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			<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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			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>

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			loading="lazy" alt="6.2.11" width="350" />
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			For problem 6.2.11
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			<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>
		@@ -69,7 +69,7 @@
			$$E = \frac{\rho h}{\varepsilon_0}$$
			</p>
			<p>
		−	Then, field generated by one of these biggest faces of the plate 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 $h-x$:
		+	Then, field generated by one of these biggest faces of the plate is the half of $E$. Let's consider now a point inside of one of these infinite plates (for example in upper one). 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., the point is in the 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>
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	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
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	<h3 id="back-link"><a href="../../#">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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.		$6.2.11$ 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>		</p>

	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" alt="6.2.11" width="350" />		loading="lazy" alt="6.2.11" width="350" />
	<figcaption>		<figcaption>
	For problem 6.2.11		For problem 6.2.11
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<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:		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>
	<p>		<p>
	@@ -69,7 +69,7 @@
	$$E = \frac{\rho h}{\varepsilon_0}$$		$$E = \frac{\rho h}{\varepsilon_0}$$
	</p>		</p>
	<p>		<p>
	Then, field generated by one of these biggest faces of the plate 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 $h-x$:		Then, field generated by one of these biggest faces of the plate is the half of $E$. Let's consider now a point inside of one of these infinite plates (for example in upper one). 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>
	<p>		<p>
	$$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$		$$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$
	</p>		</p>
	<p>		<p>
	$$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$		$$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$
	</p>		</p>
	<p>		<p>
	Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates.		Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates.
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$E_{max} = \frac{\rho h}{\varepsilon_0}$$		$$E_{max} = \frac{\rho h}{\varepsilon_0}$$
	</p>		</p>

	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	BSc. Luis Daniel Fernández Quintana<br>		BSc. Luis Daniel Fernández Quintana<br>
	Physics Department (FCNE)<br>		Physics Department (FCNE)<br>
	Universidad de Oriente, Cuba<br>		Universidad de Oriente, Cuba<br>
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