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<!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>
@@ -1598,39 +1598,51 @@ Table of contents
</svg></a><span class="headline-id">ems.ca.mi.o</span></h5>
<div class="outline-text-5" id="text-ems_ca_mi_o">
<p>
We can also play around differently. Elaborating on the grounded conducting plane above:
if we just change \(d\), the problem stays of the same nature. Now if, however, we change
the value of the charge at \(z = -d\) from \(q\) to \(q'\): what do we get ? The potential is
\[
V(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d)^2]^{1/2}}
We can also play around differently.
Elaborating on the grounded conducting plane above:
if we just change \(d\), the problem stays of the same nature.
If, however, we change
the value of the charge at \(z = -d\) from \(q\) to \(q'\),
what do we get ?
</p>
<ul class="org-ul">
<li>\frac{q'}{[x^2 + y^2 + (z+d)^2]^{1/2}} \right]</li>
</ul>
<p>
The potential is
</p>
\begin{equation*}
\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]
\end{equation*}
<p>
\]
This vanishes when
</p>
\begin{align}
\begin{equation*}
\left(\frac{q}{q'}\right)^2 \frac{ x^2 + y^2 + z^2 + d^2 + 2dz}{x^2 + y^2 + z^2 + d^2 - 2dz} = 1,
\rightarrow
x^2 + y^2 + z^2 + d^2 + 2d \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1} z = 0
\nonumber \\
\rightarrow x^2 + y^2 + (z + d\alpha)^2 = d^2 (\alpha^2 - 1), \hspace{1cm} \alpha \equiv \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1}.
\end{align}
\end{equation*}
<p>
This only makes sense if \(\alpha &gt; 1\), {\it i.e.} \(q' &lt; q\), otherwise the radius is not positive definite.
Therefore, the equipotential \(V = 0\) is a sphere of radius \(R = d\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha\).
Choosing (without loss of generality) \(q' &lt; q\), we can write
</p>
\begin{equation*}
\rightarrow x^2 + y^2 + (z + d\alpha)^2 = d^2 (\alpha^2 - 1), \hspace{1cm} \alpha \equiv \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1}.
\end{equation*}
<p>
Therefore, the equipotential \(\phi = 0\) is a sphere of radius \(R = d\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha\).
The problem is therefore equivalent to a grounded metal sphere of that radius centered at this position,
with a single point charge \(q\) at \(d\hat{z}\) !
(Griffiths pulls that out of a hat, but you now see where it comes from). Correspondence with
Griffiths Ex 3.2: his \(a-b\) is my \(2d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = 2d\),
whose solution is \(a = d[1 + \alpha]\).
with a single point charge \(q\) at \(d~\hat{z}\).
</p>
<p>
<i>N.B. Correspondence with Griffiths Ex 3.2: his \(a-b\) is my \(2d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = 2d\),
whose solution is \(a = d[1 + \alpha]\).</i>
</p>
</div>
</div>
<br><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_ca_mi_fe.html">Force and Energy&emsp;<small>[ems.ca.mi.fe]</small></a></li><li>Next:&nbsp;<a href="ems_ca_sv.html">Separation of Variables&emsp;<small>[ems.ca.sv]</small></a></li><li>Up:&nbsp;<a href="ems_ca_mi.html">The Method of Images&emsp;<small>[ems.ca.mi]</small></a></li></ul>
<br>
<hr>
<div class="license">
<a rel="license noopener" href="https://creativecommons.org/licenses/by/4.0/"
@@ -1644,7 +1656,7 @@ 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>