
% slides potentiaal derde dagdeel potentiaal loopt niet lekker.

\chapter{Electrostatics}

\section{History and basics}
The word electricity is derived from 
$\eta \lambda \epsilon \kappa \tau \rho o \nu$ 
($elektron$), the Greek word for amber. The reference to amber is not a coincidence: when an amber rod is rubbed with fur it attracts certain objects, such as a piece of paper or a hair. This was documented first by 
Thales of Miletus (600 BC).

Amber is not unique in this sense. Many other materials, such as glass, rubber, PVC and Ebony can be electrified by rubbing it with for example fur, silk or  wool. Many experiments in the 17$^{th}$ and 18$^{th}$ century were conducted to study
this phenomenon. This work led to the discovery of two kinds of electricity (positive and negative charge) that attract, while electricity of the same kind repels.
A simple experiment with a glass rod and a plastic ball as illustrated in Fig.~\ref{likech} shows that like charges repel.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/likech.ps}}
\caption{\sl A glass rod is electrically charged by rubbing it with fur. The rod is then used to charge a plastic ball. Now the two objects repel.}
\label{likech}
\end{center}
\end{figure}
Another experiment, using an additional rubber rod and ball with opposite electrical charge demonstrates that unlike charges attract, see Fig.~\ref{unlikech}. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/unlikech.ps}}
\caption{\sl The rod and ball have opposite electrical charge. The two objects attract.}
\label{unlikech}
\end{center}
\end{figure}
Well known is the glass rod that, rubbed with a silk cloth, obtains a 'positive' electrification. Famous is the Ebony rod that after being rubbed with a cat's skin is 'negative' electrified.

\section{Electrical Force and Field}
\subsection{Coulomb's law}
Charles Augustin Coulomb (1736 - 1806) was a French physicist
who studied the electrical forces in a quantitative manner utilizing a 
torsion balance. 
% A drawing of such device is shown in Fig.\ref{balance}. 
% \begin{figure}[htb]
% \begin{center}
%\mbox{\epsfxsize=8cm \epsffile{balance.ps}}
%\caption{\sl A torsion balance to study electrical forces. Shown are its
% 
%its components
%}
%\label{balance}
%\end{center}
%\end{figure}
%
 Through this experimentation, Coulomb found that the force
between two (point) charges is given by:
\begin{equation}
\vec{F}=\frac{1}{4\pi\epsilon_0} \frac{q Q}{r^2} \dakje{r}
\label{eq:coulomb}
\end{equation}
where $r=|\vec{r}|$ represents the distance between the test charge $q$ and source charge $Q$. The vector $\vec{r}$ connects the two charges. 
 The direction of the force $\vec{F}$ is given by $\dakje{r}=\frac{\vec{r}}{|\vec{r}|}$, the unit vector pointing from  $Q$ to $q$, see also Fig.\ref{fcoulomb}. Equivalently, we can write for the Coulomb force:
\begin{equation}
\vec{F}=\frac{1}{4\pi\epsilon_0} \frac{q Q}{r^2} \frac{\vec{r}}{|\vec{r}|} =
\frac{1}{4\pi\epsilon_0} \frac{q Q}{r^3} \vec{r} 
\end{equation}

\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/fcoulomb.ps}}
\caption{\sl The Coulomb force between a  source charge $Q$ and a test charge $q$.}
\label{fcoulomb}
\end{center}
\end{figure}

The SI unit of electrical charge is the Coulomb, which can be abbreviated to the unit C in equations. The factor $\frac{1}{4\pi\epsilon_0}$ is a constant term with $\epsilon_0$ called the 'electrical permittivity'. The electrical permittivity has the numerical value:
\begin{equation}
\epsilon_0 = 8.85419 \times 10^{-12} \rm{\frac{C^2}{Nm^2}}
\end{equation}
Often, Coulomb's law is also written as
\begin{equation}
\vec{F}=K \frac{q Q}{r^2} \dakje{r}
\end{equation}
with $K$ the electrical constant:
\begin{equation}
K = 8.98755 \times 10^{9} \rm{\frac{Nm^2}{C^2}}
\end{equation}

Coulomb's law is about the electrical force between two point charges.
 If we have three, four or whatever number of charges at different positions,
 the total force on a test charge becomes:
\begin{equation}
\vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} + .....
\end{equation}
with $\vec{F_i}$ the electrical force by source charge $Q_i$ on our test charge given by equation~\ref{eq:coulomb}.
Using this 'superposition principle' we can write this total force as:
\begin{equation}
\vec{F}=\sum_{i=1,N} \frac{1}{4\pi\epsilon_0} \frac{q Q_i}{|\vec{r_0}-\vec{r_i}|^2} \frac{\vec{r_0}-\vec{r_i}}{|\vec{r_0}-\vec{r_i}|}
\label{eq:superpos}
\end{equation}
with  the position of the $N$ charges $Q_i$ denoted as $\vec{r_i}$ and the position of the test charge $\vec{r_0}$. The superposition principle is illustrated in Fig.~\ref{superpos}. The vectors are defined with respect to some origin $O$.
Note that the expression for the electrical force is independent on the
choice of origin $O$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/superpos.ps}}
\caption{\sl The Coulomb force on a test charge $q$ on position $r_0$ resulting from four source charges $Q_1$,..., $Q_4$.  The connection vector, $\vec{r_0} - \vec{r_1}$ between charge $Q_1$ and the test charge $q$ is also indicated. The
vectors $\vec{r}_0$ and $\vec{r}_1$ are defined with respect to origin $O$.
}
\label{superpos}
\end{center}
\end{figure}


The physicist J.J. Thomson discovered in 1897 a new elementary particle: the
electron. The electron carries a negative electrical charge and is responsible
 for most electrical currents.
 R.A. Millikan discovered in 1909 that all electrons
carry a similar charge, $-e$, called the elementary charge.
In SI units its value is
\begin{equation}
e=1.6002 \times 10^{-19} \rm{C}
\end{equation}

We know now that in atoms negative charge is carried by electrons, while protons are positively charged. When we go back to the old experiment,
 rubbing an ebony rod with fur or cat's skin,
 the electrons from the cat's skin get transferred to the ebony rod.
 The cat's skin now has a deficiency of electrons and so is positively charged. On the other hand, the ebony rod has an excess of electrons and hence is
 negatively charged. So, in everyday life, positive electrification is due
 to the deficiency of electrons and thus not by an excess of protons.

Protons are not so elementary as electrons. 
Protons consists of quarks and gluons. The quarks carry electrical charge
of $-\frac{1}{3} e$ and $+\frac{2}{3} e$. In nature, the quarks and gluons are 
confined in other particles (like the proton). Scientific research
in the field of  
'particle physics' has led to a new 'table of elements', 
one with six quarks and six 'leptons'  which
interact by the exchange of force particles. The particle world is
 illustrated by the table in Fig.~\ref{qtable}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/qtable.ps}}
\caption{\sl The elementary particles. The leptons from left to right are:
electron-neutrino, muon-neutrino, tau-neutrino, electron, muon, tauon. 
The quarks, in the same order, are: up, charm, top, down, strange, bottom.
From left to right, the mass of the particles is increasing. For example, an up quark 'weighs' a few MeV while a top quarks weighs 175~$GeV$ (for reference: a proton weighs 1~$GeV$). This mass difference between the particles is an open
question, but the so called Higgs mechanism may be its origin.
}
\label{qtable}
\end{center}
\end{figure}
 All these particles
exist in nature, but one to remember is that our everyday world is made
 off only three particles: up and down quarks and electrons.
Electrical charge in nature comes in discrete amounts: it is quantized,
which has deep implications. 
The net charge before an  interaction is equal to the charge after an 
interaction.
  
Elementary particles like quarks and gluons are
studied in particle collisions at particle laboratories like CERN 
(Geneva) and
 Fermilab (Chicago) using large accelerators. Presently at CERN a 
new accelerator is being constructed, the Large
Hadron Collider (LHC). The apparatus with a circumference of 27~km
 accelerates protons clockwise and counter-clockwise
 to energies of 7~TeV. At a few dedicated points, interaction points,
 the protons collide.
The detectors to measure the products
 (i.c. new particles) of these collisions are large, 
typically 20 $\times$ 20 $\times$ 20~m$^3$. In the Netherlands,
 NIKHEF in Amsterdam, is the main institute where research in this field
is being conducted. NIKHEF contributes to ATLAS, a
 detector for the LHC, as shown in Fig.~\ref{ATLAS}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=9cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/ATLAS.ps}}
\caption{\sl An impression of the ATLAS detector which is being assembled.
 This detector is 44~m long!
Particles that are produced by proton collisions are measured
using several detection techniques. NIKHEF contributes to the
silicon tracking detectors and muon drift tube chambers.
See www.cern.ch for more information.}
\label{ATLAS}
\end{center}
\end{figure}
 Although ATLAS is a multipurpose detector, the focus
of the research program is on the Higgs particle.




\subsection{The electrical field}
Let's have a closer look to equation~\ref{eq:superpos} and rewrite it:
\begin{equation}
\vec{F}= q \left( \sum_{i=1,N} \frac{1}{4\pi\epsilon_0} \frac{Q_i}{|\vec{r_0}-\vec{r_i}|^2} \frac{\vec{r_0}-\vec{r_i}}{|\vec{r_0}-\vec{r_i}|} \right)
\end{equation}
The electrical field $\vec{E}$ is defined by:
\begin{equation}
\vec{F}= q \vec{E}          ,
\end{equation}
with:
\begin{equation}
\vec{E}= \sum_{i=1,N} \frac{1}{4\pi\epsilon_0} \frac{Q_i}{|\vec{r_0}-\vec{r_i}|^2}  \frac{\vec{r_0}-\vec{r_i}}{|\vec{r_0}-\vec{r_i}|}     .
\label{eq:efield} 
\end{equation}
In different textbooks you will find different notations. A common
notation is:
\begin{equation}
\vec{E}= \sum_{i=1,N} \frac{1}{4\pi\epsilon_0} 
       \frac{Q_i}{\vec{\rm{r}_i}^2} \dakje{\rm{r}_i} 
\end{equation}
In this case the vector $\vec{\rm{r_i}}$ is defined as 
$\vec{\rm{r_i}}=\vec{r_0}-\vec{r_i}$. Be aware of this freedom of
notation, which leads to many un-necessary mistakes.

A way to physically interpret the above expression for
the electrical field is that a charge $q$ senses 
an electrical field and consequently undergoes
 a force $\vec{F}= q \vec{E}$. We have introduced the field as a
relatively simple mathematical definition. However, the electrical field
is a genuine physics quantity! In electrostatic theory, the field
is present in space and you can calculate it using formula~\ref{eq:efield}
 when you know the (point) charge distribution.
 If the field is present in space,
 you may wonder, where is it made off? That is a bit harder to answer and
beyond the scope of this course.
 In particle physics, fields consist of 'force particles' that are being
exchanged when particles have an interaction.
 The electrical field consist of (virtual) photons that are exchanged
 between electrical charges.


Above, all equations are based on point charges. Since all charge are 
carried by individual particles this seems not unreasonable. However, in the
classical theory charges can be continuously distributed. This is still 
reasonable when we describe physics at a scale much (much) larger
than the size of and distances (10$^{-10}$m) between the particles. The classical 
theory describes nature in a macroscopic matter. We will discuss some
examples of continuous charge distributions.

\subsubsection{Line charge}
A line charge can be described by a charge density $\lambda(\vec{r}_l)$
 with has the unit C/m. The position vector $\vec{r}_l$ is a
coordinate defined with respect to some origin.
 To calculate the field in point $P$
 with coordinate $\vec{r}_P$,
 we integrate over the infinitesimal pieces of charge
 $\lambda(\vec{r}_l) dl$ to find the electrical field:
\begin{equation}
\vec{E}(\vec{r}_P)= \frac{1}{4\pi\epsilon_0} \int_{line} 
          \frac{ \lambda(\vec{r}_l) }{\vec{r}^2} \dakje{r} dl
\label{eq:eline}
\end{equation}
where $\vec{r}$ is the connection vector between a piece of charge and the point $P$ with coordinate $\vec{r}_P$, thus $\vec{r}=\vec{r}_P-\vec{r}_l$.
This is illustrated in Fig.~\ref{eline}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/eline.ps}}
\caption{\sl Illustration of a line charge.}
\label{eline}
\end{center}
\end{figure}

In fact, we can still interpret the integral as the sum over point
charges as we used to do in equation~\ref{eq:efield}. A point charge is
then just a piece of line charge $dq=\lambda(\vec{l}) dl$ and so we obtain:
\begin{eqnarray}
\vec{E}(\vec{r_P}) & = & \frac{1}{4\pi\epsilon_0} \int_{charge} 
          \frac{ dq }{\vec{r}^2} \dakje{r} \nonumber \\
     & = &  \frac{1}{4\pi\epsilon_0} \int_{charge} 
          \frac{ dq }{|\vec{r_P}-\vec{r}_l|^2} 
        \frac{ \vec{r}_P - \vec{r}_l }{|\vec{r}_P - \vec{r}_l|} \nonumber \\
 & \approx &
    \sum_{point charges} \frac{1}{4\pi\epsilon_0}
          \frac{Q_i}{|\vec{r}_P - \vec{r_i}|^2}
        \frac{ \vec{r}_P - \vec{r_i} }{|\vec{r}_P - \vec{r_i}|} 
\end{eqnarray}
Perhaps the interpretation of a continuous charge distribution as a collection
of point charges helps you to make the math less abstract. 


