% vim: tw=50
% 26/01/2023 09AM

\subsection{The electrostatic potential}

For general $\rho(\bf{x})$ we cannot determine
$\bf{E}(\bf{x})$ using Gauss's law alone.

\myskip
(M3') implies that $\bf{E}$ can be written in
terms of an \emph{electrostatic} (or
\emph{electric}) \emph{potential} $\Phi(\bf{x})$:
\[ \boxed{\bf{E} = -\nabla \Phi} \]
The \emph{potential difference} (or
\emph{voltage}) between two points $\bf{x}_1$ and
$\bf{x}_2$ is
\begin{align*}
    \Phi(\bf{x}_2) -\Phi(\bf{x}_1)
    &= \int \dd \Phi \\
    &= -\int_{\bf{x}_1}^{\bf{x}_2} \bf{E}(\bf{x})
    \cdot \dd \bf{x}
\end{align*}
and is path-independent because $\nabla \times
\bf{E} = \bf{0}$.

\myskip
The electric force on a particle of charge $q$,
\[ \bf{F} = q\bf{E} = -q\nabla \Phi \]
is a conservative force associated with the
potential energy
\[ \boxed{U(\bf{x}) = q\Phi(\bf{x})} \]
(M1) implies that $\Phi$ satisfies \emph{Poisson's
equation}
\[ \boxed{-\nabla^2 \Phi = \frac{\rho}{\eps_0}} \]
The solution can be written as an integral (over
all space, assuming decay at infinity)
\[ \Phi(\bf{x}) = \frac{1}{4\pi\eps_0} \int
\frac{\rho(\bf{x}'}{|\bf{x} - \bf{x}'|} \dd^2
\bf{x}' \]
This is the convolution of $\rho(\bf{x})$ with the
potential of a unit point charge,
$\frac{1}{4\pi\eps_0|\bf{x}|}$, which is the
solution of
\[ -\nabla^2 \Phi = \frac{\delta(\bf{x})}{\eps_0} \]
satisfying $\Phi \to 0$ as $|\bf{x}| \to \infty$.

\myskip
$\bf{E}$ is unaffected if we add an arbitrary
constant to $\Phi$. We usually choose this such
that $\Phi \to 0$ as $|\bf{x}| \to \infty$. If
$\rho(\bf{x})$ does not decay sufficiently
rapidly, this may not be possible, for example a
line charge $E_r \propto \frac{1}{r}$ and $\Phi
\propto \ln r$.

\subsubsection{Point Charge}

Potential due to a point charge $q$ at the origin:
\[ \Phi(\bf{x}) = \frac{q}{4\pi \eps_0|\bf{x}|} =
\frac{q}{4\pi \eps_0 r} \]

\subsubsection{Electric Dipole}

Two equal and opposite charges at different
positions. Without loss of generality, consider
charges $-q$ at $\bf{x} = \bf{0}$ and $+q$ at
$\bf{x} = \bf{d}$.
\begin{center}
    \includegraphics[width=0.15\linewidth]
    {images/c42b3c889d5a11ed.png}
\end{center}
Potential due to the dipole:
\[ \Phi(\bf{x}) = \frac{q}{4\pi\eps_0} \left(
-\frac{1}{|\bf{x}|} + \frac{1}{|\bf{x} - \bf{d}|} \right) \]
Apply Taylor's theorem for a scalar field,
\[ f(\bf{x}) = f(\bf{x}) + (\bf{h} \cdot \nabla)
f(\bf{x}) + \half (\bf{h} \cdot \nabla)^2
f(\bf{x}) + O(|\bf{h}|^3) \]
to $f(\bf{x}) = \frac{1}{|\bf{x}|} = \frac{1}{r}$:
\begin{align*}
    \Phi(\bf{x})
    &= \frac{q}{4\pi\eps_0} \left(
    -\frac{1}{r} + \frac{1}{r} - (\bf{d} \cdot
    \nabla) \frac{1}{r} + O(|\bf{d}|^2) \right) \\
    &= \frac{q}{4\pi\eps_0} \frac{\bf{d} \cdot
    \bf{x}}{|\bf{x}|^3} + O(|\bf{d}|^2)
\end{align*}
In the limit $|\bf{d}| \to 0$ with $q\bf{d}$
finite, we obtain a \emph{point dipole} with
\emph{electric dipole moment}
\[ \bf{p} = q\bf{d} ,\]
potential
\[ \Phi(\bf{x}) = \frac{\bf{p} \cdot
\bf{x}}{4\pi\eps_0 |\bf{x}|^3} \]
and electric field
\[ \bf{E} = -\nabla\Phi = \frac{3(\bf{p} \cdot
\bf{x})\bf{x} - |\bf{x}|^3
\bf{p}}{4\pi\eps_0|\bf{x}|^5} \]
In spherical polar coordinates aligned with
$\bf{p} = p\bf{e}_z$,
\begin{align*}
    \Phi
    &= \frac{p\cos\theta}{4\pi\eps_0 r^2} \\
    E_r
    &= -\pfrac{\Phi}{r}
    = \frac{2p\cos\theta}{4\pi\eps_0 r^3} \\
    E_\theta
    &= -\frac{1}{r} \pfrac{\Phi}{\theta}
    = \frac{p\sin\theta}{4\pi\eps_0 r^3} \\
    (E_\phi
    &= 0)
\end{align*}
Note
\begin{itemize}
    \item $\Phi$ and $\bf{E}$ are not spherically
        symmetric.
    \item They decrease more rapidly with $r$ than
        for a point charge.
\end{itemize}
A point dipole $\bf{p}$ at the origin corresponds
to
\begin{align*}
    \rho(\bf{x})
    &= -\bf{p} \cdot \nabla \delta(\bf{x}) \\
    \Phi(\bf{x})
    &= \bf{p} \cdot \nabla \left(
    \frac{1}{4\pi\eps_0 |\bf{x}|} \right)
\end{align*}

