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{\Huge\bf Ch. 1 - Part 1

\vspace{1in}

Lecture 1: Basic Electrostatics}

\newpage\ 

\section{Coulomb's law, electrostatic field}

The force between two charges is given by Coulomb's law 
\[
F=\kappa \,\dfrac{q_1q_2}{r^2} 
\]
where $\kappa $ depends on the choice of units. In cgs electrostatic units, $%
\kappa =1,\,$ {\it F} is in dynes, {\it r} in centimeters, $q_1$ and $q_2$ in
statcoulombs. In the MKS\ system, {\it F} is in newtons, {\it r} in meters, $q_1$
and $q_2$ in coulombs, and

\[
\kappa [\mathrm{MKS}]=\dfrac 1{4\pi \epsilon _0}=8.988\times 10^9\,\mathrm{nt%
}\cdot \mathrm{m}^2/\mathrm{coul}^2 
\]

Thus two coulombs at the distance of one meter repel each other with the
very large force of $9\times 10^9\,\mathrm{nt}$ or $9\times 10^{14}\,\mathrm{%
dy}$ (we will usually write 9 instead of 8.988... in the following).

Two coulombs at one cm repel each other with the force of 
$9\times 10^{18}\,\mathrm{dy}$. Hence

\begin{center}
$1\,\mathrm{coul}=2.998\times 10^9\,\mathrm{statcoul}$
\end{center}

\noindent ------------------------------------------------------------------

If charge {\it q} is at point {\bf x} and charge $q_1$ is at
point $\mathbf{x}_1$, the Coulomb force is directed along the vector $%
\mathbf{x}-\mathbf{x}_1$:

\begin{equation}
\mathbf{F}_{on\,q}=qq_1\,\frac{\mathbf{x}-\mathbf{x}_1}{|\mathbf{x}-\mathbf{x%
}_1|^3}  \label{Coulomb's law}
\end{equation}
This force can be regarded as the product of {\it q} times the
electric field {\bf E} produced by $q_1$ at \textbf{x}. \textbf{E}
depends on $\mathbf{r=x-x}_1$ and we can write

\[
\mathbf{E}(\mathbf{r})=q_{1\,}\frac{\mathbf{r}}{r^3}=-q_1\mathbf{\nabla }%
\frac 1r 
\]

For an arbitrary distribution of charges $q_i$ , we have

\[
\mathbf{E(x)=-}\sum_iq_i\mathbf{\nabla }\frac 1{\left| \mathbf{x-x}_i\right|
} 
\]
and for a continuous charge distribution, $\rho (\mathbf{x}^{\prime
})=\sum_iq_i\delta (\mathbf{x}^{\prime }-\mathbf{x}_i)$:

\[
\mathbf{E(x)=-}\int \rho (\mathbf{x}^{\prime })\mathbf{\nabla }\frac
1{\left| \mathbf{x-x}^{\prime }\right| }d^3x^{\prime } 
\]

\noindent =========================================

Insert on properties of $\delta $-function as needed

\noindent =========================================

\section{Gauss's law}

If the surface \textit{S} encloses the charges $q_i$%
\[
\oint_S\mathbf{E\cdot n\,}\,da=4\pi \sum_iq_i 
\]
where the normal $\mathbf{n}$ is directed outwards. This is proved by noting
that for a single charge {\it q} (see Jackson's Fig. 1.2)
\[
\mathbf{E\cdot n}=q\frac{\cos \theta }{r^2} 
\]
so that the flux of the electric field through the area element \textit{da} is 
\[
\mathbf{E\cdot n\,}da=q\frac{da\,\cos \theta }{r^2}=qd\Omega 
\]
where $d\Omega $ is the solid angle subtended by \textit{da}.

