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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-09 Wed 07:31 -->
<!-- 2022-02-09 Wed 22:40 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -408,17 +408,13 @@ Table of contents
<li>
<a href="./ems_es_ep_fp.html#ems_es_ep_fp">Field in terms of the potential</a><span class="headline-id">ems.es.ep.fp</span>
</li>
<li>
<a href="./ems_es_ep_c.html#ems_es_ep_c">Comments on the Electrostatic Potential</a><span class="headline-id">ems.es.ep.c</span>
</li>
<li class="toc-currentpage">
<a href="./ems_es_ep_ex.html#ems_es_ep_ex">Example calculations for the potential</a><span class="headline-id">ems.es.ep.ex</span>
</li>
<li>
<a href="./ems_es_ep_PL.html#ems_es_ep_PL">The Poisson Equation and the Laplace Equation</a><span class="headline-id">ems.es.ep.PL</span>
<a href="./ems_es_ep_PL.html#ems_es_ep_PL">Poisson's and Laplace's Equations</a><span class="headline-id">ems.es.ep.PL</span>
</li>
<li>
@@ -430,29 +426,8 @@ Table of contents
</details>
</li>
<li>
<details>
<summary>
<a href="./ems_es_e.html#ems_es_e">Electrostatic Energy from the Potential</a><span class="headline-id">ems.es.e</span>
</summary>
<ul>
<li>
<a href="./ems_es_e_pcd.html#ems_es_e_pcd">The Energy of a Point Charge Distribution</a><span class="headline-id">ems.es.e.pcd</span>
</li>
<li>
<a href="./ems_es_e_ccd.html#ems_es_e_ccd">The Energy of a Continuous Charge Distribution</a><span class="headline-id">ems.es.e.ccd</span>
</li>
<li>
<a href="./ems_es_e_c.html#ems_es_e_c">Comments on Electrostatic Energy</a><span class="headline-id">ems.es.e.c</span>
</li>
</ul>
</details>
</li>
<li>
@@ -1616,66 +1591,113 @@ Table of contents
</ul>
</details>
</nav>
<ul class="breadcrumbs"><li><a class="breadcrumb-link"href="ems.html">Electromagnetostatics</a></li><li><a class="breadcrumb-link"href="ems_es.html">Electrostatics</a></li><li><a class="breadcrumb-link"href="ems_es_ep.html">The Electrostatic Potential</a></li><li>Example calculations for the potential</li></ul><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_es_ep_c.html">Comments on the Electrostatic Potential&emsp;<small>[ems.es.ep.c]</small></a></li><li>Next:&nbsp;<a href="ems_es_ep_PL.html">The Poisson Equation and the Laplace Equation&emsp;<small>[ems.es.ep.PL]</small></a></li><li>Up:&nbsp;<a href="ems_es_ep.html">The Electrostatic Potential&emsp;<small>[ems.es.ep]</small></a></li></ul><div id="outline-container-ems_es_ep_ex" class="outline-5">
<ul class="breadcrumbs"><li><a class="breadcrumb-link"href="ems.html">Electromagnetostatics</a></li><li><a class="breadcrumb-link"href="ems_es.html">Electrostatics</a></li><li><a class="breadcrumb-link"href="ems_es_ep.html">The Electrostatic Potential</a></li><li>Example calculations for the potential</li></ul><ul class="navigation-links"><li>Prev:&nbsp;<a href="ems_es_ep_fp.html">Field in terms of the potential&emsp;<small>[ems.es.ep.fp]</small></a></li><li>Next:&nbsp;<a href="ems_es_ep_PL.html">Poisson's and Laplace's Equations&emsp;<small>[ems.es.ep.PL]</small></a></li><li>Up:&nbsp;<a href="ems_es_ep.html">The Electrostatic Potential&emsp;<small>[ems.es.ep]</small></a></li></ul><div id="outline-container-ems_es_ep_ex" class="outline-5">
<h5 id="ems_es_ep_ex">Example calculations for the potential<a class="headline-permalink" href="./ems_es_ep_ex.html#ems_es_ep_ex"><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><span class="headline-id">ems.es.ep.ex</span></h5>
<div class="outline-text-5" id="text-ems_es_ep_ex">
<div class="example div" id="org32cb7e0">
<div class="example div" id="org98f6377">
<p>
<b>Spherical shell: via \({\bf E}\)</b>
</p>
<p>
Consider a spherical shell of radius \(R\) with uniform surface charge density
and total charge \(q\) (so \(\sigma = q/4\pi\)). The shell is centered at the origin.
</p>
<p>
We want to know the potential throughout space. Starting from the field
as obtained from applying Gauss's law <a href="./ems_es_ef_Gl.html#Gl_i">Gl_i</a>,
</p>
<p>
<b>Example 2.6</b>: find the potential inside and outside a spherical shell or radius \(R\)
centered at origin, which carries uniform surface charge. Ref point at infinity.
<b>Solution</b>: from Gauss's law,
\[
{\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}, \hspace{5mm} r &gt; R, \hspace{2cm}
{\bf E} = 0, \hspace{5mm} r &lt; R.