\subsubsection{Field lines and equipotentials}

\emph{Electric field lines} are the integral
curves of $\bf{E}$, being tangent to $\bf{E}$
everywhere.

\myskip
Since $\nabla \cdot \bf{E} = \frac{\rho}{\eps_0}$,
field lines begin on $+$ charges and end on $-$
charges. In electrostatics, $\bf{E} = -\nabla
\Phi$, field lines are perpendicular to the
equipotential surfaces $\Phi = \text{const}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d68b78729d5d11ed.png}
\end{center}

\subsubsection{Dipole in an external field}

Consider a dipole $\bf{p}$ in an external electric
field $\bf{E} = -\nabla \Phi$ generated by
distinct charges. With $-q$ at $\bf{x}$ and $+q$
at $\bf{x} + \bf{d}$, the potential energy of the
dipole due to the external field is
\begin{align*}
    U
    &= -q\Phi(\bf{x}) + q\Phi(\bf{x} + \bf{d}) \\
    &= q(\bf{d} \cdot \nabla) \Phi(\bf{x}) +
    O(|\bf{d}|^3)
\end{align*}
In the limit of a point dipole,
\begin{align*}
    U
    &= \bf{p} \cdot \nabla \Phi \\
    &= -\bf{p} \cdot \bf{E}
\end{align*}
and sis minimised when $\bf{p}$ is aligned with
$\bf{E}$.

\subsubsection{Multipole Expansion}

For a general charge distribution $\rho(\bf{x})$
confined to a ball $\{V \colon |\bf{x}| < R\}$,
\[ \Phi(\bf{x}) = \frac{1}{4\pi\eps_0} \int_V
\frac{\rho(\bf{x}')}{|\bf{x} - \bf{x}'|}\dd^3
\bf{x}' \]
External potential $(|\bf{x}| > R$) -- expand
\begin{align*}
    \frac{1}{|\bf{x} - \bf{x}'|}
    &= \frac{1}{r} - (\bf{x}' \cdot \nabla)
    \frac{1}{r} + \half (\bf{x}' \cdot \nabla)^2
    \frac{1}{r} + O(|\bf{x}'|^3) \\
    &= \frac{1}{r} \left[ 1 + \frac{\bf{x}' \cdot
    \bf{x}}{r^2} + \frac{3(\bf{x}' \cdot \bf{x})^2
    - |\bf{x}'|^2 |\bf{x}|^2}{2r^4} + O \left(
    \frac{R^3}{r^3} \right) \right]
\end{align*}
Leads to the \emph{multipole expansion} of the
potential (see Example Sheet 1, Question 11)
\[ \Phi(\bf{x}) = \frac{1}{4\pi\eps_0} \left(
\frac{Q}{r} + \frac{\bf{p} \cdot \bf{x}}{r^3} +
\half \frac{Q_{ij} x_i x_j}{r^5} + \cdots \right) \]
First three multipole moments:
\begin{itemize}
    \item total charge (monopole - scalar) $Q =
        \int_V \rho(\bf{x}) \dd^3 \bf{x}$
    \item electric dipole moment (vector) $\bf{p}
        = \int_V \bf{x} \rho(\bf{x}) \dd^3
        \bf{x}$.
    \item electric quadripole moment (traceless
        symmetric 2nd order tensor)
        \[ Q_{ij} = \int_V (3x_i x_j - |\bf{x}|^2
        \delta_{ij})\rho(\bf{x}) \dd^3 \bf{x} \]
\end{itemize}
For $r \gg R$, $\Phi$ and $\bf{E}$ look
increasingly like those of a point charge $Q$,
unless $Q = 0$, in which case they look like those
of a point dipole, unless $\bf{p} = \bf{0}$, etc
(See Example Sheet 1, Question 11).