\subsubsection{Example of a line charge}
\label{diline}
As an example we will calculate the electrical field in a point $P$ 
with $z=z_P$
 of a piece of wire with length $L$ centered on the $x$ axis. The wire carries an uniform charge
density $\lambda$.  The configuration is shown in Fig.~\ref{exline}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/lijnlading.eps}}
\caption{\sl A wire of length $L$ with uniform charge distribution. Also indicated is the contribution to the electrical field in point $P$ of a piece of line $dx$.}
\label{exline}
\end{center}
\end{figure}

When we consider the contribution, $d\vec{E}$, of a piece of charge $\lambda dx$ at position $x$ to the electrical field in $P$ we see that there are two components, an $x$ and $z$ component:
\begin{eqnarray}
dE_x&=&
     \frac{1}{4\pi\epsilon_0} \frac{\lambda dx}{z_P^2 + x^2 }sin(\alpha) \nonumber \\
dE_z&=&\frac{1}{4\pi\epsilon_0} \frac{\lambda dx}{z_P^2 + x^2}
                                                          cos(\alpha)
\end{eqnarray}
 Look at the symmetry of this problem. 
The components in the $x$ direction cancel ($E_x=0$) and  thus we only have to integrate the $z$ contributions. So, we need to know $cos(\alpha)$, which we can
get from the figure (convince yourself!): $cos(\alpha)= \frac{z_P}{\sqrt{z_P^2 + x^2}} $. Now we can integrate all contributions:
\begin{eqnarray}
E_z&=&\int dE_z         \nonumber  \\
   &=&
        \int_{-L/2}^{+L/2} \frac{1}{4\pi\epsilon_0}          
        \frac{\lambda dx}{z_P^2 + x^2} \frac{z_P}{\sqrt{z_P^2 + x^2}}
  \nonumber \\
   &=&
        \int_{-L/2}^{+L/2} \frac{1}{4\pi\epsilon_0}          
        \frac{z_P \lambda dx}{(z_P^2 + x^2)^{\frac{3}{2}}} 
  \nonumber \\
   &=&  \frac{\lambda}{4\pi\epsilon_0} 
       \frac{ x}{z_P \sqrt{z_P^2 + x^2}}|^{+L/2}_{-L/2}  \nonumber     \\
   &=& 
      \frac{\lambda}{4\pi\epsilon_0} \frac{ L }{z_P \sqrt{z_P^2 + (L/2)^2}}
\end{eqnarray}
Does the result make sense? Well, we know that the field of a point charge
is linear with charge $Q$ and drops quadratically with  the distance.
If we look from very large distance to the line charge, $z_P >> L$, so that
all the charge on the line appears concentrated in a point, we find that
\begin{equation}
  E_z=  \frac{1}{4\pi\epsilon_0} \frac{\lambda L}{z^2_P}=  \frac{1}{4\pi\epsilon_0} \frac{Q_{total}}{z^2_P}
\end{equation}
which is the field of a point charge $Q_{total}=\lambda L$, as expected.


\subsubsection{Surface charge}
An illustration of a surface charge is depicted in Fig.~\ref{esurf}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/esurf.ps}}
\caption{\sl Illustration of a surface charge.}
\label{esurf}
\end{center}
\end{figure}
A surface charge is described by a charge density $\sigma(\vec{r}_s)$ with has
the unit C/m$^2$. When $\vec{r}_s$ lies on the surface, $\sigma(\vec{r}_s)$
 has
some value that represents the charge density. When $\vec{r}_s$ lies outside
the surface, $\sigma(\vec{r}_s)=0$~C/m$^2$. 
To calculate the electrical field we integrate over
the contribution of the infinitesimal pieces of charge $dq=\sigma(\vec{r}_s)
 do$.
The electrical field in point $P$ is given by:
\begin{equation}
\vec{E}(\vec{r_P})= \frac{1}{4\pi\epsilon_0} \int_{surface} 
          \frac{ \sigma(\vec{r}_s) }{\vec{r}^2} \dakje{r} do
\label{eq:esurf}
\end{equation}
Now $\vec{r}$ is defined as the connection vector between point $\vec{r}_P$
and the location of some infinitesimal piece of surface, thus 
$\vec{r}=\vec{r}_P - \vec{r}_s$ .

Like the integration over a line, the integration over a surface is
 a difficult task. Only in special cases 
 we can actually perform the integration. So, don't worry
for the moment if you do not see how in general you could use
the above expression. Finally we remark that we have been 'sloppy' with
our notation: a surface is two dimensional, thus one would expect
a double integral with two integration variables. Well, get used to it:
 in different textbooks you will find at least as many different
notations. Be prepared and just 'bluf' through it.



\subsubsection{Volume charge}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/evol.ps}}
\caption{\sl Illustration of a volume charge.}
\label{evol}
\end{center}
\end{figure}
An illustration of a volume charge is depicted in Fig.~\ref{evol}.
The volume charge density $\rho(\vec{r}_v)$ is the most general continuous
charge density and has the unit C/m$^3$. The contribution of the 
infinitesimal pieces of charge $dq=\rho(\vec{r}_v) dv$ lead to the following
expression:
\begin{equation}
\vec{E}(\vec{r}_P)= \frac{1}{4\pi\epsilon_0} \int_{volume} 
          \frac{ \rho(\vec{r}_v) }{\vec{r}^2} \dakje{r} dv
\label{eq:evol}
\end{equation}
Now $\vec{r}$ is defined as the connection vector between point $\vec{r}_P$
and the location of some infinitesimal piece of volume $dv$ at position 
 $\vec{r}_v$.
Applications of this expression come later. Note that we again have been
sloppy with the notation. A volume (in this report) has three dimensions
and thus three integration variables. For example when we integrate
a function $f$ over a volume in Cartesian coordinates:
\begin{equation}
\int_{volume} f dv  = \int^{\infty}_{x=-\infty}
                    \int^{\infty}_{y=-\infty}
                    \int^{\infty}_{z=-\infty} f(x,y,z) dx dy dz 
\end{equation}
and in spherical coordinates:
\begin{equation}
\int_{volume} f dv  = \int^{2 \pi}_{\phi=0}
                      \int^{\pi}_{\theta=0}
                    \int^{\infty}_{r=0} f(r,\phi,\theta) r^2 sin(\theta) d\phi d\theta dr 
\end{equation}
If this is 'abracadabra' to you, first work through a 
textbook on mathematics\footnote{
 Please, let the authors of this report know
 when integration using spherical coordinates 
 are unknown to you at the time of the lectures. Don't hesitate! Additionally.
in the slides that are discussed during the sessions, added as Appendix  you can find more explanation and
 many examples.}.

% However, at the moment we don't care too much 
% about these mathematical details and opt for the
% short notation where possible.


\subsection{Knowledge and Skills}
The knowledge and skills 
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
Charge can be positive and negative. Electrons carry the elementary charge
 $-e$, with $e=1.6002 \times 10^{-19}$~C. 
The total charge is conserved.
\item
The electrical or Coulomb force between two electrically charged objects
is given by:
\begin{equation}
\vec{F}=\frac{1}{4\pi\epsilon_0} \frac{q Q}{r^2} \dakje{r} \nonumber
\end{equation}
Make sure you understand the notation!
\item
The electrical field of a point charge $q$ is given by:
\begin{equation}
\vec{E}=\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \dakje{r} \nonumber
\end{equation}
\item
The electrical field of a volume charge is:
\begin{equation}
\vec{E}(\vec{r}_P)= \frac{1}{4\pi\epsilon_0} \int_{volume} 
          \frac{ \rho(\vec{r}_v) }{\vec{r}^2} \dakje{r} dv \nonumber
\end{equation}
You understand all the vectors in this notation and you can make
a drawing that illustrates this expression.
You can also write down the formulas for the 
electrical field of a line charge and a surface charge.
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.


\section{Electrical flux and Gauss' Law}
\label{section:gauss}

\subsection{Electrical field lines and electrical flux}
Figure~\ref{efieldvec} graphically displays the electrical field 
corresponding to the electrical field of a point charge. To make this drawing a grid was chosen and on each grid-point the electrical field is represented by an arrow.
 The length of the arrow corresponds to the magnitude of the field and, as can
be seen in the figure, decreases quadratically as appropriate. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/efieldvec.eps}}
\caption{\sl The electrical field of a point charge depicted by vectors.}
\label{efieldvec}
\end{center}
\end{figure}

Traditionally, the electrical field is depicted by field lines as shown in Fig.~\ref{efield}. The density of the lines represent the magnitude of the field.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/efield.eps}}
\caption{\sl The electrical field of a point charge depicted by field lines.}
\label{efield}
\end{center}
\end{figure}
Note that such drawings are two
 dimensional projections and thus that the density
of the lines drops with the circumference of a circle, $2\pi r$. In three
dimensions the line-density drops with the surface of a sphere, $4 \pi r^2$

Have another look to the field lines in Fig.~\ref{efield} and imagine a spherical surface around a point charge. The number of field lines through the surface
is constant, thus independent on the size (or better radius) of the sphere.
You find an animation of this phenomena on the web-page.

The number of field lines can be expressed by the electrical flux:
\begin{equation}
\Phi=\int_{surface} \vec{E} \dot d\vec{o}
\end{equation}
This the flux through a surface. The infinitesimal surface element 
 $d\vec{o}$ is a vector with magnitude $do$ and its direction
 perpendicular to
the surface. We can write
\begin{equation}
 \vec{E} \dot d \vec{o} = \vec{E} \dot \dakje{n} do= 
  |\vec{E}| cos(\phi) do
\end{equation}
 with $\dakje{n}$ the normal on the surface\footnote{In textbooks several
notations are used for the normal vector, a few common notations
are: $\dakje{n}=\vec{n}=\vec{e}_n=\dakje{e}_n$.}  and 
 $\phi$ represents the angle between the electrical field and 
normal direction of the surface. 


Figure~\ref{fluxes} shows three examples of the electrical flux through a
surface for different angles between the field and surface.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/fluxes.ps}}
\caption{\sl The electrical flux through a surface $S$ for different angles between the field and surface. The normal $\dakje{n}$ is also indicated.}
\label{fluxes}
\end{center}
\end{figure}
With the electrical field being constant, the flux through each surface $S$ in Fig.~\ref{fluxes}a-c is:
\begin{eqnarray}
\rm{a:} \Phi_S &=& |\vec{E}| cos(0)  S=|\vec{E}| S \nonumber \\
\rm{b:} \Phi_S &=& |\vec{E}| cos(\pi/2)  S=0 \nonumber \\
\rm{c:} \Phi_S &=& |\vec{E}| cos(\phi)  S 
\end{eqnarray}

\subsection{Gauss' Law}
Gauss' Law states that the electrical flux through a closed surface, independent of its shape, 
 equals the total charge enclosed by this surface multiplied by
the factor $\frac{1}{\epsilon_0}$.
Thus:
\begin{equation}
\Phi=\oint_{closed-surface}\vec{E} \dot d\vec{o} = 
   \sum_{charges-enclosed} \frac{Q_i}{\epsilon_0}
\end{equation}
The circle in the integral symbol $\oint$ indicates that you have to integrate over a closed surface, without skipping any parts. 
As in many textbook, we often use the normal symbol, $\int$, and mention specifically whether the integral runs over a full or restricted domain.

The relation above allows us to calculate the electrical field inmany cases in an remarkable elegant way. But first we will deduce, or better,
verify the validity of Gauss' Law for a simple case. 
Therefore we consider a point charge, $Q$, 
and calculate the electrical flux through a spherical surface with radius $R$. This configuration
is illustrated in Fig.~\ref{gausspoint}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/gausspoint.ps}}
\caption{\sl Illustration of the electrical flux from a point charge $Q$ 
through a spherical surface $S$. Everywhere, the electrical field points
along the normal on the surface, i.c. the radial direction. Some infinitesimal piece of surface $do$ on the sphere is also indicated. }
\label{gausspoint}
\end{center}
\end{figure}
The flux is given by:
\begin{equation} 
\Phi = \int_{surface} \vec{E} \dot d\vec{o} =
\int_{surface} \vec{E} \dot \dakje{n}  do
\label{eq:gp}
\end{equation}
where $\dakje{n}$ is the normal vector on the surface.
Now we use the argument that on sphere, both $\vec{E}$ and $\dakje{n}$ point in 
the radial direction, hence
\begin{equation} 
\Phi = \int_{surface} |\vec{E}| cos(\alpha_{\vec{E},\dakje{n}}) do = \int_{surface} |\vec{E}| do
\end{equation}
The electrical field at the surface follows from equation~\ref{eq:efield},
 $|\vec{E}|=E_r(R)=\frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$.
We find
\begin{eqnarray}
\Phi &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \int_{surface} do \nonumber \\
     &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} 4 \pi R^2 \nonumber \\
     &=& \frac{Q}{\epsilon_0} 
\end{eqnarray}
 independent of the location on the surface, in accordance with Gauss' Law.

\subsubsection{Intermezzo: a more mathematical approach}
To evaluate (i.c. to get rid off) the dot-product $\vec{E} \dot d\vec{o}$
 we used the argument that both vectors point in the radial direction. 
Perhaps this step is hard to digest and you want to see what is behind it. Well, let's give it a try. If we had started using Cartesian coordinates we would have written:
\begin{eqnarray} 
\Phi & = & \int_{surface} \vec{E} \dot d\vec{o} \nonumber \\
     & = &  \int_{surface} (E_x \dakje{x} + E_y \dakje{y} + E_z \dakje{z}) \dot
 (n_x \dakje{x} + n_y \dakje{y} + n_z \dakje{z}) do 
\end{eqnarray}
with $\dakje{n}=(n_x, n_y, n_z)$ the normal vector on the spherical surface.
 The unit vectors
in the $x$, $y$ and $z$ direction are represented as $\dakje{x}$, 
$\dakje{y}$ and $\dakje{z}$ respectively. Another common notation
for these unit vectors are $\dakje{i}$, $\dakje{j}$ and $\dakje{k}$.

