<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<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.">
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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.">
<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>
$6.2.10.$ 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>
@@ -53,7 +53,7 @@
<center>
<figure>
<img src="statement.png"
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loading="lazy" width="350" />
+
loading="lazy" width="250" />
<figcaption>
For problem $6.2.10$
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
Consider the following figure...
</p>
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<center>
<figure>
<img src="figure.png"
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loading="lazy" width="350" />
+
loading="lazy" width="300" />
<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>
@@ -82,14 +82,14 @@
$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
</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
+
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
For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$.
+
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 $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
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<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="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="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: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">
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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.">
<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>
<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>
$6.2.10.$ 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 $\pm\sigma$?
$6.2.10.$ 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>
</p>
@@ -53,7 +53,7 @@
<center>
<center>
<figure>
<figure>
<img src="statement.png"
<img src="statement.png"
loading="lazy" width="350" />
loading="lazy" width="250" />
<figcaption>
<figcaption>
For problem $6.2.10$
For problem $6.2.10$
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
Consider the following figure...
Consider the following figure...
</p>
</p>
@@ -66,7 +66,7 @@
<center>
<center>
<figure>
<figure>
<img src="figure.png"
<img src="figure.png"
loading="lazy" width="350" />
loading="lazy" width="300" />
<figcaption>
<figcaption>
Field Analysis
Field Analysis
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$.
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>
<p>
<p>
For region 1 (see part $a$ of above figure):
For region 1 (see part $a$ of above figure):
</p>
</p>
<p>
<p>
@@ -82,14 +82,14 @@
$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
$$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$
</p>
</p>
<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
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
For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$.
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>
<p>
<p>
Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$
<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>
