Added Luis's English solution of 6.5.2

Luisito правка от 21 июн. 2024 г., 02:57

правка #1478 GitHub 41c1d9a позже →

en/6.5.2.md +137 −0

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		+
		+	<!DOCTYPE html>
		+	<html lang="en">
		+
		+	<head>
		+	<meta charset="UTF-8" />
		+	<meta name="viewport" content="width=device-width, initial-scale=1.0" />
		+	<meta http-equiv="content-language" content="en">
		+	<script async src="https://www.googletagmanager.com/gtag/js?id=G-DDMB38YMLD"></script>
		+
		+	<title>Savchenko Solutions</title>
		+	<meta property="og:image" content="img/logo.png">
		+	<meta property="og:title" content="Savchenko Solutions"/>
		+	<meta name="description"
		+	content="A website with solutions to physics problems from Savenko Textbook" />
		+	<meta name="keywords" content="physics, olympiads, ipho, savchenko" />
		+	<link rel="stylesheet" href="https://savchenko-physics.github.io/css/css-latex/style.css" />
		+	<link rel="icon" type="image/x-icon" href="https://savchenko-physics.github.io/img/favicon.ico">
		+	</head>
		+
		+	<body id="top">
		+	<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>
		+	$6.5.2$
		+	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>
		+	<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>
		+	</article>
		+	</main>
		+
		+	<script>
		+	MathJax = {
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			<script async src="https://www.googletagmanager.com/gtag/js?id=G-DDMB38YMLD"></script>

			<title>Savchenko Solutions</title>
			<meta property="og:image" content="img/logo.png">
			<meta property="og:title" content="Savchenko Solutions"/>
			<meta name="description"
			content="A website with solutions to physics problems from Savenko Textbook" />
			<meta name="keywords" content="physics, olympiads, ipho, savchenko" />
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			</head>

			<body id="top">
			<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>
			$6.5.2$
			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>
			<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>
			</article>
			</main>

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