The expression in Cartesian coordinates is  for our case just terrible
 because the physical symmetry of the original problem is obscured.
Obviously we need to work in spherical coordinates. Then, formally we obtain:
\begin{eqnarray} 
\Phi &=& \int_{surface} \vec{E} \dot d\vec{o} \nonumber \\
     &=& \int_{surface} (E_{\phi} \dakje{\phi} + E_{\theta} \dakje{\theta} +  E_r \dakje{r}) \dot (n_{\phi} \dakje{\phi} + n_{\theta} \dakje{\theta} +  n_r \dakje{r})  do \nonumber \\
     &=& \int_{surface} (E_{\phi} n_{\phi} \dakje{\phi} + E_{\theta} n_{\theta} \dakje{\theta} +  E_r n_r \dakje{r}) do 
\end{eqnarray}
With $\dakje{\phi}$, $\dakje{\theta}$ and $\dakje{r}$ the unit direction
vectors in the $\phi$, $\theta$ and $r$ direction respectively.
It looks abstract, (it is), but just try to see through the notation
and realize that the vectors are written out in components. 

We then use $\vec{E}=(0 \dakje{\phi}, 0 \dakje{\theta}, E_r \dakje{r})$ with
 $E_r= \frac{Q}{4\pi \epsilon_0 r^2} $. In fact $E_r$ is a function and depends in general on $\vec{r}$ or in this case for a point charge just on $r$. 
Hence, on the surface $E_r=E_r(r=R)$. The normal vector $\dakje{n}$ on the surface points in the radial direction and has the unit length: $\dakje{n}=\dakje{r}$. When we substitute these expressions and reshuffle, we acquire: 
\begin{eqnarray}
\Phi &=& \int_{surface} \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  do \nonumber \\
     &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi}    R^2 sin(\theta)  d\phi d\theta \nonumber \\
     &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} R^2 \int_{\theta=0}^{\pi}    
       (\phi)|^{2\pi}_0  sin(\theta)  d\theta  \nonumber  \\
    &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 2\pi   (-cos(\theta))|_0^{\pi} \nonumber    \\
    &=&   \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 4\pi     \nonumber \\
    &=&   \frac{Q}{\epsilon_0}    
\end{eqnarray}
Probably this intermezzo did not increase your confidence in Gauss' Law. It showed however the basic steps which many textbooks tend to skip. In the following
we often skip such steps; they contain too much mathematical detail
and so distract from our physics case.
 What was the physics case? Well,  
 the symmetry in the behavior of the electrical field, namely the $1/r^2$ 
dependence, and that of the size of the spherical surface growing with $r^2$
 leads to a cancellation.

\subsection{Validity of Gauss' Law}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/blop.ps}}
\caption{\sl (a) Point charge Q in a closed surface.  (b) Point charge Q outside a closed surface.}
\label{blop}
\end{center}
\end{figure}
We checked Gauss' Law for a spherical surface with a point charge located 
at its center. Figure~\ref{blop}a shows a weird shaped closed surface with
a point charge somewhere in its volume.
 Also for this shape Gauss' Law is valid; 
it is the 'dot' product which does the job. Put the point charge in the origin.
Its electrical field has only radial components. The dot-product with the
normal vector of the surface kills all non-radial contributions. In addition, the distance of the surface with respect to the charge is irrelevant: the field
drops with $1/r^2$ and the radial projection of the surface grows with $r^2$.
Also, we could have put many point charge at different location within the closed surface. Obviously for each point charge individually the contribution
to the flux is its charge (divided by $\epsilon_0$), and the total flux
is given by the sum of all charges (/$\epsilon_0$).

Figure~\ref{blop}b shows a closed surface with a charge located outside
 its volume. Use the above arguments to deduce (qualitatively) that
 $\Phi=0$ in this case.

\subsection{Applications of Gauss's law}
'Gauss' Law is always valid, but not always useful'. What does this mean?
 Well, 
Gauss' Law can be applied in  some cases to evaluate the 
electrical field in an elegant way. 
 Below, we describe a few of these configurations.

\subsubsection{The line charge}
\label{gaussline}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/gline.ps}}
\caption{\sl A line charge (dark gray). A Gaussian cylinder (light gray) with height $h$ and radius $r$ is also drawn. }
\label{gline}
\end{center}
\end{figure}
Figure~\ref{gline} shows an infinitely long and infinitely thin line charge with uniform charge
 density $\lambda$. The optimal
 Gaussian surface is a cylinder. To obtain this insight, first deduce the shape
of the electrical field. It needs some practice to acquire a feeling for this,
but we can give some hints.
\begin{itemize}
\item
 Imagine the line is build from
small point charges $d\lambda$. Every point charge produces a electrical field point in the radial (spherical-wise) direction. 
 Consider a position somewhere near the line. At this position you 'feel'
an equal amount of field lines from above as from below. Hence, the field in the $z$ direction cancels. Obviously there are no components in the $\phi$ direction. There can only be a component pointing away from the line charge.
\item
Look at the symmetry of the line charge. Suppose there is a field component
in the $z$ direction. Remember that the line is infinitely long and imagine that you mirror the configuration in the $r\phi$ plane. This does not change the physical configuration, but the 'would be' $z$ component of the field has changed sign. There is only one possibility: there is no $z$ component.
\end{itemize}
We conclude that the electrical field has only a  radial (cylinder-wise) component. Therefore we try a cylindrical Gaussian surface with arbitrary height 
$h$ and radius $r$. The flux through the cylinder is given by the sum
of the contributions of the curved body and the ends:
\begin{equation}
\Phi=\oint_{cylinder} \vec{E} \dot d\vec{o} = \int_{ends}\vec{E} \dot d\vec{o} + \int_{curved-body}\vec{E} \dot d\vec{o}
\end{equation}
The normal vector on the ends of the cylinder point (only) in the $z$ direction. Hence, the dot-product filters out the $z$ component of the electrical field, which is zero. So, we are left with the contribution of the flux through the curved body of the cylinder. The normal vector is radial (cylinder-wise) and thus filters out the one and only radial component of the electrical field, ($E_r(r)$).We obtain:
\begin{equation}
\Phi=\int_{curved-body} \vec{E} \dot d\vec{o}=\int_{curved-body} E_r(r) do=
\int_{\phi=0}^{2\pi} \int_{z=0}^{z=h} E_r(r) dz r d\phi
\end{equation}
The electrical field is independent on the integration variables and we may write:
\begin{equation}
\Phi= E_r(r) \int_{\phi=0}^{2\pi} \int_{z=0}^{z=h} dz r d\phi = E_r(r) 2\pi r h
\end{equation}
Now we apply Gauss' Law:
\begin{equation}
\Phi= E_r(r) 2\pi r h = \frac{Q_{enclosed}}{\epsilon_0} =\frac{\lambda h }{\epsilon_0} 
\end{equation}
For the electrical field of a line charge we find that:
\begin{eqnarray}
E_z &=& 0   \nonumber \\
E_{\phi} &=& 0 \nonumber \\
E_r(r) &=& \frac{\lambda }{2\pi \epsilon_0 r} 
\end{eqnarray}

If the line has a finite length $L$ this method can not be applied. However, it then still provides a good estimate of the field close to the wire, i.c. $r<<L$. If you are not convinced, use the result of the direct integration in Section~\ref{diline} to show this. Also when the charge density is not constant in the $z$ direction we can
not simply apply Gauss' law. When the line has a finite thickness with radius $\rho$ this method can still be applied as we will see later. 

\subsubsection{A flat surface charge}
\label{fsurf}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/gplate.ps}}
\caption{\sl A flat surface charge (dark-gray). A Gaussian cubical surface is also indicated. }
\label{gplate}
\end{center}
\end{figure}
Figure~\ref{gplate} shows an, infinitely large and infinitely thin, surface charge with uniform charge density $\sigma$.
What are the components of the electrical field?
\begin{itemize}
\item
 Imagine the plate constituted of small point charges $d\sigma=\sigma do$. Every point charge produces an electrical field point in the radial (spherical-wise) direction. 
 Consider a position somewhere near the plate. At this position you 'feel'
an equal amount of field lines from above as from below and from left as from right. Hence, the field in the $x$ and $z$ direction is zero. 
There can only be a component perpendicular to the surface charge.
\item
We can also used arguments based on symmetry. 
Suppose there is a field component in the $x$ and/or $z$ direction. 
Turn the configuration around its $y$ axis. The plate is infinitely large and
thus remains physically the same. The would be components would have changed
direction, while the physics is invariant.
\item
In addition, a shift of the plate in the $x-z$ plane does not change the electrical configuration. This implies that the electrical field does not depend
on $x$ or $z$ and thus only  depends on $y$.
\end{itemize}
We conclude that the electrical field has only a component in the $y$
 direction, opposite for the region $\pm y$.
We try a cubical Gaussian surface (a pill-box) with ribs sized 
$a$. The flux through the sides with normal vector in the $x$ and $z$ direction
is zero. We only need to calculate the flux through
 the top-covers with normal vector in the $y$ direction. The normal vector
in the $-y$ region is opposite to that in the $+y$ region, but the electrical field direction also swaps. Hence,
\begin{equation}
\Phi=\int_{top-covers} \vec{E} \dot d\vec{o}=2 \int_{top-cover} E_y(y) do = 2 a^2 E_y(y)
\end{equation}
Make sure you understand the steps above. 
From Gauss' Law follows
\begin{equation}
\Phi= 2 E_y(y) a^2= \frac{Q_{enclosed}}{\epsilon_0}= \frac{a^2 \sigma}{\epsilon_0}   
\end{equation}
And for the size of the electrical field we obtain $E_y= \frac{\sigma}{2 \epsilon_0}$ and  thus
\begin{equation}
 \vec{E}= \frac{\sigma}{2 \epsilon_0} \dakje{y}.
\label{eplate}
\end{equation} 
Note that the field is constant, but point in the opposite direction for positive and negative $y$ values respectively.
 
In general, thus for non-uniform charge densities this method cannot be used, unless it is known that the
electrical field has only a $\dakje{y}$ component. Then, the result will look like $\vec{E} (x,z) \sim \sigma (x,z) \dakje{y} $.

\subsubsection{A spherical surface charge}
\label{secsurfsphere}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/surfsphere.ps}}
\caption{\sl Illustration of a spherical surface charge density $\sigma$ with radius $R$.}
\label{surfsphere}
\end{center}
\end{figure}
We consider a spherical charged surface (or shell) with radius $R$
 and surface charge density
 $\sigma$ as illustrated in Fig~\ref{surfsphere}.

What are the components of the electrical field? 
Suppose there are non-radial (spherical-wise) components.
 Rotate the configuration around its center such that the non-radial components change direction. Realize that the physical
configuration is invariant allowing no non-radial components. The same argument can be used to deduce that the radial component of the field only depends on $r$. Hence, $\vec{E}=E_r(r) \dakje{r}$.
We did not specify whether we discussed the field inside or outside the shell.
 Well, it doesn't matter. Both inside and outside the shell we can use the
above arguments.

For the electrical flux inside the shell follows:
\begin{equation}
\Phi=\int_{spherical-surface} \vec{E} \dot d\vec{o}=E_r(r) \int_{spherical-surface} do =E_r(r) 4\pi r^2
\end{equation}
There is no enclosed charge, so we have:
\begin{equation}
\Phi =E_r(r) 4\pi r^2 = 0
\end{equation}
There is only one possibility: $E(r)=0$ inside the surface.

Outside the shell, $r>R$, we find the same expression for the flux. The 
enclosed charge is $Q=\int_{surface} \sigma do= \sigma 4\pi R^2$. We obtain:
\begin{equation}
\Phi=E_r(r) 4\pi r^2 = \frac{1}{\epsilon_0} \sigma 4\pi R^2
\end{equation}
For the electrical field follows
\begin{equation}
\vec{E}= E_r(r) \dakje{r} =  \frac{ \sigma R^2 }{\epsilon_0 r^2} \dakje{r}
             \,\,\,\,\, r>R
\label{eshell}
\end{equation}

In fact, the electrical field outside the shell is  identical to
that of a point charge in the center with the same charge as present on  the
shell. This can be easily shown.  We substitute $\sigma=\frac{Q}{4\pi R^2}$ 
in equation~\ref{eshell} and find:
\begin{equation}
\vec{E}= \frac{Q}{4 \pi \epsilon_0 r^2} \dakje{r}
\end{equation}
which is the field of a point charge, as expected.


\subsubsection{A massive spherical charge}
\label{msc}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/gsphere.ps}}
\caption{\sl A spherical volume with radius $R$ carrying a uniform charge density $\rho$ (dark-gray). Two Gaussian spherical surfaces are indicated. One inside and one outside the charge density. }
\label{gsphere}
\end{center}
\end{figure}
Figure~\ref{gsphere} shows a spherical volume charge density $\rho$. The volume has radius $R$.
Like in the previous example, we have spherical symmetry.
The electrical field has only a radial component and depends only on the
radial distance: $\vec{E}=E_r(r) \dakje{r}$.

