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Update 2022-03-22 10:53

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Jean-Sébastien
2022-03-22 10:53:22 +01:00
parent f8fdc5f8b4
commit 2d1d8d39a4
209 changed files with 1240 additions and 1226 deletions
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build/ems_es_ef_ccd.html
+13 -13
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-15 Tue 08:10 -->
<!-- 2022-03-22 Tue 10:52 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1619,14 +1619,14 @@ calculated from Coulomb's law using the superposition principle. Since each inf
volume element \(d\tau' = dx' dy' dz'\) contains a charge \(dq' = \rho({\bf r}') d\tau'\), we have
</p>
<div class="eqlabel" id="org3f167fe">
<div class="eqlabel" id="org0685ec2">
<p>
<a id="E_vcd"></a><a href="./ems_es_ef_ccd.html#E_vcd"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<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>
</p>
<div class="alteqlabels" id="orgea0c302">
<div class="alteqlabels" id="org2a6f970">
<ul class="org-ul">
<li>Gr4 (2.8)</li>
</ul>
@@ -1634,7 +1634,7 @@ volume element \(d\tau' = dx' dy' dz'\) contains a charge \(dq' = \rho({\bf r}')
</div>
</div>
<div class="main div" id="orgb8d314c">
<div class="main div" id="org6569d43">
<p>
</p>
@@ -1654,14 +1654,14 @@ Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}
\(\sigma({\bf r})\), we have over an infinitesimal area \(da'\) a charge \(dq' = \sigma({\bf r}') da'\), so
</p>
<div class="eqlabel" id="org88e8806">
<div class="eqlabel" id="org463d4a8">
<p>
<a id="E_scd"></a><a href="./ems_es_ef_ccd.html#E_scd"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<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>
</p>
<div class="alteqlabels" id="orgb53c96f">
<div class="alteqlabels" id="org1997b74">
<ul class="org-ul">
<li>Gr4(2.7)</li>
</ul>
@@ -1669,7 +1669,7 @@ Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}
</div>
</div>
<div class="main div" id="org9052c26">
<div class="main div" id="org0795fa2">
<p>
</p>
@@ -1686,14 +1686,14 @@ Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}
Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf r}')\),
</p>
<div class="eqlabel" id="org047ff2c">
<div class="eqlabel" id="org5ea882d">
<p>
<a id="E_lcd"></a><a href="./ems_es_ef_ccd.html#E_lcd"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<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>
</p>
<div class="alteqlabels" id="org285c4e1">
<div class="alteqlabels" id="org18ca9b1">
<ul class="org-ul">
<li>Gr (2.6)</li>
</ul>
@@ -1701,7 +1701,7 @@ Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf
</div>
</div>
<div class="main div" id="orgc32983b">
<div class="main div" id="org8e461ef">
<p>
</p>
@@ -1715,7 +1715,7 @@ Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf
</div>
<div class="example div" id="org8bade7b">
<div class="example div" id="orge5b772b">
<p>
<b>Example</b>
</p>
@@ -1749,7 +1749,7 @@ most easily by observing that \(\frac{d}{dx} \left( \frac{x}{\sqrt{z^2 + x^2}} \
= \frac{1}{\sqrt{z^2 + x^2}} - \frac{x^2}{(z^2 + x^2)^{3/2}} = \frac{z^2}{(z^2 + x^2)^{3/2}}\),
leading to
</p>
<aside id="org726fb53">
<aside id="org371c5b7">
<p>
You could alternately proceed by using changes of variables \(y = zx\) followed by \(y = \tanh \alpha\):
\(\int_{-L}^L \frac{dx}{(z^2 + x^2)^{3/2}} = \frac{1}{z^2}
@@ -1806,7 +1806,7 @@ 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: 2022-03-15 Tue 08:10</p>
<p class="date">Created: 2022-03-22 Tue 10:52</p>
<p class="validation"></p>
</div>