Added Luis's English solution of 6.2.10

Luisito правка от 27 сент. 2024 г., 17:18

правка #8857 GitHub 16a128a позже →

en/6.2.10.md +121 −0

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		+	<header style="text-align:center;">
		+	<div id = "logo">
		+	<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>
		+	</p>
		+	</header>
		+
		+	<h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$?
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="statement.png"
		+	loading="lazy" width="350" />
		+	<figcaption>
		+	For problem $6.2.10$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<h3>Solution</h3>
		+	<p>
		+	Consider the following figure...
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="figure.png"
		+	loading="lazy" width="350" />
		+	<figcaption>
		+	Field Analysis
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$.
		+	</p>
		+	<p>
		+	For region 1 (see part $a$ of above figure):
		+	</p>
		+	<p>
		+	$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
		+	</p>
		+	<p>
		+	and for $x$-direction, $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. So, as $$E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2}$$, and taking in account that $$\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}}$$, it is obtained
		+	</p>
		+	<h4>Answer 1</h4>
		+	<p>
		+	$$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$$
		+	</p>
		+	<p>
		+	For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$.
		+	</p>
		+	<p>
		+	Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
		+	</p>
		+
		+	<h4>Answer 2</h4>
		+	<p>
		+	$$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$
		+	</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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			<meta name="description" content="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.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="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.">
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			<meta property="og:description" content="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.">
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			</head>
			<body style="">
			<header style="text-align:center;">
			<div id = "logo">
			<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>
			</p>
			</header>

			<h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$?
			</p>
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" width="350" />
			<figcaption>
			For problem $6.2.10$
			</figcaption>
			</figure>
			</center>
			<h3>Solution</h3>
			<p>
			Consider the following figure...
			</p>
			<center>
			<figure>
			<img src="figure.png"
			loading="lazy" width="350" />
			<figcaption>
			Field Analysis
			</figcaption>
			</figure>
			</center>
			<p>
			It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$.
			</p>
			<p>
			For region 1 (see part $a$ of above figure):
			</p>
			<p>
			$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
			</p>
			<p>
			and for $x$-direction, $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. So, as $$E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2}$$, and taking in account that $$\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}}$$, it is obtained
			</p>
			<h4>Answer 1</h4>
			<p>
			$$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$$
			</p>
			<p>
			For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$.
			</p>
			<p>
			Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
			</p>

			<h4>Answer 2</h4>
			<p>
			$$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$
			</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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