The electrical flux outside the sphere, $r>R$, is given by:
\begin{equation}
\Phi=\int_{spherical-surface} \vec{E} \dot d\vec{o}=E_r(r) \int_{spherical-surface} do =E_r(r) 4\pi r^2
\end{equation}
Outside the spherical charge the enclosed charge is
 $\int_{volume} \rho dv= \rho \frac{4}{3}\pi R^3$.
For the electrical field we find:
\begin{equation}
\vec{E}= \frac{\rho R^3}{3 \epsilon_0 r^2} \dakje{r} \,\,\,\,\, r>R
\label{eq:epshere}
\end{equation}

Inside the spherical charge, $r<R$, we have in principal two contributions
to the electrical field. One contribution from the inner sphere (surrounded by
the Gaussian surface) with
radius $r$ and a contribution of the shell between $r$ and $R$. In
the previous section we calculated that the electrical field contribution
inside a charged shell is zero. This implies that we only have to account
for the contribution from the inner sphere.
The Gaussian spherical surface encloses a charge $\rho \frac{4}{3}\pi r^3$.
For the electrical flux we find:
\begin{equation}
E_r(r) 4\pi r^2 =\frac{\rho}{\epsilon_0}  \frac{4}{3}\pi r^3
\end{equation}
This leads to an electrical field of
\begin{equation}
\vec{E}= \frac{\rho r}{3 \epsilon_0} \dakje{r} \,\,\,\,\, r<R
\end{equation}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/esphere.ps}}
\caption{\sl The size of the electrical field of a spherical uniform charge density as function of $r$. }
\label{esphere}
\end{center}
\end{figure}
Figure~\ref{esphere} shows the magnitude of the electrical field as function of $r$. Starting from the center, the field grows linearly with $r$ till the 
 surface of the spherical charge is reached. Then it drops with $1/r^2$, similar to
the field of a point charge in the origin.


\subsection{Knowledge and Skills}
The knowledge and skills  
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
Gauss' Law:
\begin{equation}
\int_{closed-surface}\vec{E} \dot d\vec{o} = 
   \sum_{charge-enclosed} \frac{Q_i}{\epsilon_0} =
   \frac{1}{\epsilon_0}  \int_{volume} \rho dv \nonumber
\end{equation}
which is always valid.
\item
In electrical configuration with a symmetry between the field and
the charge distribution, Gauss' Law can be used to 
determine the electrical field.
\item
You can apply Gauss' Law for a line, flat surface, and spherical charge.
You know how to use Cartesian, cylinder and spherical coordinates to
perform surface and volume integrals.
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.


\section{More on electrical field equations}
Gauss' Law is a so called field equation. It uses an (surface) integral
and is therefore called an integral equation. There exists also
a differential form of this law, which we derive in this section.

However, Gauss' Law does not specify all properties of the electrical field.
One more field equation is required. The integral
form of the second field equation is based on the loop integral of the
 electrical field as we will see below.



\subsection{Flux and divergence}
The divergence of the electrical field is defined as
\begin{equation}
\divnodot \dot \vec{E}=( \partial_x  , \partial_y   , \partial_z ) \dot (E_x, E_y, E_z) =
\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} 
\end{equation}
We will introduce this quantity in a natural way and describe its link 
with Gauss' Law.


\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/box.ps}}
\caption{\sl An infinitesimal box with volume= $dxdydz$ in an electrical field $\vec{E} (x,y,z)$. The covers a to f are indicated. Also, the field vectors on cover
a and b are shown. }
\label{box}
\end{center}
\end{figure}
Figure~\ref{box} shows a infinitesimally small box, which is placed in an unspecified 
electrical field $\vec{E} (x,y,z)$. We work in Cartesian coordinates and calculate the electrical
 flux through the box. The flux $\int_{cover} \vec{E} \dot d\vec{o}$ through each of the covers a to f  of the box is given by:
\begin{eqnarray}
\rm{a:}& & -E_x(x,y,z) dydz \nonumber \\
\rm{b:}& &  E_x(x+dx,y,z) dydz \nonumber \\
\rm{c:}& & -E_y(x,y,z) dxdz \nonumber \\
\rm{d:}& &  E_y(x,y+dy,z) dxdz \nonumber \\
\rm{e:}& & -E_z(x,y,z) dxdy \nonumber \\
\rm{f:}& &  E_z(x,y,z+dz) dxdy
\end{eqnarray}
where $E_x (x,y,z)$ and so on are defined on the centre of the corresponding cover.
Note the relative minus sign for opposite covers which comes from the opposite
direction of the normal vectors on these covers.
Now we add all contributions to obtain the flux through the box:
\begin{eqnarray}
\int_{box} \vec{E} \dot d\vec{o} =
 & & [E_x(x+dx,y,z) - E_x(x,y,z)] dydz \label{c1} \nonumber \\
 & & [E_y(x,y+dy,z) - E_y(x,y,z)] dxdz \label{c2} \nonumber \\
 & & [E_z(x,y,z+dz) - E_z(x,y,z)] dxdy \label{c3} 
\end{eqnarray}
Remember the rule of elementary calculus that $df=f(x+dx)-f(x)$. The above equation can be rewritten as:
\begin{equation}
\int_{box} \vec{E} \dot d\vec{o} = dE_x dydz + dE_y dxdz + dE_z dxdy
\end{equation}
Now multiply the part with $E_x$, $E_y$ and $E_z$ with $\frac{dx}{dx}$, $\frac{dy}{dy}$ and $\frac{dz}{dz}$ respectively, which is (for physicists) mathematically equivalent to multiplying with unity.
We find
\begin{eqnarray}
\int_{box} \vec{E} \dot d\vec{o} &=& \frac{dE_x}{dx} dxdydz + 
                       \frac{dE_y}{dy} dxdydz + \frac{dE_z}{dz} dxdydz 
\nonumber \\
		       &=& \left[ \frac{dE_x}{dx} +
                       \frac{dE_y}{dy} + \frac{dE_z}{dz} \right] dxdydz \nonumber \\
		       &=&   \divnodot \dot \vec{E} \, \rm{volume_{box}}  
\end{eqnarray}
Hence, we derived a relation between the flux through infinitesimal box
and the divergence of the electrical field. The relation is however valid for
any volume. To see this, glue boxes together to make you any volume
 as illustrated in Fig.~\ref{boxes}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/boxes.ps}}
\caption{\sl Any volume consists of a collection of infinitesimal boxes.}
\label{boxes}
\end{center}
\end{figure}
 We obtain:
\begin{equation}
\label{gtheorem}
\int_{surface} \vec{E} \dot d\vec{o} = \int_{volume} [ \divnodot \dot \vec{E} ] dv
\end{equation}
Although we derived this expression for the electric field, it is valid for
any vector field and was first derived by Gauss. 
We will refer to this expression as Gauss' Theorem.

\subsection{Gauss' Law and Gauss' Theorem}
Gauss' Law for the electrical flux:
\begin{equation}
\int_{surface} \vec{E} \dot d\vec{o} = \frac{1}{\epsilon_0} Q_{enclosed}
   = \frac{1}{\epsilon_0} \int_{volume} \rho dv
\end{equation}
can be combined with Gauss' theorem:
\begin{equation}
\int_{surface} \vec{E} \dot d\vec{o} = \int_{volume}  \divnodot \dot \vec{E}  dv
\end{equation}
This leads to the following expression:
\begin{equation}
\int_{surface} \vec{E} \dot d\vec{o} = \int_{volume} \divnodot \dot \vec{E}  dv
 = \frac{1}{\epsilon_0} \int_{volume} \rho dv
\end{equation}
which implies
\begin{equation}
 \divnodot \dot \vec{E} = \frac{\rho}{\epsilon_0} 
\end{equation}
This relation is Gauss' Law in differential form. It locally relates the
charge density and the electrical field. 

\subsection{ Gauss' Law for a charged sphere}
Given the electrical field, we can 'simply' determine the charge density using
Gauss' Law (in differential form). We start with the known field
of a uniformly charged sphere with density $\rho$.
 The electrical field inside the sphere is:
\begin{equation}
\vec{E}= \frac{\rho r}{3 \epsilon_0} \dakje{r}=  
         \frac{\rho}{3 \epsilon_0} \vec{r}  \,\,\,\,\, r<R
\end{equation}
Now take the divergence.
\begin{eqnarray}
\divnodot \dot \vec{E} &=& \frac{\rho}{3 \epsilon_0} \divnodot \dot \vec{r}=
 \frac{\rho}{3 \epsilon_0} (\partial_x , \partial_y, \partial_z) \dot (x,y,z) \\
&=&
 \frac{\rho}{3 \epsilon_0} (\partial_x x + \partial_y y + \partial_z z)
=  \frac{\rho}{3 \epsilon_0} 3= \frac{\rho}{\epsilon_0} 
\end{eqnarray}
Yes!

Outside the charge sphere $\rho=0$, the electrical field is $ \vec{E}= \frac{\rho R^3}{3 \epsilon_0 r^3} \vec{r} $. We  calculate the divergence
 of $\frac{\vec{r}}{r^3}$:
\begin{eqnarray}
\divnodot \dot  \frac{\vec{r}}{r^3} & =&  
  \divnodot \dot [ \frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}} (x,y,z) ]  \nonumber \\
 & =& \divnodot \dot     ( \frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}} , 
               \frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}} , 
            \frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}} )    \nonumber  \\
 & =&      \frac{\partial}{\partial x} \frac{ x}{(x^2+y^2+z^2)^{\frac{3}{2}}}
          +   \frac{\partial}{\partial y} 
          \frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}} + 
   \frac{\partial}{\partial z}  \frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}} 
                                                    \nonumber    \\
 & =&   \left[ \frac{\partial x}{\partial x}\right] 
      \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}} +
       x \left[ \frac{\partial}{\partial x} 
            \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}}\right] +
       \partial_y .... + \partial_z .....        \nonumber    \\
  & =&  1  \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}} +
      x \left[-\frac{3}{2} 2x    \frac{ 1}{(x^2+y^2+z^2)^{\frac{5}{2}}}\right]
      +    \partial_y .... + \partial_z .....             \nonumber \\
 &= &  \frac{ 3}{(x^2+y^2+z^2)^{\frac{3}{2}}} - 3(x^2+y^2+z^2)  
   \frac{ 1}{(x^2+y^2+z^2)^{\frac{5}{2}}}  \nonumber \\
 & =& 0
\end{eqnarray}
Not convinced? Try it yourself!
 
\subsection{The loop integral of the electrical field}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/path.ps}}
\caption{\sl An illustration of a path integral in the electrical field of a point charge  $q$.}
\label{path}
\end{center}
\end{figure}
Another important characteristic of the electric field emerges when we consider
the path integral of the field of a point charge $q$:
\begin{equation}
\int_a^b \vec{E} \dot d\vec{l} = \frac{q}{4\pi \epsilon_0} \int_a^b \frac{1}{r^2} \dakje{r} \dot d\vec{l} = \int_a^b E_r \dakje{r} \dot d\vec{l}
\end{equation}
The electrical field is pointing purely radially. No matter what the exact path is followed from $a$ to $b$, the dot-product filters out the radial component ($\dakje{r} \dot d\vec{l}=dr$). Thus
we can replace the path integral by:
\begin{equation}
\int_{r_a}^{r_b} E_r \dot dr = \frac{q}{4\pi \epsilon_0} \int_{r_a}^{r_b} \frac{1}{r^2} dr = \frac{q}{4\pi \epsilon_0} \frac{-1}{r}|_{r_a}^{r_b} =   \frac{q}{4\pi \epsilon_0} (\frac{1}{r_a} -  \frac{1}{r_b})
\end{equation}
For a closed path $r_a=r_b$ we obtain:
\begin{equation}
\int \vec{E} \dot d\vec{l} = 0
\end{equation}
Thus, independent of the path we followed, the integral of a closed path of the
electric field is zero. Using the superposition principle we can argue that this relation derived for a point charge is valid for any charge density.

The expression $\int \vec{E} \dot d\vec{l} = 0$ is the second electrical field
 equation in integral form and has no historical name.
 The differential form $\rot{\vec{E}}=\vec{0}$ we will derive later.

\subsection{Knowledge and Skills}
The knowledge and skills  
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You understand Gauss' theorem and the relation between the electrical field
an the charge density:
\begin{equation}
\int_{surface} \vec{E} \dot d\vec{o} = \int_{volume} \divnodot \dot \vec{E}  dv
 = \frac{1}{\epsilon_0} \int_{volume} \rho dv   \nonumber
\end{equation}
\item
You can apply the field equation:
\begin{equation}
 \divnodot \dot \vec{E} = \frac{\rho}{\epsilon_0}  \nonumber 
\end{equation}
which is Gauss' Law in differential form.
\item
You can derive and use the loop integral of the electrical field:
\begin{equation}
\int \vec{E} \dot d\vec{l} = 0, \nonumber
\end{equation}
independent of the path we followed.
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.



\section{The electric potential}
In this Section we will introduce the electric potential. It turns
out that the electric potential is a powerful quantity to
calculate the electric field of complex charge configurations.
 However, within the scope of this report we have to limit to  
more straightforward but elegant examples. 

\subsection{Work in a gravitational field}
To refresh your memory we first consider work and potential energy
in a gravitational field. The work, $W_{person}$ when you lift an object with mass $m$ from
the ground to height $h$ is
\begin{equation}
W_{person}=\int_0^h \vec{F} \dot d\vec{l}
\end{equation}
with the force,  $|\vec{F}|=m g$, the mass times the gravitational constant.
To lift the object we need to apply a force $\vec{F}=-\vec{F}_G=mg \dakje{l}$ and
thus the the total amount of work needed is:
\begin{equation}
W_{person}=mgh
\end{equation}
The potential energy of the object equals $U=W_{person}=mgh$. In
the following Section we apply this principle to the electrical
field.


