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Update 2022-02-14 20:42

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Jean-Sébastien
2022-02-14 20:42:37 +01:00
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commit 09a8ba5fb6
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build/ems_es_ep_ex.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1597,7 +1597,7 @@ Table of contents
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a><span class="headline-id">ems.es.ep.ex</span></h5>
<div class="outline-text-5" id="text-ems_es_ep_ex">
<div class="example div" id="org4356f0f">
<div class="example div" id="org3f10f04">
<p>
<b>Spherical shell: via \({\bf E}\)</b>
</p>
@@ -1645,7 +1645,7 @@ and using spherical symmetry, we get:
</div>
<div class="example div" id="orgebd5896">
<div class="example div" id="orgd2f7e0b">
<p>
<b>Spherical shell: direct calculation</b>
</p>
@@ -1664,7 +1664,7 @@ a direct integral, starting from <a href="./ems_es_ep_d.html#p_scd">p_scd</a>
<p>
Exploiting spherical symmetry, let us orient \({\bf r}\) along the \(\hat{\bf z}\) axis
so \({\bf r} = z \hat{\bf z}\). Our integration point \({\bf r}'\) is at position
\(R \sin \theta' \cos \phi' \hat{{\bf x}} + R \sin \theta' \sin \phi' \hat{\bf y} + R\cos \theta' \hat{\bf z}\) so
\(R \sin \theta' \cos \varphi' \hat{{\bf x}} + R \sin \theta' \sin \varphi' \hat{\bf y} + R\cos \theta' \hat{\bf z}\) so
</p>
\begin{equation*}
@@ -1674,11 +1674,11 @@ so \({\bf r} = z \hat{\bf z}\). Our integration point \({\bf r}'\) is at positio
<p>
Using the element of surface area (in spherical coordinates)
\(R^2 \sin \theta' d\theta' d\phi'\), so
\(R^2 \sin \theta' d\theta' d\varphi'\), so
</p>
\begin{align}
4\pi \varepsilon_0 \phi(z) &amp;= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\phi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
4\pi \varepsilon_0 \phi(z) &amp;= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\varphi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
&amp;= 2\pi R^2 \sigma \int_0^{\pi} d\theta' \frac{\sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
&amp;= 2\pi R^2 \sigma \left. \left( \frac{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}}{Rz} \right)\right|_0^{\pi} \nonumber \\
&amp;= \frac{2\pi R\sigma}{z} \left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz} \right) \nonumber \\
@@ -1722,7 +1722,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
</div>
<div id="postamble" class="status">
<p class="author">Author: Jean-Sébastien Caux</p>
<p class="date">Created: 2022-02-13 Sun 21:20</p>
<p class="date">Created: 2022-02-14 Mon 20:35</p>
<p class="validation"></p>
</div>