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\subsection{Electrostatic Energy}

The work done against the electric force $\bf{F} =
q\bf{E}$ in bringing a particle of charge $q$ from
infinity (where we assume $\Phi = 0$) to $\bf{x}$
is
\[ \int_\infty^{\bf{x}} \bf{F} \cdot \dd \bf{x}
= + q \int_\infty^{\bf{x}} \nabla \Phi \cdot \dd
\bf{x} = q\Phi(\bf{x}) .\]
Consider assembling a configuration of $N$ point
charges one by one. Particle $i$ of charge $q_i$
is brought from $\infty$ to $\bf{x}_i$ while the
previous particles remain fixed.
\begin{itemize}
    \item Particle 1: No work is involved: $W_1 =
        0$.
    \item Particle 2:
        \[ W_2 = q_2 \left( \frac{q_1}
        {4\pi\eps_0|\bf{x}_1 - \bf{x}_2|} \right) \]
    \item Particle 3:
        \[ W_3 = q_3 \left(
        \frac{q_1}{4\pi\eps_0|\bf{x}_3 -
        \bf{x}_1|} + \frac{q_2}{4\pi\eps_0|\bf{x}_3 -
        \bf{x}_2|} \right) \]
        etc.
\end{itemize}
Total work done:
\[ U = \sum_{i = 1}^N W_i = \sum_{i = 2}^N
\sum_{j = 1}^{i - 1} \frac{q_i
q_j}{4\pi\eps_0|\bf{x}_i - \bf{x}_j|} \]
Can be rewritten as
\[ U = \half \sum_{i = 1}^N \sum_{\substack{j =
1\\j\neq i}} \frac{q_i q_j}{4\pi\eps_0|\bf{x}_i -
\bf{x}_j|} \]
or alternatively
\[ \boxed{U = \half \sum_{i = 1}^N q_i
\Phi(\bf{x}_i)} \]
Generalise to a continuous charge distribution
$\rho(\bf{x})$ occupying a finite volume $V$:
\[ U = \half \int_V \rho(\bf{x}\Phi(\bf{x} \dd^3
\bf{x} \]
\[ \boxed{U = \half \int_V \rho\Phi \dd V} \]
Using (M1) we have
\begin{align*}
    U
    &= \half \int_V (\eps_0 \nabla \cdot \bf{E})
    \Phi \dd V \\
    &= \frac{\eps_0}{2} \int_V (\nabla \cdot
    (\Phi\bf{E}) - \bf{E} \cdot \nabla \Phi) \dd V
    \\
    &= \frac{\eps_0}{2} \int_S \Phi \bf{E} \cdot
    \dd \bf{S} + \int_V \frac{\eps_0
    |\bf{E}|^2}{2} \dd V
\end{align*}
Let $S = \partial V$ be a sphere of radius $R \to
\infty$. Then $\Phi = O(R^{-1})$ and $\bf{E} =
O(R^{-2})$ on $S$, while the area of $S$ is
$O(R^2)$, so $\int_S$ is $O(R^{-1})$ and $\to 0$
as $R \to \infty$. Thus
\[ \boxed{U = \int \frac{\eps_0|\bf{E}|^2}{2} \dd V} \]
integrated over all space.

\myskip
This implies that \emph{energy is stored in the
electric field even in a vaccum}.

\myskip
Any of the expressions for $U$ suggests that the
self-energy of a point charge is infinite. We can
discuss this as it is unchanging and causes no
force.

\subsection{Conductors}

In a \emph{conductor} such as a metal, some
charges (usually electrons) can move freely. In
electrostatics we require
\[ \boxed{\bf{E} = \bf{0}, \qquad \Phi =
\text{constant}} \]
inside a conductor, hence $\rho = 0$. Otherwise
free charges would move in response to the
electric force and a current would flow.

\myskip
A surface charge density $\sigma$ can exist on the
surface of a conductor, which is an equipotential.