\subsection{Potential energy in an electric field}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/qpath.ps}}
\caption{\sl A test charge $q$ in the field of a source charge $Q$ is brought in from infinity to  $P$.}
\label{qpath}
\end{center}
\end{figure}
Consider a (test) charge $q$ in the electrical field of a
 point charge $Q$ at position $P$, see also Fig.~\ref{qpath}. The electrical force on the test charge is $\vec{F}_{elec}=q \vec{E}$. The (minimum) force you must exert on $q$ to move it opposite to the electrical field is -$q \vec{E}$.
 The potential energy of the configuration is defined as
the minimal work needed (for you) to bring the test charge $q$ from
 infinity to $P$.
\begin{equation}
U_P=W_{person}= - \int_{\infty}^P q \vec{E} \dot d\vec{l} 
\end{equation}

%Where did the minus sign go?
% First the integration variable $\vec{l}$ ran from $\inf$ to $P$.
% In the last step, we changed the variable to $\vec{r}$, the radial distance
% with respect to the source charge $Q$ which points opposite to $\vec{l}$. 

It must be emphasized that the path followed $d\vec{l}$ in principle has a radial and non radial components (the $\dakje{\phi}$ and $\dakje{\theta}$ direction).
However, as before we argue that the electrical field has only a radial 
component and we can write:
\begin{equation}
U_P= -q  \int_{\infty}^{P}  \vec{E} \dot d\vec{l} =
 -q  \int_{\infty}^{P}  E_r \dakje{r} \dot d\vec{l} = 
 -q  \int_{\infty}^{P=r_P}  E_r dr 
\end{equation}
Now substitute the electrical field of a point charge and obtain:
\begin{equation}
U_P= \frac{-qQ}{4\pi \epsilon_0} \int_{\infty}^{r_P} \frac{1}{r^2} dr
 = \frac{qQ}{4\pi \epsilon_0}  \frac{1}{r} |^{r_P}_{\infty}
=  \frac{qQ}{4\pi \epsilon_0}   \frac{1}{r_P}
\end{equation}
A definition of the 'potential' $V$ is the potential energy of a charge unit 
 in the field of the source charge $Q$:
\begin{equation}
V_P=\frac{U_P}{q}=  \frac{Q}{4\pi \epsilon_0}  \frac{1}{r_P}
\label{eq:potq}
\end{equation}
where the reference point or 'gauge-point' is implicitly taken at infinity.


To calculate the potential for a collection of charge, we  simply extend
 equation~\ref{eq:potq} and integrate over all point charges in the collection:
\begin{equation}
\label{vcollection}
V_P= \int_{charge} \frac{dq}{4\pi \epsilon_0} \frac{1}{r}
\end{equation}
and for a continuous charge density $\rho$ we write:
\begin{equation}
\label{vdensity}
V_P=\int_{volume} \frac{\rho(r)}{4\pi \epsilon_0} \frac{1}{r} d\vec{r}
\end{equation}


\subsection{The definition of the electric potential}
The most general definition of the potential is
\begin{equation}
\label{vfrome}
V_P= - \int_{gauge}^{P}  \vec{E} \dot d\vec{l}  
\end{equation}
where the gauge-point can be chosen where-ever you want.
You can verify yourself that for a point charge and gauge-point at infinity you find back equation~\ref{eq:potq}. 


We have defined the potential starting the potential energy of a simple charge configuration. Fine, the potential and the potential energy are related; that is useful to know. But, what about the general definition of the potential? It depends in general on the free choice of a gauge-point. How can such freedom be useful for describing physics?  
Indeed, the potential itself has no physical interpretation. However,
potential difference and as we will see soon, 
the gradient of the potential are relevant 
  physics quantities.

\subsection{The potential and the electrical field}
 From the previous section we know how to calculate the potential from the electrical field. Is it also possible to determine the electrical field given the
potential. To find this relation, consider the difference in potential between points $A$ and $B$:
\begin{equation}
V_{AB}=V_{B}-V_{A}=\int_B^{\infty} \vec{E} \dot d\vec{l}
 - \int_A^{\infty} \vec{E} \dot d\vec{l}= -\int_A^{B} \vec{E} \dot d\vec{l}
\label{pdive}
\end{equation}
where we swapped the upper and lower boundary of the integral, so
keep track of the 'plus and minus' signs!
Note that the potential difference $V_{AB}$ is uniquely defined,
 independent of the gauge-point. 

The potential is just a scalar function. Hence,
\begin{eqnarray}
V_B-V_A=\int^B_A dV &=& \int^B_A \frac{\partial V}{\partial x} dx +
                             \frac{\partial V}{\partial y} dy +
                             \frac{\partial V}{\partial z} dz \nonumber     \\
                     &=&  \int^B_A
			  \left(   \frac{\partial V}{\partial x} ,
                             \frac{\partial V}{\partial y} ,
                             \frac{\partial V}{\partial z} \right) \dot
                         \left(  dx, dy, dz \right) 
\end{eqnarray}  
With the standard definition of the gradient:
\begin{equation}
\divnodot = \left(   \frac{\partial }{\partial x} ,
                             \frac{\partial }{\partial y} ,
                             \frac{\partial }{\partial z} \right) 
\end{equation}
we find:
\begin{eqnarray}   
        V_B-V_A           &=&  \int^B_A
                             \divnodot V \dot
                         \left(  dx, dy, dz \right) \nonumber \\
                     &=&  \int^B_A  \divnodot V \dot d\vec{l} \label{pdivg}
\end{eqnarray}     
Combining Equations~\ref{pdive} and ~\ref{pdivg} leads to:
\begin{equation}
\vec{E}=-\divnodot V
\end{equation}
Thus the gradient of $V$ has a physical interpretation; it is the electrical field (with a minus sign). This closes the circle. We can now determine the potential from the electric field and vice versa.

For completeness we express Gauss' Law in terms of the electric potential.
For that we need the Laplacian operator, $\divnodot^2$, defined by:
\begin{equation}
\divnodot^2 = \left(   \frac{\partial^2 }{\partial x^2} +
                             \frac{\partial^2 }{\partial y^2} +
                             \frac{\partial^2 }{\partial z^2} \right).
\end{equation}
Now we go back to Gauss' Law and derive:
\begin{eqnarray}
 \divnodot \dot \vec{E} & =      & \divnodot \dot (-\divnodot V) \\ \nonumber
                   & = & -\divnodot^2 V = \frac{\rho}{\epsilon_0}
\label{eq:poisson}   
\end{eqnarray}
which is usually called  'Poisson's equation'. 
Let's throw in one more definition. In the absence of any charge,
Poisson's equation becomes:
\begin{equation}
   \divnodot^2 V =0
\label{eq:laplace}
\end{equation}
which is called Laplace's equation.

\subsubsection{The electrical potential and field of a point charge}
We determined the potential for a point charge to be:
\begin{equation}
V=\frac{Q}{4\pi \epsilon_0} \frac{1}{r}
\end{equation}
Now we check whether the expression  $\vec{E}=-\divnodot V$ returns the correct electrical field.
We start with:
\begin{eqnarray}
\divnodot \frac{1}{r} &=& \divnodot \frac{1}{\sqrt{x^2+y^2+z^2}} \nonumber \\
   &=& \frac{\partial }{\partial x} \frac{1}{\sqrt{x^2+y^2+z^2}} \dakje{x}
   + \frac{\partial }{\partial y} ..... \dakje{y}
   + \frac{\partial }{\partial z} ..... \dakje{z} \nonumber \\
   &=& - \frac{x}{(x^2+y^2+z^2)^\frac{3}{2}} \dakje{x}
       - \frac{y}{(x^2+y^2+z^2)^\frac{3}{2}} \dakje{y}
       - \frac{z}{(x^2+y^2+z^2)^\frac{3}{2}} \dakje{z} \nonumber \\
   &=& - \frac{1}{(x^2+y^2+z^2)^\frac{3}{2}} (x,y,z) =
     -\frac{\vec{r}}{r^3} =  -\frac{\dakje{r}}{r^2} 
\end{eqnarray}
and conclude that
\begin{equation}
\divnodot V=\frac{Q}{4\pi \epsilon_0} \divnodot \frac{1}{r} = - \frac{Q}{4\pi \epsilon_0}  \frac{1}{r^2} \dakje{r}= - \vec{E}
\end{equation}

Most expressions in electrodynamics involve spherical symmetric function. For such a function, $f(r)$, you can quickly determine the gradient using the following relation.
\begin{equation}
\divnodot f(r)=\frac{df}{dr} \dakje{r}
\end{equation}

\subsubsection{The potential of a charged sphere}
In Section~\ref{msc} we determined the electrical field inside and outside
a massive spherical charge with radius $R$ and charge density $\rho$.
\begin{eqnarray}
\vec{E} &=& \frac{\rho r}{3 \epsilon_0} \dakje{r} \,\,\,\,\, r<R \nonumber \\
\vec{E} &=& \frac{\rho R^3}{3 \epsilon_0 r^2} \dakje{r} \,\,\,\,\, r>R
\end{eqnarray}
As an example we calculate the potential inside the spherical charge.
\begin{equation}
V(r)=-\int_{\infty}^{r} \vec{E} \dot d\vec{l}=
 -\int_{\infty}^{R} \vec{E} \dot d\vec{l} -\int_{R}^{r} \vec{E} \dot d\vec{l}
\end{equation}
We substitute the expression for the electrical field and find:
\begin{eqnarray}
V(r) & = & -\int_{\infty}^{R}    \frac{\rho R^3}{3 \epsilon_0 r^2} dr   -\int_{R}^{r}  \frac{\rho r}{3 \epsilon_0} dr \nonumber \\
     & = & \frac{\rho R^3}{3 \epsilon_0 r} |_{\infty}^{R}   -
           \frac{\rho r^2}{6  \epsilon_0} |_R^r     \nonumber \\
     & = & \frac{\rho R^3}{3 \epsilon_0 R}    -
           \frac{\rho r^2}{6  \epsilon_0 }
         +  \frac{\rho R^2}{6  \epsilon_0}   \nonumber   \\
     & = & \frac{\rho R^2}{2 \epsilon_0} -
           \frac{\rho r^2}{6  \epsilon_0} \label{vinside}
\end{eqnarray}
At this point it may be not yet clear why the potential is relevant anyway. Don't worry about that now and make sure you understand the mathematics. 


\subsubsection{The potential of a circular charge}
We introduced the potential and perhaps the idea came to your mind 'where do we need that for'. Well, to calculate the electrical field of complex charge configuration can be a difficult task, even numerically~\footnote{there will be an exciting exercise in the second year course on 'Numerical Physics'}. Starting with a calculation of the potential and then calculate the electrical field is often much
easier.

We want to calculate the electrical field at a distance $d$ of a circular 
charge with radius $R$ as indicated in Fig.~\ref{ring}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/ring.ps}}
\caption{\sl A circular charge positioned parallel to the $XY$ plane, centered at $z=d$ with radius $R$ carries a uniform charge density $\lambda$. Point $P$, that coincides with the origin is also indicated.}
\label{ring}
\end{center}
\end{figure}
The charge is positioned parallel to the $XY$ plane,
 centered at $z=d$ with uniform charge density $\lambda$. To calculate the field at distance $d$, which is the origin in our case, we first calculate
the electrical potential using equation~\ref{vcollection}:
\begin{equation}
V_P=\int_{circular-charge} \frac{dq}{4\pi \epsilon_0} \frac{1}{r} 
\end{equation}
with $r=\sqrt{R^2+d^2}$ the distance to a piece of charge $dq=\lambda R d\phi$.
Rather straightforward we obtain:
\begin{equation}
V_P=\int_{0}^{2\pi} \frac{\lambda}{4\pi \epsilon_0} \frac{R}{\sqrt{R^2+d^2}}
 d\phi=  \frac{\lambda}{2 \epsilon_0} \frac{R}{\sqrt{R^2+d^2}}
\end{equation}
When we define the total charge $Q=2\pi R \lambda$, the above expression
becomes:
\begin{equation}
V_P= \frac{Q}{4 \pi \epsilon_0} \frac{1}{\sqrt{R^2+d^2}}
\end{equation}

We can now obtain the electrical field  
from the potential using $\vec{E}=-\divnodot V$. The general expression for
$V$ somewhere along the $z$ axis is:
\begin{equation}
V(z)= \frac{Q}{4 \pi \epsilon_0} \frac{1}{\sqrt{R^2+(z-d)^2}}
\end{equation}
From the symmetry of this configuration we know that the field in the $x$ or $y$ direction is zero and thus only the derivative to $z$ has to be considered.
For the electrical field follows:
\begin{eqnarray}
E_z &=& - \frac{\partial V }{\partial z} \dakje{z} \nonumber \\
  &=&  - \frac{Q}{4 \pi \epsilon_0}   \frac{\partial  }{\partial z}  \frac{1}{\sqrt{R^2+(z-d)^2}}  \dakje{z}\\ \nonumber
 &=& \frac{Q}{4 \pi \epsilon_0}  \frac{(z-d)}{(R^2+(z-d)^2)^{3/2}} \dakje{z}
\end{eqnarray}
Of course, this result can also be obtained from direct integration, but
that requires a somewhat more complicated integration. Just do it and check the result!


