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Update 2022-02-14 06:33

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Jean-Sébastien
2022-02-14 06:33:37 +01:00
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-10 Thu 08:32 -->
<!-- 2022-02-13 Sun 21:20 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1607,31 +1607,49 @@ looking at combinations of point charges.
<p>
Let's consider the simplest electrostatic problem above a single point charge:
a system of two point charges. For definiteness, we put a charge \(q\) at coordinate
\(d \hat{z}\), and a charge \(-q\) at \(-d \hat{z}\). By superposition, we have that
\[
V(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d)^2]^{1/2}}
\(d ~\hat{z}\), and a charge \(-q\) at \(-d ~\hat{z}\).
</p>
<ul class="org-ul">
<li>\frac{q}{[x^2 + y^2 + (z+d)^2]^{1/2}} \right]</li>
</ul>
<p>
\label{Gr(3.9)}
\]
Now it's obvious that \(V = 0\) when \(z = 0\). Therefore, in the region \(z &gt; 0\), this problem
is completely equivalent to a second problem: a point charge \(q\) at \(d \hat{z}\),
{\it and a grounded conductor on the whole plane \(z = 0\)}. Yet another equivalent
problem in the region \(z &lt; 0\) is that of a charge \(-q\) at \(-d \hat{z}\) with a grounded
By superposition, we have that
</p>
<div class="eqlabel" id="org50ef5fe">
<p>
<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">
<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="orga1a93a0">
<ul class="org-ul">
<li>Gr (3.9)</li>
</ul>
</div>
</div>
\begin{align}
\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]
\tag{p_dip_z}\label{p_dip_z}
\end{align}
<p>
Now it's obvious that \(\phi = 0\) when \(z = 0\). Therefore, in the region \(z &gt; 0\), this problem
is completely equivalent to a second problem: a point charge \(q\) at \(d ~\hat{z}\),
and a grounded conductor on the whole plane \(z = 0\).
Yet another equivalent
problem in the region \(z &lt; 0\) is that of a charge \(-q\) at \(-d ~\hat{z}\) with a grounded
conductor on the plane \(z = 0\).
</p>
<p>
We can go one step further. In the two point charges problem (dipole), we could put
a conductor on {\it any} of the equipotential lines, and still have an explicit solution
a conductor on <i>any</i> of the equipotential lines, and still have an explicit solution
for the potential in terms of the potential from the original two point charges.
</p>
<p>
In the Image problem, we might want to know what the induced surface charge is.
Our next step in such image problems is to determine what the induced surface charge is.
</p>
</div>
@@ -1644,6 +1662,8 @@ In the Image problem, we might want to know what the induced surface charge is.
<li><a href="ems_ca_mi_o.html">Other Image Problems</a><span class="headline-id">ems.ca.mi.o</span></li>
</ul>
<br><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_ca_fe_uP.html">Uniqueness of Solution to Poisson's Equation&emsp;<small>[ems.ca.fe.uP]</small></a></li><li>Next:&nbsp;<a href="ems_ca_mi_isc.html">Induced Surface Charges&emsp;<small>[ems.ca.mi.isc]</small></a></li><li>Up:&nbsp;<a href="ems_ca.html">Calculating or Approximating the Electrostatic Potential&emsp;<small>[ems.ca]</small></a></li></ul>
<br>
<hr>
<div class="license">
<a rel="license noopener" href="https://creativecommons.org/licenses/by/4.0/"
@@ -1657,7 +1677,7 @@ 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: 2022-02-10 Thu 08:32</p>
<p class="date">Created: 2022-02-13 Sun 21:20</p>
<p class="validation"></p>
</div>