\myskip
Taking $\bf{n}$ to point out of the conductor, the
condition
\[ [\bf{n} \cdot \bf{E}] = \frac{\sigma}{\eps_0} \]
implies
\[ \boxed{\bf{n} \cdot \bf{E} =
\frac{\sigma}{\eps_0}} \]
immediately outside the conductor.

\myskip
The constant potential of a conductor can be set
by connecting it to a battery or another
conductor. An \emph{earthed} (or \emph{grounded})
conductor is connected to the ground, usually
taken as $\Phi = 0$.

\myskip
To find $\Phi(\bf{x})$ and $\bf{E}(\bf{x})$ due
to a charge distribution $\rho(\bf{x})$ in the
presence of conductors with surfaces $S_i$ and
potentials $\Phi_i$, we solve Poisson's equation
\[ -\nabla^2 \Phi = \frac{\rho}{\eps_0} \]
with Dirichlet boundary conditions
\[ \Phi = \Phi_i \qquad \text{on} \qquad S_i .\]
The solution depends linearly on $\rho$ and
$\{\Phi_i\}$.

\subsubsection*{Example}

A point charge $q$ at position $(0, 0, h)$ in a
half-space ($z > 0$) bounded by an earthed
conducting wall ($\Phi = 0$ on $z = 0$).

\myskip
By the method of images, the solution in $z > 0$
is identical to that of a dipole, with image
charge $-q$ at $(0, 0, -h)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9ca9fb2ca14c11ed.png}
\end{center}
The wall coincides with an equipotential of the
dipole. The induced surface charge density on the
wall can be worked out from
\[ \frac{\sigma}{\eps_0} = \bf{n} \cdot \bf{E} =
E_z = -\frac{2qh}{4\pi\eps_0(r^2 + h^2)^{3/2}} \]
($r = \sqrt{x^2 + y^2}$).
The total induced surface charge is
\begin{align*}
    \int_0^\infty \sigma 2\pi r \dd r
    &= -qh \int_0^\infty \frac{r\dd r}{(r^2 +
    h^2)^{3/2}} \\
    &= -q
\end{align*}
(equal to the image charge).

\subsubsection*{Capacitors}

A simple \emph{capacitor} consists of two
separated conductors carrying charges $\pm Q$.

\myskip
If the potential difference (voltage) between them
is $V$, then the capacitance is defined by
\[ \boxed{C = \frac{Q}{V}} \]
and depends only on the geometry, because $\Phi$
depends on linearly on $Q$.

\myskip
For example, two infinite parallel plates
separated by distance $d$. Let the plate surfaces
at $z = 0, d$ have surface charge densities $\pm
\sigma$. Then $\bf{E} = E \bf{e}_z$ with $E =
\sigma / \eps_0 = \text{const}$ for $0 < z < d$
(and $\bf{E} = \bf{0}$ elsewhere). $\Phi = -Ez +
\text{const}$ and $V = Ed$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/86200fbca14d11ed.png}
\end{center}
The same solution holds approximately for parallel
plates of area $A \gg d^2$ if end-effects are
neglected. So
\[ C = \frac{Q}{V} \approx \frac{\sigma A}{Ed}
\approx \frac{\eps_0 A}{d} \]
Electrostatic energy stored in the capacitor:
\[ U = \int \frac{\eps_0|\bf{E}|^2}{2} \dd V
\approx \frac{\eps_0 E^2}{2} Ad \approx \half CV^2 \]
In general
\[ \boxed{U = \half CV^2 = \frac{Q^2}{2C}} \]
The work done in moving an element of charge
$\delta Q$ from one plate to another is $\delta W
= V \delta Q$, so the total work done is
\[ \int_0^Q \frac{Q'}{C} \dd Q'
= \frac{Q^2}{2C} \]
Or use
\begin{align*}
    U
    &= \half \int \rho \Phi \dd V \\
    &= \half Q \Phi_+ - \half Q \Phi_- \\
    &= \half QV
\end{align*}