\subsubsection{The electrical field of a dipole}
Another example where the electrical potential makes life easier
is the calculation of the field of an electrical dipole.
 Figure~\ref{dipole} shows a electrical dipole configuration, consisting of two opposite point charges at a distance 2$d$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/dipole.ps}}
\caption{\sl An electrical dipole consisting of  two opposite point charges at distance 2$d$.}
\label{dipole}
\end{center}
\end{figure}
The potential in a point $P(r,\theta)$ has a contribution from the positive and negative point charge. Thus,
\begin{equation}
V_P(r,\theta)=V_+ + V_- = \frac{q}{4\pi \epsilon_0} \frac{1}{r_+} +
 \frac{-q}{4\pi \epsilon_0} \frac{1}{r_-}
\end{equation}
with $r_{\pm}$ the distance between the corresponding point charge and point $P$. When $r>>d$ we can make the approximation:
\begin{equation}
r_{\pm} = r \mp d cos(\theta)
\end{equation}
Verify this yourself. For the potential in $P$ we obtain:
\begin{equation}
V_P(r,\theta) = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{r-d cos(\theta)} -
       \frac{1}{r+d cos(\theta)} \right) 
\end{equation}
We can simplify this expression, using a 'cunning' trick. Multiply the numinators and denominators with $r+d cos(\theta)$ and $r-d cos(\theta)$ respectively and
realize that $(r+d cos(\theta)) (r-d cos(\theta))=r^2 - d^2 cos^2(\theta)$.
Remember that we work in approximation $r>>d$ and thus 
 $r^2-d^2 cos^2(\theta) \simeq r^2$. The leads to
\begin{eqnarray}
V_P(r,\theta) &=& \frac{q}{4\pi \epsilon_0} \frac{2 d cos(\theta)}{r^2} \nonumber \\ 
&=&  \frac{2q d cos(\theta)}{4 \pi \epsilon_0 r^2} \nonumber \\ 
&\equiv&  \frac{p cos(\theta)}{4 \pi \epsilon_0 r^2} =  
\frac{\vec{p} \dot \dakje{r} }{4 \pi \epsilon_0 r^2} = 
\frac{\vec{p} \dot \vec{r} }{4 \pi \epsilon_0 r^3}  
\end{eqnarray}
where $\vec{p}=2 q \vec{d}$. The quantity $\vec{p}$ is called the dipole moment. Mathematical dipoles have distance $d\rightarrow 0$ and
 $q \rightarrow \infty$, such that $p=2qd$ remains constant.


Now we determine the electrical field by taking the gradient of the potential:
\begin{eqnarray}
\vec{E}_P &=&-\divnodot V_P=-\divnodot \left(\frac{\vec{p} \dot \vec{r} }
                {4 \pi \epsilon_0 r^3} \right)         \nonumber     \\
&=&-\frac{1}{4 \pi \epsilon_0}  \left( \frac{\partial}{\partial x} 
 \frac{xp_x + yp_y +zp_z}{ r^3}, ..........., ..........\right) \nonumber   \\
&=&-\frac{1}{4 \pi \epsilon_0}  \left(  \frac{p_x}{ r^3}
  -\frac{3x(xp_x + yp_y +zp_z)}{ r^5} , ..........., ..........\right) \nonumber \\
&=&\frac{1}{4 \pi \epsilon_0}  \frac{-\vec{p}+3 \dakje{r}(\dakje{r} \dot \vec{p}) }{ r^3}  
\end{eqnarray}
The drop of the magnitude of the field with $r^3$ is characteristic for
 a dipole field.
             


\subsubsection{The potential of a line charge}
Several examples convinced us starting with the electrical
 potential simplifies the calculation of the electrical field. The following
example shows that we have to be careful.
 
We calculate the electric potential of an infinitely long line charge
 in point $P$ by integrating over
all charge on the line ( see Equation~\ref{vcollection}):
\begin{equation}
V_P=  \int_{line}  \frac{dq }{4 \pi \epsilon_0 r}  
\end{equation}
We put the line on the z-axis and thus $dq=\lambda dz$. Hence,
\begin{eqnarray}
V_P &=&  \int_{-\infty}^{+\infty}  \frac{\lambda }{4\pi \epsilon_0 \sqrt{r_P^2+z^2}} dz  \\
    &=&  \frac{\lambda }{4\pi \epsilon_0} ln(z +\sqrt{r_P^2+z^2})  |_{-\infty}^{+\infty}  = \rm{undefined!} 
\end{eqnarray}
How can this be? Well, remember our definition of the potential in
 equation~\ref{vfrome} and also look at equation~\ref{eq:potq}.
 We have taken the gauge-point of the potential at infinity, which has been
 a reasonable choice
for charge configurations with a potential vanishing at infinity.
However, the potential of the infinitely long line charge doesn't.

In this case we have to obtain the potential
 by integration of the electrical field.
In Section~\ref{gaussline} we calculated the electrical field at radial
 distance $r$ of an infinitely long line charge:
\begin{equation}
 \vec{E}(\vec{r}) = \frac{\lambda }{2\pi \epsilon_0 r} \dakje{r} 
\end{equation}
and thus
\begin{equation}
V_P= - \int_{gauge}^{P}  \vec{E} \dot d\vec{r} = -   \frac{\lambda }{2\pi \epsilon_0} ln(r) |_{gauge}^{P} 
\end{equation}
When we choose gauge-point $r=1$ we obtain:
\begin{equation}
V_P= - \frac{\lambda }{2\pi \epsilon_0} ln(r)  
\end{equation}
But isn't it bizarre that we can just choose a gauge? Perhaps it is, but
 remember that the potential is not a physical quantity. 
The electrical field is a  physical quantity and also the potential
 difference in two points, which is even easily  measurable, and these quantities 
remain independent of the gauge-point.


\subsection{The energy of a charge configuration}
We have already discussed that the work needed to bring one point charge, $q$
 from infinity to a point $P$ in the field of another point charge equals:
\begin{equation}
U = W= - \int_{\infty}^P q \vec{E} \dot d\vec{l} = \int_{\infty}^P q \divnodot V \dot d\vec{l} = q V_P
\end{equation}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/qcol.ps}}
\caption{\sl Illustration of a collection of point charges.
}
\label{qcol}
\end{center}
\end{figure}

Figure~\ref{qcol} shows a collection of point charges.
To determine the energy of a charge collection (of point charges) we have to 
calculate how much work is required to assemble such collection.
The first charge takes no force (there is no field yet) and thus physically no work is done. To place the second charge, we require $W_2=q_2 V(r_{12})=\frac{q_2}{4\pi \epsilon_0} \frac{q_1}{r_{12}}$, 
where $r_{12}$ represents the distance between charge $q_1$ and $q_2$. 
When we bring in the third charge we feel the field of the first an second charge, thus:
\begin{equation}
W_{3}=q_3 V(r_{13}) + q_3 V(r_{23}) = q_3 \left( \frac{q_1}{4 \pi \epsilon_0 r_{13}}  +  \frac{q_2}{4 \pi \epsilon_0 r_{23}}  \right)
\end{equation}
The total energy of our collection so-far is
\begin{equation}
W_{123}=W_1 + W_2 + W_3 = 0 + \frac{q_1 q_2}{4 \pi \epsilon_0 r_{12}}  +  
                              \frac{q_1 q_3}{4 \pi \epsilon_0 r_{13}}  +
			      \frac{q_2 q_3}{4 \pi \epsilon_0 r_{23}} 
\end{equation}
For a collection of $N$ point charges we find:
\begin{equation}
W_{N}=\frac{1}{2} \sum_{i=1}^N \sum_{j \neq i} \frac{q_i q_j}{4 \pi \epsilon_0 r_{ij}}= \frac{1}{2} \sum_{i=1}^{N} q_i V(r_i)          
\label{wn}
\end{equation}
Note that all combinations of $q_i$ and $q_j$ appear twice, which
is accounted for by the factor $\frac{1}{2}$. Another remark we have to make
is that the self-energy to make the point charges is completely ignored
in the above expressions. 

\subsection{The energy of continuous charge distribution}
For a volume charge density we can generalize Equation~\ref{wn} and obtain:
\begin{equation}
W=\frac{1}{2} \int_{volume} \rho V dv
\label{wv}
\end{equation}
Consider the following question of a smart student.
 There appears a contribution from the charge $\rho dv$, 
sensing its own potential in infinitesimal volume $dv$. Such contribution from
the self energy is not present in equation~\ref{wn}. 
Wouldn't this contribution lead to unphysical results?
 To check the size of this contribution, we imagine that
our continuous charge distribution consists of infinitesimal charged spheres.  
We calculate the self energy of a uniformly charged sphere with (infinitesimal)
 radius $R$, due to its own potential (see equation~\ref{vinside}):
\begin{eqnarray}
W_{self}&=& \frac{1}{2} \int_{sphere} \rho V_{inside} dv =
  \frac{1}{2} \int_{0}^{R} \rho  (   \frac{\rho R^2}{2 \epsilon_0} -
 \frac{\rho r^2}{6  \epsilon_0} ) 4\pi r^2 dr \nonumber \\ 
 &=&\frac{1}{2} \int_{0}^{R}   ( 
   \frac{4\pi \rho^2 R^2 r^2}{2 \epsilon_0} -
    \frac{4\pi \rho^2 r^4}{6  \epsilon_0} )  dr \nonumber \\
 &=&\frac{1}{2}  ( 
   \frac{4\pi \rho^2 R^5} {6 \epsilon_0} -
    \frac{4\pi \rho^2 R^5}{30  \epsilon_0} )  \nonumber \\
 &=&  \frac{4\pi \rho^2 R^5} {15 \epsilon_0} 
\end{eqnarray}
Remember that our sphere is infinitesimal ($R \rightarrow 0$) and thus 
the contribution $W_{self}=0$. Hence, equation~\ref{wv} correctly
represents the energy of a continuous charge distribution.
 

\subsection{The energy in the electrical field}
Starting point is the energy of the continuous charge distribution.
\begin{equation}
W=\frac{1}{2} \int_{volume} \rho V dv
\end{equation}
Substitute Gauss's law $\divnodot \dot \vec{E}=\rho / \epsilon_0 $:
\begin{equation}
W=\frac{\epsilon_0}{2} \int_{volume} (\divnodot \dot \vec{E})  V dv
\label{wdivev}
\end{equation}
We can further simplify this expression. Therefore we first consider the
expression:
\begin{eqnarray}
\divnodot \dot ( \vec{E} V) &=& \frac{\partial (E_x V)}{\partial x} + ...
       \nonumber \\
        &=& (\frac{\partial E_x}{\partial x}) V +   (\frac{\partial V}{\partial x}) E_x  + ......  \nonumber  \\
   &=& (\divnodot \dot \vec{E}) V + \vec{E} \dot (\divnodot V)
\label{divev}
\end{eqnarray}
We return to the energy (equation~\ref{wdivev}) and use 
equation~\ref{divev} to write:
\begin{eqnarray}
W&=&\frac{\epsilon_0}{2} \int_{volume} \left(\divnodot \dot ( \vec{E} V) - \vec{E}
         \dot (\divnodot V) \right)dv \nonumber  \\
 &=&\frac{\epsilon_0}{2} \int_{volume} \left(\divnodot \dot ( \vec{E} V) + \vec{E} \dot \vec{E}  \right) dv \nonumber \\
 &=&  \frac{\epsilon_0}{2}  \int_{volume} 
     \left(\divnodot \dot ( \vec{E} V)  \right) dv +
      \frac{\epsilon_0}{2} \int_{volume}  
      \left( \vec{E}^2 \right) dv
\end{eqnarray}
Using Gauss' Theorem~\ref{gtheorem}, we obtain:
\begin{equation}
W=\frac{\epsilon_0}{2}  \int_{surface}  \left( \vec{E} V \right) d\vec{o}
   + \frac{\epsilon_0}{2}  \int_{volume} \vec{E}^2 dv   
\end{equation}
The integral of the surface equals zero. Why?
 Suppose we consider the field of a point charge. The electrical field drops
 with $1/r^2$ and the potential with $1/r$. 
We can write:
\begin{equation}
  \int_{surface} d\vec{o} \dot  \vec{E} V \approx  \int_{surface} d\vec{o} \dot  1/r^3 \approx  \int_{surface} d\phi d\theta sin(\theta) 1/r
\end{equation}
 We should consider a surface enclosing all space, 
thus $r \rightarrow \infty$ and thus
\begin{equation} 
 \int_{surface} d\phi d\theta sin(\theta)  1/r=0
\end{equation}
Finally, we obtain for the energy of the electrical field:
\begin{equation}
W=\frac{\epsilon_0}{2}  \int_{volume} \vec{E}^2 dv   
\label{uee}
\end{equation}

\subsubsection{$\vec{E}$, V and Energy of a spherical surface charge}
In Section~\ref{secsurfsphere} we calculated the electrical field of
a charged spherical surface (or shell) with radius $R$ and charge 
density $\sigma$.
We found that outside the shell:
\begin{equation}
\vec{E}= E_r \dakje{r}  =   \frac{ \sigma R^2 }{\epsilon_0 r^2} \dakje{r}
            \,\,\,\,\, r>R
\end{equation}
Inside the the sphere no charged is enclosed and thus $E_r=0$.

