Update 2022-02-08 17:21

This commit is contained in:
Jean-Sébastien
2022-02-08 17:21:33 +01:00
parent 077433c40a
commit 3454aba504
207 changed files with 1882 additions and 1097 deletions
+18 -14
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
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<head>
<!-- 2022-02-08 Tue 06:55 -->
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<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -272,6 +272,10 @@ Table of contents
</summary>
<ul>
<li>
<a href="./in_t_l.html#in_t_l">Section and equation labelling</a><span class="headline-id">in.t.l</span>
</li>
<li>
<a href="./in_t_c.html#in_t_c">Contextual colors</a><span class="headline-id">in.t.c</span>
</li>
@@ -736,7 +740,7 @@ Table of contents
</li>
<li>
<a href="./emsm_esm_d.html#emsm_esm_d">Dielectrics</a><span class="headline-id">emsm.esm.d</span>
<a href="./emsm_esm_di.html#emsm_esm_di">Dielectrics</a><span class="headline-id">emsm.esm.di</span>
</li>
<li>
@@ -1624,14 +1628,14 @@ 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="org93fd4f2">
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<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">
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</svg></a>
</p>
<div class="alteqlabels" id="org2fcc353">
<div class="alteqlabels" id="orgaddc7cb">
<ul class="org-ul">
<li>Gr4 (2.8)</li>
</ul>
@@ -1639,7 +1643,7 @@ volume element \(d\tau' = dx' dy' dz'\) contains a charge \(dq' = \rho({\bf r}')
</div>
</div>
<div class="main div" id="orgf2f415b">
<div class="main div" id="org81b4885">
<p>
</p>
@@ -1659,14 +1663,14 @@ 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="org6008079">
<div class="eqlabel" id="orgd0ad464">
<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="orgc6962f2">
<div class="alteqlabels" id="orgfba2df1">
<ul class="org-ul">
<li>Gr4(2.7)</li>
</ul>
@@ -1674,7 +1678,7 @@ Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}
</div>
</div>
<div class="main div" id="org433e39b">
<div class="main div" id="org743c5ef">
<p>
</p>
@@ -1691,14 +1695,14 @@ 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>
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<div class="eqlabel" id="orgadea8dd">
<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"/>
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</svg></a>
</p>
<div class="alteqlabels" id="orgf23e73b">
<div class="alteqlabels" id="orgcdc346f">
<ul class="org-ul">
<li>Gr (2.6)</li>
</ul>
@@ -1706,7 +1710,7 @@ Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf
</div>
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<div class="main div" id="orgd91bb3b">
<div class="main div" id="org2974477">
<p>
</p>
@@ -1720,7 +1724,7 @@ Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf
</div>
<div class="example div" id="orgac43438">
<div class="example div" id="org2edb900">
<p>
<b>Example</b>
</p>
@@ -1754,7 +1758,7 @@ 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="orgdec9553">
<aside id="orge50da61">
<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}
@@ -1798,7 +1802,7 @@ whereas for short distances \(z \ll L\) the field looks like that of an infinite
<hr><div id="postamble" class="status">
<p class="author">Author: Jean-Sébastien Caux</p>
<p class="date">Created: 2022-02-08 Tue 06:55</p>
<p class="date">Created: 2022-02-08 Tue 17:21</p>
<p class="validation"><a href="https://validator.w3.org/check?uri=referer">Validate</a></p>
</div>