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Jean-Sébastien
2022-02-14 20:42:37 +01:00
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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1601,8 +1601,8 @@ Table of contents
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 \(q'\),
If we however change
the value of the charge at \(z = -d/2\) from \(q\) to \(q'\),
what do we get ?
</p>
@@ -1610,32 +1610,32 @@ what do we get ?
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]
\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]
\end{equation*}
<p>
This vanishes when
</p>
\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,
\left(\frac{q}{q'}\right)^2 \frac{ x^2 + y^2 + z^2 + (d/2)^2 + dz}{x^2 + y^2 + z^2 + (d/2)^2 - dz} = 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
x^2 + y^2 + z^2 + (d/2)^2 + d \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1} z = 0
\end{equation*}
<p>
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}.
\rightarrow x^2 + y^2 + (z + d\alpha/2)^2 = \frac{d^2}{4} (\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 \(z = -d\alpha\).
Therefore, the equipotential \(\phi = 0\) is a sphere of radius \(R = \frac{d}{2}\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha/2\).
The problem is therefore equivalent to a grounded metal sphere of that radius centered at this position,
with a single point charge \(q\) at \(d~\hat{z}\).
with a single point charge \(q\) at \(\frac{d}{2}~\hat{z}\).
</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>
<i>N.B. Correspondence with Griffiths Ex 3.2: his \(a-b\) is my \(d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = d\),
whose solution is \(a = \frac{d}{2}[1 + \alpha]\).</i>
</p>
</div>
</div>
@@ -1656,7 +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: 2022-02-13 Sun 21:20</p>
<p class="date">Created: 2022-02-14 Mon 20:35</p>
<p class="validation"></p>
</div>