What is the potential as function of $r$? 
We use $V(r)=-\int_{\infty}^{r} \vec{E} \dot d\vec{l}$ and write:
\begin{equation}
V(r)  =  - \int_{\infty}^{r} dr \frac{ \sigma R^2 }{\epsilon_0 r^2} = \frac{\sigma R^2}{\epsilon_0 r} \,\,\,\,\,\, r>R
\end{equation}
Now, calculate the potential inside the sphere (where the electrical field is zero):
\begin{equation}
V(r)  = -\int_{\infty}^{R} \vec{E} \dot d\vec{l} - \int_{R}^{r} \vec{E} \dot d\vec{l} = V(R) + 0 = \frac{\sigma R}{\epsilon_0} \,\,\,\,\,\, r<R
\end{equation}
You could always check the results for the potential by calculating the electrical field using $\vec{E}= - \divnodot V$.
The results for the electrical field are also shown in Fig~\ref{evsurfsphere},
where you can see that inside the surface the electrical field becomes zero,
while the potential is constant, $V(R)$.   
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/evsurfsphere.ps}}
\caption{\sl The electrical field and the potential of a spherical surface charge} 
\label{evsurfsphere}
\end{center}
\end{figure}

Now we can calculate the energy of this configuration in two ways. We start with the expression for the energy based on the potential (equation~\ref{wv}).
The volume charge density $\rho$ is zero everywhere, except
on the spherical surface where it is $\sigma$. Hence,
\begin{eqnarray}
W &=&\frac{1}{2} \int_{volume} \rho V dv \rightarrow \frac{1}{2} \int_{surface} \sigma V(R) do\\
 & =&  \frac{1}{2} \int_{surface} \sigma \frac{\sigma R}{\epsilon_0} do = \frac{1}{2} 4\pi R^2 \sigma \frac{\sigma R}{\epsilon_0}=2\pi\frac{\sigma^2R^3}{\epsilon_0}
\end{eqnarray}  
The other expression for the energy with the electrical field in quadrature (equation~\ref{uee}) should yield the same result. Let's check that.
\begin{eqnarray}
W & =& \frac{\epsilon_0}{2}  \int_{volume} \vec{E}^2 dv = 
\frac{\epsilon_0}{2}  \int_{r>R} \vec{E}^2 dv \nonumber \\
 & =& 
\frac{\epsilon_0}{2}  \int_{r>R} (\frac{R^2 \sigma}{r^2 \epsilon_0})^2 dv =
\frac{4\pi \epsilon_0}{2}  \int_{r>R} dr (\frac{R^2 \sigma}{\epsilon_0})^2 \frac{1}{r^2} \nonumber \\
& =& \frac{4\pi \epsilon_0}{2} (\frac{R^2 \sigma}{\epsilon_0})^2 \frac{-1}{r}|_R^{\infty} = 2\pi\frac{\sigma^2R^3}{\epsilon_0}
\end{eqnarray}
as expected.


\subsection{Knowledge and Skills}
The knowledge and skills  
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You understand the definition (and the derivation) of the potential: 
\begin{equation}
V_P= - \int_{gauge}^{P}  \vec{E} \dot d\vec{l}  \nonumber
\end{equation}
For a point charge, the potential is given by:
\begin{equation}
V_P=\frac{U_P}{q}=  \frac{Q}{4\pi \epsilon_0}  \frac{1}{r_P} \nonumber
\end{equation}
\item
you can also calculate the electrical field given the potential, using
$ \vec{E}=-\divnodot V$.
\item
you understand how we derived the energy of a charge collection:
\begin{equation}
W_{N}=\frac{1}{2} \sum_{i=1}^N \sum_{j \neq i} \frac{q_i q_j}{4 \pi \epsilon_0 r_{ij}}= \frac{1}{2} \sum_{i=1}^{N} q_i V(r_i)  
\end{equation}
\item
For a continuous charge density the energy is given by:
\begin{equation}
W=\frac{1}{2}  \int_{volume} \rho V  dv   \nonumber
\end{equation}
or, alternatively:
\begin{equation}
\label{eq:wfromee}
W=\frac{\epsilon_0}{2}  \int_{volume} \vec{E}^2 dv   \nonumber
\end{equation}
which you can apply for a spherical charge density.
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.



\section{Electrical fields in matter}
So-far, we considered electrical fields in vacuum. Usually we started with some (symmetrical) charge configuration and then we calculated the electrical field, its potential or energy. What if we place (electrically neutral) objects
in the electrical  field? What happens inside those objects and what is
the effect on the electrical field in and outside the object? In this
Section we will discuss these questions and more. 

\subsection{The Conductor}
What is a conductor? For our purposes a conductor is an object that conducts
electrical currents because the negative charge carriers (electrons) can move
freely inside the material. The number of the free charge carriers is
unlimited. Materials that approach these ideal properties are metals
like iron, copper and gold. If the conductor is electrically neutral, it contains an equal amount of negative and positive charge. The positive charge is
always bound and thus cannot move freely through the material.

Given the above we can deduce what happens when a conductor is placed inside
an electrical field as illustrated in Fig.~\ref{conductor}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/conductor.ps}}
\caption{\sl 
A conductor is placed in an extern electrical field.} 
\label{conductor}
\end{center}
\end{figure}
\begin{itemize}
\item
What is the electrical field inside a conductor? Suppose there is an electrical
 field in the conductor. Then, the free electrons would be subjected 
to the electrical force and start moving. Well, they may do for a short
time when the field is 
just turned on, but we discuss only electrostatic situations.
There is only one stable solution to our question and that is that there
is no electrical field inside a conductor!
\item
How can the field be zero inside a conductor? Well, suppose the electrical
field is just turned on. Then some free electrons will be attracted
by the electrical force and flow to the surface of the conductor (they can not
escape). This process goes on till the electrical field inside the
conductor has vanished, or better, the original field gets canceled by the
field of the free charge sitting on the surface and the nett positive
charge that keeps its original position~\footnote{This is always possible
in one, and only one, way based on the 'uniqueness theorem'.
The derivation of this theorem is beyond the scope of this course.}.
 In general, the nett positive charge
will appear on the opposite surface with respect to the side of the free
negative charge.
\item
What is the charge density inside the conductor? We know now that the field
inside the conductor is zero. Hence, $\divnodot \dot \vec{E}=\rho / \epsilon_0 =0$
The charge density $\rho$ inside the conductor must be zero. Thus, any
nett negative or positive charge must sit on the surface(s) of the conductor.
\item 
What about the electrical field on the surface? Suppose there is an electrical field along the surface of the conductor. The free charge carriers will immediately 'respond' and flow into a configuration with no electrical
charge along the surface. Hence, there are no electrical field components along the surface of a conductor. Now suppose there is electrical field perpendicular
to the surface. The charge sitting just on the surface
 will be attracted or repelled by the field, but it cannot move inward or 
outward the conductor. This is a stable situation and thus
 there can be an electrical field just outside an conductor
 perpendicular to its surface.
\item
Is there a potential in a conductor?
No, there can't be, because $V(a) - V(b) = \int_a^b \vec{E} \dot d\vec{l}=0$. 
 At any place inside or at the surface of a conductor 
the potential is constant.
\end{itemize}

\begin{table}[thb]
\begin{center}
\begin{tabular}{|c|l|l|} \hline
\# & characteristic &                           why   \\ \hline
1 & $\vec{E}=0$ inside a conductor      &    otherwise charge starts moving. \\ 
2 & $\rho=0$ inside a conductor         &  $\divnodot \dot \vec{E}=0=\rho / \epsilon_0$. \\
3 & charge sits on the surface          & where-else could it be? \\
4 & $V$ is constant inside a conductor  & $V(a) - V(b) = \int_a^b \vec{E} \dot d\vec{l}=0$. \\
5 & Field lines leave the conductor  & otherwise the surface-charge starts \\ 
 & perpendicular.    &  moving. \\ \hline
\end{tabular}
\caption{\sl The conductor rules. }
\label{t:conductor}
\end{center}
\end{table}
Table~\ref{t:conductor} summarizes the electric characteristics of the
 conductor.
So, we know now what happens if we place a conductor inside an
 electrical field. Well, not really. The (original) field outside the
conductor also changes and we have not discussed that. This is a though 
problem to solve without a general solution.
 When there is an obvious symmetry in the configuration
we can give the solution. It will take a lot of practice
to adopt a 'feeling' for these configurations.
Anyway, the examples in the following paragraph help you to get started.

\subsubsection{A conducting plate in a uniform electrical field.}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=12cm \epsffile{conplatecharge.ps}}
\caption{\sl a) 
An infinitely large conducting plate is placed in an external electrical field
 in the $z$ direction. b) The physical equivalent charge configuration on the
plate.} 
\label{conplate}
\end{center}
\end{figure}
We start with a given uniform electrical field $E_{external}$
 in the horizontal ($z$) direction and then
place an infinitely large conducting plate as shown in Fig.~\ref{conplate}a.
After a few nanoseconds an electrostatic configuration exists.
What is the resulting electrical field? 

Start by applying the rules of conductors (Table~\ref{t:conductor}) to get the nett charge configuration. Then, with the charge configuration we can calculate the electrical field everywhere using standard techniques  as we have been doing in the previous Sections.

So, we know that the nett charge will sit on the
surface (rule~2~and~3). We can replace the plate by the electrical 
configuration as shown in Fig.~\ref{conplate}b. This situation is physically
equivalent. Of course we also need to know the amount of charge, or better, the
charge density $\sigma_-$ and $\sigma_+$. We started with a neutral plate, so for sure   $\sigma=\sigma_+=- \sigma_-$. According to rule~1, the field inside
the conductor shall be zero. Hence, in the region where the plate
 resided, the electrical fields from the charge
densities have to cancel the external field:
\begin{equation}
\vec{E}_{in-plate}=\vec{0}= \vec{E}_{external} + \vec{E}_{(-)} + \vec{E}_{(+)}  \end{equation}
 We have already  calculated the electrical field of a flat surface charge: 
$E_{\pm}=\frac{\sigma_{\pm}}{2 \epsilon_0}$, which we substitute in
the above expression:
\begin{equation}
\vec{0} = |\vec{E}_{external}| \dakje{z} + \frac{-\sigma}{2\epsilon_0} \dakje{z} - \frac{\sigma}{2 \epsilon} \dakje{z} 
\end{equation}
from which follows:
\begin{equation}
 \sigma =  \epsilon_0  |\vec{E}_{external}|
\end{equation}
Note that the (vector) contribution of the positive charge has acquired 
a minus sign, because it points in the negative $z$ direction, 
which is also illustrated
in Fig.~\ref{confield} (for the sake of the argument we made the external
field vector fit just inside the plate). Inspect the figure yourself and
verify that the field inside the plate gets canceled. Outside the plate the contributions of the positive and negative charge density
cancel and hence the field outside still equals $E_{external}$!
\begin{figure}[h]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{confield.ps}}
\caption{\sl 
The electrical field contributions inside the plate form the external field
and charge densities $\sigma_+$ and $\sigma_-$ respectively.  
} 
\label{confield}
\end{center}
\end{figure}

Finally, we remark that inducing two layers of opposite charge on a plate
requires work. The work is performed in the small timespan before reaching
the equilibrium.  The external field gets the 'energy bill'. 
From equation~\ref{uee} we can seen that the energy difference comes form
the volume inside the plate where the original field has vanished.
 Thus, for a plate that has surface $S$ and thickness $d$, the energy difference is:
\begin{equation}
U  = \frac{\epsilon_0}{2}  \int_{volume-plate} \vec{E}_{external}^2 dv = 
 \frac{\epsilon_0 S d }{2}  \vec{E}_{external}^2  
\end{equation}
(we ignore edge effects.) 
You can verify yourself that this equals the work needed to separate
the positive an negative surface charge.

% Somebody or something has to yield the force $F$ to separate
%the negative and positive charge. To caculate the requires work we start
%we just the plate without induced charge and imagine that the positive and
%negative layer are already present on the left side (see figure~\ref{forceplate}) and that somebody/thing pulls out the negative layer to the right through
%the electrical field of the positive layer.
%Hence, the work done per surface unit for a plate with thickness $d$ is 
%\begin{equation}
%W=\int dW=\int_0^d  dF \dot  d\vec{z} =\int_0^d \sigma_- \vec{E}_+ d\vec{z}= 
%\int_0^d -\sigma  \frac{\sigma}{2 \epsilon} \dakje{z} d\vec{z}=
% \frac{-\sigma^2}{2 \epsilon} d
%\end{equation}

 

\subsubsection{A grounded conducting plate}
We can also consider a grounded plate. This means that the plate is connected to the earth. For this purpose the earth should be seen as an unlimited source 
of charges. Thus the earth can pump electrons in or out the plate
at no cost without acquiring any net charge itself. Furthermore, the
potential of the earth is usually defined to be zero and grounded objects are at the same potential. 

 Figure~\ref{groundplate} shows a grounded plate (as you can see from the piece of wire that ends in the special symbol for ground),  together with its charge distribution.
\begin{figure}[h]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/groundplate.ps}}
\caption{\sl 
A grounded conducting plate is placed in an extern electrical field. Note
the international symbol for 'grounding' indicated in the figure.
} 
\label{groundplate}
\end{center}
\end{figure}

The surface charge $\sigma_{grounded}$, needed to cancel the electrical
 field inside  the plate,
 can be easily calculated using similar steps as above:
\begin{equation}
 \sigma_{grounded} =  -2 \epsilon_0  |\vec{E}_{external}|
\end{equation}
But why reaches the plate an equilibrium without any
positive net (surface)charge as in the previous example. 
Well, as mentioned above, to make a positive and negative layers of 
charge work is performed, which requires a force.
 In the 'grounded' case  all the necessary electrons are provided for free
 by the earth.  Note that the field outside the plate has 
drastically changed. In fact, the field becomes zero
everywhere left from the negative surface charge and it doubles on
the right side. This also affects the energy in the field
dramatically. Of course,
the earth is not really an unlimited source of electrons and we have to know 
all the details to calculate the complete energy balance. Hence, in practice
this cannot be done. In the scope of this course, our 'responsibility' ends
 with checking that all rules for conductors are satisfied. Check that
yourself. 




