<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
−
<meta name="description" content="">
+
<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 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.">
<meta property="og:image" content="img/logo.png">
−
<meta property="og:description" content="">
+
<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.">
<title>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.</title>
$6.2.11$ 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>
−
−
<center>
−
<figure>
−
<img src="statement.png"
−
loading="lazy" alt="6.2.11" width="350" />
−
<figcaption>
−
For problem 6.2.11
−
</figcaption>
−
</figure>
−
</center>
−
+
$6.2.11.$ 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>
+
<center>
+
<figure>
+
<img src="statement.png"
+
loading="lazy" width="350" />
+
<figcaption>
+
For problem $6.2.11$
+
</figcaption>
+
</figure>
+
</center>
<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:
+
The absolute value of the total charge of each plate is $\rho S h$, where $S$ is the surface of the plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is:
</p>
<p>
$$E = \frac{\rho h}{\varepsilon_0}$$
</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 (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 $h-x$:
+
Then, the field generated by one of these biggest faces of the plate is half of $E$. Let's consider now a point inside of one of these infinite plates (for example, in the upper one). The resultant field over this point is the vectorial sum of the fields generated by a lower negative plate, a plate with thickness $x$ (where $x$ is the distance from the boundary between plates and the point), and a upper plate with thickness $h-x$:
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="">
<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 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.">
<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
<meta property="og:description" content="">
<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.">
<title>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.</title>
$6.2.11$ 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" 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:
The absolute value of the total charge of each plate is $\rho S h$, where $S$ is the surface of the plate's biggest faces. So, applying Gauss Law, the absolute value of electric field intensity generated by this plate is:
</p>
</p>
<p>
<p>
$$E = \frac{\rho h}{\varepsilon_0}$$
$$E = \frac{\rho h}{\varepsilon_0}$$
</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 (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 $h-x$:
Then, the field generated by one of these biggest faces of the plate is half of $E$. Let's consider now a point inside of one of these infinite plates (for example, in the upper one). The resultant field over this point is the vectorial sum of the fields generated by a lower negative plate, a plate with thickness $x$ (where $x$ is the distance from the boundary between plates and the point), and a upper plate with thickness $h-x$:
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