\noindent ---------------------------------

For a continuous distribution 
\[
\oint_S\mathbf{E\cdot n\,}\,da=4\pi \int_V\rho (\mathbf{x)}d^3x
\]
where $V$ is the volume enclosed by \textit{S}. Using the divergence theorem 
\[
\int_V\mathbf{\nabla \cdot E}\,\,d^3x=\oint_S\mathbf{E\cdot n\,}\,da
\]
we obtain the differential form of Gauss's law 
\[
\mathbf{\nabla \cdot E}=4\pi \rho 
\]

\noindent ---------------------------------

\subsection{Applications of Gauss's law}

These are left by Jackson as Problems, and can be found in more elementary
books. There are three basic types:

\begin{itemize}
\item  Problems with spherical symmetry, which are solved by noting that on
a sphere of radius {\it r} the field is radial and given in magnitude by 
\[
4\pi r^2E=4\pi Q
\]

where {\it q} is the total charge inside the sphere. To the outside world, it is
as if all the charge were in the center of the sphere. The analogous result
for the gravitational field is due to Newton.

\item  Problems with cylindrical symmetry, which similarly give 
\[
2\pi \rho E=4\pi \lambda 
\]
where $\lambda $ is the total charge per unit length inside the cylinder of
radius $\rho .$

\item  ``Pillbox'' arguments (see Jackson's Fig. 1.4). The best known of
these leads to the conclusion that the field at the surface of a conductor
(which is normal to the surface) is related to the surface charge density $%
\sigma $ by $E=4\pi \sigma $.
\end{itemize}

\section{The potential and Poisson's equation}

The field of a unit point charge is $\mathbf{E}=-\mathbf{\nabla (}1/r)$. The
field of a set of charges can also be written as a gradient: 
\[
\mathbf{E(x)=-}\sum_iq_i\mathbf{\nabla }\frac 1{\left| \mathbf{x-x}_i\right|
}=\mathbf{-\nabla }\Phi (\mathbf{x)} 
\]
where $\Phi $ is the electrostatic potential: 
\[
\Phi (\mathbf{x)}=\sum_iq_i\frac 1{\left| \mathbf{x-x}_i\right| }=\int \rho (%
\mathbf{x}^{\prime })\frac 1{\left| \mathbf{x-x}^{\prime }\right|
}d^3x^{\prime } 
\]
For any function $\Phi $, $\mathbf{\nabla \times }\left( \mathbf{\nabla }%
\Phi \right) =0$. Hence 
\[
\mathbf{\nabla \times E}=0, 
\]

The physical meaning of $\Phi (\mathbf{x)}$ is as follows: if there are no
charges at infinity, so that $\Phi (\infty )=0$, then $\Phi (\mathbf{x)}$ is
the work required to bring a unit charge from $\infty $ to $\mathbf{x}$ (the
other charges being fixed). More generally, the work required to bring a
charge {\it q} from point A to point B is 
\[
W=-\int_A^Bq\mathbf{E\cdot }d{\/}\mathbf{l}=q\left[ \Phi (B)-\Phi
(A)\right] 
\]

The work done depends only on the end points, not on the path; hence the net
work in going around a closed path is zero; the electric field is \textit{%
conservative. }These properties are just a consequence of $\mathbf{\nabla
\times E}=0,$as can also be seen from \textit{Stokes' theorem:} 
\[
\oint_C\mathbf{E\cdot }d{\/}\mathbf{l}=\int_S\mathbf{\nabla \times E\cdot n\,}%
da=0 
\]
where \textit{S} is any surface bounded by the closed contour \textit{C}.

\noindent ---------------------------

The esu unit of potential is the $\mathrm{statvolt=statcoul/cm=erg/statcoul}$%
.

The MKSA unit is the $\mathrm{volt=J/coul}=10^7\mathrm{erg}/(3\cdot 10^9%
\mathrm{statcoul})$. Thus 
\[
\mathrm{volt=}\frac{\mathrm{statvolt}}{300} 
\]

\noindent ---------------------------

Starting from Coulomb's law we have derived the two \textit{differential
field equations of electrostatics} 
\begin{eqnarray*}
\mathbf{\nabla \times E} &=&0 \\
\mathbf{\nabla \cdot E} &=&4\pi \rho
\end{eqnarray*}

The most general solution of the first equation can be written $\mathbf{E}=-%
\mathbf{\nabla }\Phi $, where at this point $\Phi $ is unspecified.
Inserting in the second equation, we find that $\Phi $ must satisfy the 
\textit{Poisson equation:} 
\begin{equation}
\nabla ^2\Phi =-4\pi \rho  \label{Poisson}
\end{equation}