\]
Thus: using spherical symmetry, for \(r &gt; R\):
</p>
<p>
Adopting our usual convention of putting the reference point at infinity,
and using spherical symmetry, we get:
</p>
<ul class="org-ul">
<li>for \(r &gt; R\):</li>
</ul>
<p>
\[
V(r &gt; R) = -\int_{{\cal O}: \infty}^{\bf r} {\bf E} \cdot d{\bf l} = -\frac{1}{4\pi \varepsilon_0} \int_{\infty}^r q \frac{dr'}{r'^2}
= \frac{1}{4\pi \varepsilon_0} \frac{q}{r}, \hspace{5mm} r &gt; R.
\phi(r &gt; R) = -\int_{{\cal O}: \infty}^{\bf r} {\bf E} \cdot d{\bf l} = -\frac{1}{4\pi \varepsilon_0} \int_{\infty}^r q \frac{dr'}{r'^2}
= \frac{1}{4\pi \varepsilon_0} \frac{q}{r},
\]
For \(r &lt; R\), the integrand vanishes for the part \(r &lt; R\), and
</p>
<ul class="org-ul">
<li>for \(r &lt; R\): the integrand vanishes for the part \(r &lt; R\), and</li>
</ul>
<p>
\[
V(r &lt; R) = V(R) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}.
\phi(r &lt; R) = \phi(R) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}.
\]
</p>
</div>
<div class="example div" id="org126b36e">
<div class="example div" id="org36fa2ff">
<p>
<b>Spherical shell: direct calculation</b>
</p>
<p>
We can also compute the potential for the uniformly-charged shell using
a direct integral, starting from <a href="./ems_es_ep_d.html#p_scd">p_scd</a>
</p>
<p>
<b>Example 2.7</b>: find the potential of a uniformly charged spherical shell of radius \(R\) (same problem as 2.6, but now done with \ref{Gr(2.30)}).
<b>Solution</b>: Use
\[
V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int da' \frac{\sigma}{|{\bf r} - {\bf r}'|}.
\phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int da' \frac{\sigma}{|{\bf r} - {\bf r}'|}.
\]
Put \({\bf r}\) on \(\hat{\bf z}\) axis so \({\bf r} = z \hat{\bf z}\), use law of cosines:
</p>
\begin{equation}
|{\bf r} - {\bf r}'| = R^2 + z^2 - 2Rz \cos \theta'
\end{equation}
<p>
Element of surface area: \(R^2 \sin \theta' d\theta' d\phi'\), so
Exploiting spherical symmetry, let us orient \({\bf r}\) along the \(\hat{\bf z}\) axis
so \({\bf r} = z \hat{\bf z}\). Our integration point \({\bf r}'\) is at position
\(R \sin \theta' \cos \phi' \hat{{\bf x}} + R \sin \theta' \sin \phi' \hat{\bf y} + R\cos \theta' \hat{\bf z}\) so
</p>
\begin{equation*}
|{\bf r} - {\bf r}'|^2 = R^2 \sin^2 \theta' + (R \cos \theta' - z)^2
= R^2 + z^2 - 2Rz \cos \theta'
\end{equation*}
<p>
Using the element of surface area (in spherical coordinates)
\(R^2 \sin \theta' d\theta' d\phi'\), so
</p>
\begin{align}
4\pi \varepsilon_0 V(z) &amp;= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\phi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
4\pi \varepsilon_0 \phi(z) &amp;= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\phi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
&amp;= 2\pi R^2 \sigma \int_0^{\pi} d\theta' \frac{\sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\
&amp;= 2\pi R^2 \sigma \left. \left( \frac{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}}{Rz} \right)\right|_0^{\pi} \nonumber \\
&amp;= \frac{2\pi R\sigma}{z} \left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz} \right) \nonumber \\
&amp;= \frac{2\pi R \sigma}{z} \left( \sqrt{(R + z)^2} - \sqrt{(R - z)}^2 \right).
&amp;= \frac{2\pi R \sigma}{z} \left( \sqrt{(R + z)^2} - \sqrt{(R - z)^2} \right).
\end{align}
<p>
Take positive root:
We must obviously choose the positive root \(\sqrt{(R + z)^2} = R+z\) and thus
</p>
\begin{align}
V(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (z-R)) = \frac{R^2 \sigma}{\varepsilon_0 z}, \hspace{1cm} \mbox{outside}, \nonumber \\
V(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (R-z)) = \frac{R \sigma}{\varepsilon_0}, \hspace{1cm} \mbox{inside}.
\phi(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (z-R)) = \frac{R^2 \sigma}{\varepsilon_0 z}, \hspace{1cm} \mbox{outside}, \nonumber \\
\phi(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (R-z)) = \frac{R \sigma}{\varepsilon_0}, \hspace{1cm} \mbox{inside}.
\end{align}
<p>
In terms of total charge on shell, \(q = 4\pi R^2 \sigma\), \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{z}\) outside,
and \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\) inside.
In terms of total charge on the shell, \(q = 4\pi R^2 \sigma\),
we thus obtain
\(\phi(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{z}\) outside,
and \(\phi(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\) inside which indeed
correspond to our previous solution via \({\bf E}\).
</p>
</div>
@@ -1698,7 +1720,7 @@ 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: 2022-02-09 Wed 07:31</p>
<p class="date">Created: 2022-02-09 Wed 22:40</p>
<p class="validation"></p>
</div>