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\newpage
\section{Electrostatics}

In a time-independent situation, Maxwell's equations reduce to
\begin{align*}
    \nabla \cdot \bf{E}
    &= \frac{\rho}{\eps_0}
    &
    \nabla \times \bf{E}
    &= \bf{0} \\
    \nabla \cdot \bf{B}
    &= 0
    &
    \nabla \times \bf{B}
    &= \mu_0 \bf{J}
\end{align*}
Since $\bf{E}$ and $\bf{B}$ are decoupled, we can study them separately.

\myskip
\emph{Electrostatics} is the study of the electric
field generated by a stationary charge
distribution.
\[ \nabla \cdot \bf{E} = \frac{\rho}{\eps_0}
\tag{M1} \]
\[ \nabla \times \bf{E} = \bf{0} \tag{M3'} \]

\subsection{Gauss's Law}

Consider a surface $S$ enclosing a volume $V$.
Integrate (M1) over $V$ and use the divergence
theorem to obtain \emph{Gauss's law}
\[ \boxed{\int_S \bf{E} \cdot \dd \bf{S} =
\frac{Q}{\eps_0}} \]
where $Q = \int_V \rho \dd V$ is the total charge
in $V$. Gauss's law is the integral version of
(M1) and is valid generally.
\[ \text{electric flux} \propto \text{total charge
enclosed} \]
In special situations we can use Gauss's law
together with symmetry to deduce $\bf{E}$ from
$\rho$, by choosing the \emph{Gaussian surface}
$S$ appropriately.

\subsubsection{Spherical symmetry}

Consider a spherically symmetric charge
distribution, $\rho(r)$ in spherical coordinates,
with total charge $Q$ contained within an outer
radius $R$. To have spherical symmetry, the
electric field should have the form
\[ \bf{E} = E(r) \bf{e}_r \]
This will satisfy (M3'), as required. To find
$E(r)$, apply Gauss's law to a sphere of radius
$r$. If $r > R$ then
\begin{align*}
    \int_S \bf{E} \cdot \dd \bf{S}
    &= E(r) \int_S \bf{e}_r \cdot \dd \bf{S} \\
    &= E(r) \int_S \dd S \\
    &= E(r) 4 \pi r^2 \\
    &= \frac{Q}{\eps_0}
\end{align*}
\begin{center}
    \includegraphics[width=0.6\textwidth]
    {images/38d62169-13be-410e-9178-fae699fe5931.png}
\end{center}
Thus
\[ \bf{E} = \frac{Q}{4\pi \eps_0 r^2} \bf{e}_r \]
So the external electric field of a spherically
symmetric body depends only on the total charge.
The Lorentz force on a particle of charge $q$ in
$r > R$ is
\[ \bf{F} = q\bf{E} = \frac{Qq}{4\pi \eps_0
r^2} \bf{e}_r \]
This is the \emph{Coulomb force} between charged
particles. The force is repulsive if the charges
have the same sign ($Qq > 0$) and attractive if
they have opposite signs ($Qq < 0$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e84b9ffe9bca11ed.png}
\end{center}
In the limit $R \to 0$, we obtain the electric
field of a \emph{point charge} $Q$, corresponding
to $\rho = Q\delta(\bf{x})$.

\subsubsection*{Comparison with gravity}

There is a close analogy between the Coulomb force
and the gravitational force between massive
particles,
\[ \bf{F} = -\frac{GMm}{r^2} \bf{e}_r \]
Both involve an inverse-square law and the product
of the charges / masses. However:
\begin{itemize}
    \item While gravity is always attractive,
        electric forces can be repulsive or
        attractive.
    \item Gravity is very much weaker.
\end{itemize}
For example, for two protons the ratio of the
electric to gravitational forces is
\[ \frac{e^2}{4\pi \eps_0 G m_{\rho}^2} \approx
10^{36} \]
On the atomic scale, gravity is irrelevant. But
$+$ and $-$ charges balance so accurately that on
the planetary scale, gravity is dominant.

\subsubsection{Cylindrical symmetry}

Consider a cylindrically symmetric charge
distribution, $\rho(r)$ in \emph{cylindrical}
polar coordinates, with total charge $\lambda$ per
unit length contained within an outer radius $R$.
To have cylindrical symmetry,
\[  \\bf{E} = E(r) \bf{e}_r \]
(Will satisfy (M3') as required.) To find $E(r)$,
apply Gauss's law to a cylinder of radius $r$ and
arbitrary length $L$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/99e771b09bcc11ed.png}
\end{center}
If $r > R$ then
\begin{align*}
    \int_S \bf{E} \cdot \dd \bf{S}
    &= E(r) \int_S \bf{e}_r \cdot \dd \bf{S} \\
    &= E(r) \int_{\substack{S\\\text{(curved
    part)}}} \dd S \\
    &= E(r) 2 \pi r L \\
    &= \frac{\lambda L}{\eps_0}
\end{align*}
Thus
\[ \bf{E} = \frac{\lambda}{2\pi \eps_0 r} \bf{e}_r \]
In the limit $R \to 0$ we obtain the electric
field of a \emph{line charge} $\lambda$ per unit
length, corresponding to
\[ \rho = \lambda \delta(x) \delta(y) \]

\subsubsection{Planar symmetry}

Planar charge distribution: $\rho(z)$ in Cartesian
coordinates, with total charge $\sigma$ per unit
area contained within a region $-d < z < d$ of
thickness $2d$. Assume reflectional symmetry, then
$\rho(z)$ is even. To have planar symmetry,
$\bf{E} = E(z) \bf{e}_z$. Will satisfy (M3') as
required. Reflectional symmetry implies $E(-z) =
-E(z)$. To find $E(z)$ for $z > 0$, apply Gauss's
law to a ``Gaussian pillbox'' of hight $2z$ and
arbitrary area $A$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/506ecfd29bcd11ed.png}
\end{center}
If $z > d$ then
\begin{align*}
    \int_S \bf{E} \cdot \dd \bf{S}
    &= E(z)A - E(-z)A \\
    &= 2E(z)A \\
    &+ \frac{\sigma A}{\eps_0}
\end{align*}
Thus
\[ \bf{E} = \begin{cases}
    \frac{\sigma}{2\eps_0} \bf{e}_z & z > d \\
    -\frac{\sigma}{2\eps_0}\bf{e}_z & z < -d
\end{cases} \]
In the limit $d \to 0$ we obtained the electric
field of a \emph{surface charge} $\sigma$ per unit
area, corresponding to
\[ \rho = \sigma \delta(z) \]

\subsubsection{Surface charge and discontinuity}

Let $\bf{n}$ be a unit vector normal to the
charged surface, pointing from region 1 to region
2.
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/9800005a9bcd11ed.png}
\end{center}
In our example, $\bf{n} = \bf{e}_z$. The
discontinuity in $\bf{E}$ is given by
\[ [\bf{n} \cdot \bf{E}] = \frac{\sigma}{\eps_0}
\tag{1} \]
where $\sigma$ is the surface charge density and
\[ [X] = X_2 - X_1 \]
denotes a discontinuity. The tangential components
are continuous:
\[ [\bf{n} \times \bf{E}] = \bf{0} \tag{2} \]