Conversely, one can show directly that the only solution of the Poisson
equation that vanishes at infinity is the familiar 
\begin{equation}
\Phi (\mathbf{x)}=\int \rho (\mathbf{x}^{\prime })\frac 1{\left| \mathbf{x-x}%
^{^{\prime }}\right| }d^3x^{\prime }  \label{Sol}
\end{equation}
To prove that (\ref{Sol}) is a solution of (\ref{Poisson}), we make use of 
\[
\nabla ^2\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=-4\pi \delta (%
\mathbf{x-x}^{\prime }\mathbf{),} 
\]
which is really nothing but Poisson's equation when $\rho $ is a point
charge. If this equation is granted, then 
\[
\nabla ^2\Phi (\mathbf{x)}=\int \rho (\mathbf{x}^{\prime })\nabla ^2\left(
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }\right) d^3x^{^{\prime
}}=-4\pi \int \rho (\mathbf{x}^{\prime })\delta (\mathbf{x-x}^{\prime }%
\mathbf{)}d^3x^{\prime } 
\]
gives immediately $\nabla ^2\Phi (\mathbf{x)=}$ $-4\pi \rho (\mathbf{x})$.

Then, putting $\mathbf{r}=$ $\mathbf{x-x}^{\prime }$, all we need to show is
that 
\[
\nabla ^2\frac 1r=-4\pi \delta (\mathbf{r)} 
\]
The easiest way to proceed is to note that for $r\neq 0$ 
\[
\nabla ^2\frac 1r=\frac 1{r^2}\frac \partial {\partial r}r^2\frac \partial
{\partial r}\frac 1r=0 
\]
but for $r=0$ we have a singularity with integrated strength 
\[
\int \nabla ^2\frac 1rd^3x=\int_S\left( \mathbf{\nabla }\frac 1r\right)
\cdot \mathbf{n\,}da 
\]
by the divergence theorem. The integral is conveniently evaluated on a
sphere of finite radius {\it r}, and it gives $4\pi R^2\left( -1/R^2\right)
=-4\pi $.

Another way to handle the singular behavior is to replace \textit{1/r} with a
``regularized'' function, such as 
\[
\frac 1{\sqrt{r^2+a^2}} 
\]
and then let $a\rightarrow 0$. This is done in Jackson and may appeal to the
rigor-minded. ``Regularization'' procedures of this type are a way of life
in quantum electrodynamics, so Jackson takes the opportunity to show one
here.

------------------------------------

\section{Fields of point, line, and surface charges}

\begin{itemize}
\item  Point charge $q\delta (\mathbf{r)}=q\delta (x)\delta (y)\delta (z)$

\begin{eqnarray*}
\mathbf{E}=q\frac{\mathbf{r}}{r^3} \\
\Phi =\frac qr
\end{eqnarray*}

\item  Straight line charge $\lambda \delta (\mathbf{\rho })=\lambda \delta
(x)\delta (y)$%
\[
\mathbf{E}=2\lambda \,\frac{\mathbf{\rho }}{\rho ^2}
\]
\[
\Phi =-2\lambda \ln \rho =-\lambda \ln \rho ^2
\]

\item  Plane charge $\sigma \delta (x)$%
\begin{eqnarray*}
E_x=2\pi \sigma \limfunc{sign}(x) \\
\Phi =-2\pi \sigma \left| x\right| 
\end{eqnarray*}

That is, the potential has a cusp and the field is discontinuous. For a
non-planar charge layer, there is the same discontinuity, but $\Phi $ is not
symmetric for $\pm x.$

\item  Plane dipole layer $-D\delta $\/$^{^{\prime }}(x)$%
\begin{eqnarray*}
E_x=-4\pi D\delta (x) \\
\Phi =2\pi D\limfunc{sign}(x)
\end{eqnarray*}
That is, the potential is discontinuous and the field vanishes, except
inside the dipole layer, where it cannot be seen. For a non-planar dipole
layer, there is the same discontinuity, but $\Phi $ is not antisymmetric for 
$\pm x.$ A simple result applies for any dipole layer of constant $D$ (see
Jackson's Fig. 1.7): $-\Phi /D$ is equal to the solid angle subtended by the
layer.
\end{itemize}

\end{document}
                                                                    
