$, so
+ $$F_e = Q E =\frac{Q^2}{2\varepsilon_0 S} \;(2)$$
+ Putting $(2)$ into $(1)$
+ $$P = P_0 + \frac{Q^2}{2\varepsilon_0 S} \;(3)$$
+ Taking account that the system is good heat conductor, we can suppose that thermal equilibrium is instantaneously established or that the container acts as a thermos, keeping the temperature unchanged.Then, applying Boyle-Mariotte law
+ $$P_0 S L = P S l$$
+ $$\frac{L}{l} = \frac{P}{P_0} \;(4)$$
+ Substituting $(3)$ into $(4)$
+ $$\boxed{\frac{L}{l} = 1 + \frac{Q^2}{2P_0\varepsilon_0 S^2}}$$
</p>
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+ <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>
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