Update 2022-02-14 20:42
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@@ -1,7 +1,7 @@
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<!-- 2022-02-13 Sun 21:20 -->
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<!-- 2022-02-14 Mon 20:35 -->
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<meta charset="utf-8">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<title>Pre-Quantum Electrodynamics</title>
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@@ -1598,6 +1598,11 @@ Table of contents
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</svg></a><span class="headline-id">ems.ca.mi</span></h4>
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<div class="outline-text-4" id="text-ems_ca_mi">
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<ul class="org-ul altsecnrs">
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<li>Gr 3.2</li>
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</ul>
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<p>
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The fact that conductors are equipotentials means that we can fomally solve
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loads of electrostatic problems involving conductors of various shapes, by
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@@ -1606,22 +1611,24 @@ looking at combinations of point charges.
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<p>
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Let's consider the simplest electrostatic problem above a single point charge:
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a system of two point charges. For definiteness, we put a charge \(q\) at coordinate
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\(d ~\hat{z}\), and a charge \(-q\) at \(-d ~\hat{z}\).
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a system of two point charges \(\pm q\) separated by distance \(d\).
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For definiteness, we put a charge \(q\) at coordinate
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\(\frac{d}{2} ~\hat{z}\), and a charge \(-q\) at \(-\frac{d}{2} ~\hat{z}\).
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</p>
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<p>
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By superposition, we have that
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</p>
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<div class="eqlabel" id="org50ef5fe">
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<div class="eqlabel" id="org2ce446c">
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<p>
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<a id="p_dip_z"></a><a href="./ems_ca_mi.html#p_dip_z"><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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<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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<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"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="orga1a93a0">
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<div class="alteqlabels" id="orgb1af7bf">
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<ul class="org-ul">
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<li>FLS II (6.8)</li>
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<li>Gr (3.9)</li>
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</ul>
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@@ -1629,16 +1636,16 @@ By superposition, we have that
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</div>
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\begin{align}
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\phi(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d)^2]^{1/2}} - \frac{q}{[x^2 + y^2 + (z+d)^2]^{1/2}} \right]
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\phi(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d/2)^2]^{1/2}} - \frac{q}{[x^2 + y^2 + (z+d/2)^2]^{1/2}} \right]
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\tag{p_dip_z}\label{p_dip_z}
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\end{align}
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<p>
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Now it's obvious that \(\phi = 0\) when \(z = 0\). Therefore, in the region \(z > 0\), this problem
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is completely equivalent to a second problem: a point charge \(q\) at \(d ~\hat{z}\),
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is completely equivalent to a second problem: a point charge \(q\) at \(\frac{d}{2} ~\hat{z}\),
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and a grounded conductor on the whole plane \(z = 0\).
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Yet another equivalent
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problem in the region \(z < 0\) is that of a charge \(-q\) at \(-d ~\hat{z}\) with a grounded
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problem in the region \(z < 0\) is that of a charge \(-q\) at \(-\frac{d}{2} ~\hat{z}\) with a grounded
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conductor on the plane \(z = 0\).
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</p>
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@@ -1677,7 +1684,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
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</div>
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<div id="postamble" class="status">
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<p class="author">Author: Jean-Sébastien Caux</p>
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<p class="date">Created: 2022-02-13 Sun 21:20</p>
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<p class="date">Created: 2022-02-14 Mon 20:35</p>
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<p class="validation"></p>
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</div>
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