\subsubsection{A spherical conducting shell with a point charge inside}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/conshell.ps}}
\caption{\sl 
A positive point charge is positioned in a conducting shell with inner radius $R_i$ and
out outer radius $R_o$.} 
\label{conshell}
\end{center}
\end{figure}
We consider an electrically neutral conducting shell with an inner and outer radius of $R_i$ and $R_o$ respectively. We have put a positive point charge $q$ in its center
as shown in Fig.~\ref{conshell}.  What is the charge density in the shell and what is the electrical field everywhere in space?

It is amazing that we can answer this question given such
little information for such a complicated configuration. But we can, again
just by using the rules for conductor (and all the stuff we learned before).
Rule nr.~1 tells us that in the meat of the shell (that is between $R_i$ and $R_o$)
 the electrical field
of the point charge should cancel. This can  (only) be achieved by
induced charge on the inner and outer surface.
since we started with an electrically neutral object,
 there must be a similar amount of induced negative and positive 
charge,$Q_i=-\Q_o$ on 
the inner and outer surface respectively. 

The question is now how to calculate the $\sigma_i=\frac{Q_i}{4 \pi R_i^2}$ (or equivalently $\sigma_o$).
Well, we have calculated the electrical field of a spherical surface charge
density before (see equation~\ref{eshell}):
\begin{eqnarray}
 \vec{E}  &= &  \frac{ \sigma R^2 }{\epsilon_0 r^2} \dakje{r} \,\,\,\,\, r>R \\
          &= &  \frac{ Q }{4\pi \epsilon_0 r^2} \dakje{r} \,\,\,\,\, r>R
\end{eqnarray}
and $\vec{E}=0$ inside the shell (between $R_i$ and $R_o$).
 In the shell, we add the contributions of the point charge, the negative charge density on the inner shell and the positive
 one on the outer shell, we find inside the conductor:
\begin{eqnarray}
\vec{0}&=&\vec{E}_{q} + \vec{E}_{Q_i} + \vec{E}_{Q_o} \nonumber \\
  &=& \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r^2} + \frac{Q_i}{r^2} + 0 
  \right) \dakje{r}
\end{eqnarray}

We conclude that $Q_i=-q$ (and $Q_o=-Q_i=q$).
 Note that the charge densities on the
inner and outer surface, besides the sign,  are also in magnitude not
the same: $\sigma_i=-q/(4\pi R_i^2)$ and $\sigma_o=q/(4\pi R_o^2)$.
Check yourself that all rules for conductors are fulfilled.

What are the consequences for the electrical field everywhere in space
outside the conductor?
Well, the field  is just the field of the point
 charge.
 We could say that the presence of the conductor has not disturbed
 the outer field at all!

Finally we remark that we have been able to calculate the induced charge 
relatively easily because off the large symmetry in this problem. Most notably is that all contributions to
 the field, i.c.: that of the point charge, the surface charge at $R_i$
 and the charge  at $R_o$ all have radially pointing contributions.
For a weird shaped conductor in an external field you can try, just for fun,
to make a sketch of the charge distribution on its surface.



\subsubsection{Method of images}
A powerful technique to calculate the electrical field in many situations
is the 'method of images' using a 'mirror charge'. This method relies on
the fact (not derived here) that the electric potential is uniquely
defined $inside$  some volume 
\begin{itemize}
\item[] with a given charge density inside the volume, and,
\item[] with given the potential at its boundary surface.
\end{itemize}
In such case we may change the original charge configuration outside the
 volume
by an alternative configuration respecting the boundary conditions. Inside
the volume the electrical potential of the alternative configuration
equals that of the original configuration. You may have to read the
previous sentence twice. Outside this volume, the potential of the
original and alternative configuration generally completely differ.
In the following example we will see the strength of this method.

We consider a grounded plate and place a point charge $q$ at
a distance $d$ as illustrated in Fig.~\ref{qmirror}.
The point charge induces charge on the plate surface with an a priori unknown
distribution. For this example the plate is infinitely large and
 infinitely thin
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/qmirror.ps}}
\caption{\sl 
A grounded conducting plate is placed in the field of a point charge $q$.
The point charge is positioned at $z=d$ and the right surface of the plate
is located at $z=0$.
} 
\label{qmirror}
\end{center}
\end{figure}
What is the electrical field everywhere? Well, on the right side of the plate
($z>0$) we have a volume with a given charge density (the point charge) and
we know the potential at the boundaries: $V=0$ at the plate ($z=0$) and at infinity.

Hence, it is allowed to use the technique of the mirror charge. 
We have to keep the point charge, but can remove the plate
if we can find a charge configuration that gives the same potential $V=0$ at
the boundary $z=0$ and $z=+\infty$.
As the name of the technique suggests, put a mirror charge ($-q$) at $z=-d$
which leads to the configuration shown in Fig.~\ref{qmirroralt}. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/qmirroralt.ps}}
\caption{\sl 
An alternative charge configuration, existing of a electrical dipole.
} 
\label{qmirroralt}
\end{center}
\end{figure}
Are the boundary conditions fulfilled? At $z=0$ the contributions
to the electric potential of the two charges cancel and at infinity 
(by the way: only positive $z$ is relevant in this case)
 the potential is also still zero. 
The resulting configuration is an electrical dipole which we have seen
before. Hence, we know the potential for $z>0$ now. But what about
the potential in the region $z<0$. This is outside the our volume with
known boundary conditions. It is certainly wrong to use the potential
for a dipole in this region!

The trick is to go back to the original configuration and first
 determine the charge density on the plate. For $z>0$ we know the
 potential is given by:
\begin{equation}
V(x,y,z)= \frac{1}{4\pi \epsilon_0} 
     \left( \frac{q}{\sqrt{x^2+y^2+(z-d)^2}} -
      \frac{q}{\sqrt{x^2+y^2+(z+d)^2}} \right)
\end{equation}
The corresponding electrical field can be calculated using
 $\vec{E}(x,y,z) =- \divnodot V(x,y,z)$. For the field at $z=0$ we find:  
\begin{equation}
\vec{E}(x,y,0) =\frac{-q}{4\pi \epsilon_0} 
     \left( \frac{2d}{(x^2+y^2+(d)^2)^{\frac{3}{2}}}  \right) \dakje{z}
\end{equation}
as you can check yourself.
Now we know the electrical field (for $z>0$) which allows us to
 the determine the charge density on the plate using the relation  
 $\frac{\sigma}{2 \epsilon_0} = \vec{E}(z=0) \dot \dakje{z}$
 (see equation~\ref{eplate}). Hence,
\begin{equation}
\sigma (x,y)= \frac{-q}{2\pi} 
     \left( \frac{d}{(x^2+y^2)^{\frac{3}{2}}} \right),
\end{equation}
which together with the the point charge at $z=d$ determines the physical
 situation, enabling us to calculate the field everywhere.

We have discussed when and how you can use the method of images and we will conclude this section by summarizing this method:
\begin{itemize}
\item Check the following. the potential should be known at the boundaries. Inside the boundaries, the charge density should be known as well.
\item Find an alternative charge configuration outside the
 boundaries that respects (together with the charge density inside) the original boundary conditions.
\item Calculate the potential of the alternative configuration.  
\item If required, you can calculate the original charge configuration using
the potential (or electric field).
\end{itemize}
There is no general concept that returns the alternative configuration
and it probably became clear to you that it needs some experience to find 
such configuration.



\subsection{Capacitors}
\label{parplates}
We take two parallel plates of conducting material and charge the plates with opposite charge as indicated in Fig.~\ref{platecond}. The plates are separated
by a distance $d$. We ignore the thickness of the plates and assume that the plates have a large surface $A$ with respect to $d$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/platecond.ps}}
\caption{\sl 
Two parallel plates of conducting material. The plate have opposite charge $Q$ and are separated by a distance $d$. The contribution to the electrical field
of each plate individual is also indicated.
} 
\label{platecond}
\end{center}
\end{figure}
We have calculated the electrical field for an infinitely large flat surface
 charge already in Sec.~\ref{fsurf}, which provides a good approximation
  for the present configuration with $A>>d$.
We have to add the contribution of the two plates. In between the plates, we obtain: 
\begin{equation}
\vec{E}= \frac{+Q}{2 A \epsilon_0} \dakje{z} - \frac{-Q}{2 A \epsilon_0} \dakje{z}= \frac{Q}{ A \epsilon_0} \dakje{z}  
\end{equation}
The electrical 
field is constant and points away from the positive charged plate. The
contribution of the negative charge obtain an additional minus sign, because 
if it would be a positive charge  the field would point in the
negative $z$ direction. Make sure you understand this argument. 

The potential difference between the plates is given by:
\begin{equation}
V = V_+ - V_- =-\int_-^+ \vec{E} \dot dz = 
- \frac{1}{ A \epsilon_0} Q  \int_d^0  dz = \frac{d}{ A \epsilon_0}  Q   
\end{equation}
What we have shown now is that the potential difference between the conductors
is proportional to $Q$, which is generally valid for capacitors. 

Is the behavior $V\sim Q$ independent on the shape and size of the
 conductors? Yes. 
Suppose you take two conductors of arbitrary shape and size as illustrated
in Fig.~\ref{capacitor}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/capacitor.ps}}
\caption{\sl 
A capacitor, consisting of two conductors with bizarre shapes.
} 
\label{capacitor}
\end{center}
\end{figure}
The electrical field is proportional to the charge density in, or better $on$, 
the conductors. The charge density is proportional to $Q$. The potential $V$ is proportional
to $E$ and thus also to $Q$.

\subsubsection{Capacitance}
We have seen that $V \sim  Q$. Now we introduce a constant of proportionality
called capacitance, $C$, such that
\begin{equation}
C = \frac{Q}{V}
\end{equation}
The capacitance depends completely on the
geometry of the electrical configuration.
For the configuration with the two plates, we have shown that
  $C=\frac{A \epsilon_0}{d}$. The larger $A$ and the smaller $d$ the more
charge can be stored in the configuration for the same potential $V$.
 The unity of
capacitance is called Farad (=Coulomb/Volt), denoted by F.
 In practice $C$, measured in Farad is numerically small.
 For our plate configuration with $d=1$~mm and $A=1$~m, $C=9 \times 10^{-11}$~F.
In the newspapers a few years ago, there was an item about a small capacitor
with $C=1$~F, but I haven't figured out yet how to make that. Let me
know if you find the 'trick' on the web.

\subsubsection{The energy of a capacitor}
We start with an uncharged capacitor and  move electrons from one 
conductor to the other to charge it up. The electrons 'sense' the
electrical force in this process. Hence, moving them requires energy.

The energy needed to bring some charge $dq$ to the other conductor is
$dU=V(q) dq$. With $V(q)$ the potential difference between the two
conductors as function of the already moved charge $q$. Since the capacitance
$C$ is a purely geometrical quantity we can write $V(q)=q/C$ and thus
 $dU=\frac{q dq}{C}$. The total energy of a charge capacitor is then
given by:
\begin{equation}
U=\int dU=\int_0^Q \frac{q dq}{C}= \frac{1}{2} \frac{q^2}{C} |_0^Q
=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} C V^2
\end{equation}
with $Q$ and $V$ the final charge and potential
 of the capacitor respectively.

\subsubsection{Examples of Capacitors}
\subsubsection{Cylindrical configuration}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/ccil.ps}}
\caption{\sl 
Two coaxial conductors. The inner and outer conductors have radii $a$ and $b$ respectively. The conductors have opposite total charge $Q$.
} 
\label{ccil}
\end{center}
\end{figure}
Figure~\ref{ccil} shows a cylindrical configuration, consisting of two
coaxial conductors with length $L$. The inner conductor has a radius $a$. The outer radius is $b$ and is much smaller than
the length. The conductors have opposite total charge $Q$. 
What is the capacitance of this configuration?
\begin{itemize}
\item
First, determine the electrical field in the space between the two conductors.
 We may assume that $L>>b$ which implies that away from the edges 
the field is radial. The charge density on the inner conductor is 
$\sigma_i=\frac{Q}{2\pi a L}$.
  We calculate the flux through an imaginary small
 cylindrical Gaussian surface and obtain:
\begin{equation}
E_r 2 \pi r l = \frac{\sigma_i 2 \pi a l }{\epsilon_0}
\end{equation}
Note that in between the conductors 
there is no contribution to the flux (and field) from the outer conductor.
The electrical field is $\vec{E}=\frac{\sigma_i a}{\epsilon_0 r} \dakje{r}$
\item
The potential difference between the two conductors is given by:
\begin{equation}
V=-\int_b^a \vec{E} \dot d\dakje{r} =
  -\int_b^a \frac{\sigma_i a}{\epsilon_0 r} dr =
   -\frac{\sigma_i a}{\epsilon_0} ln(r) |_b^a =
   \frac{\sigma_i a}{\epsilon_0} ln(b/a)
\end{equation}
\item
Now we know the potential and it is trivial to obtain the
capacitance:
\begin{equation}
C=\frac{Q}{V}= \frac{2\pi a L \sigma_i \epsilon_0}{\sigma_i a ln(b/a)}
=\frac{2\pi L \epsilon_0}{ln(b/a)}
\end{equation}
\end{itemize}
Like the plate-capacitor, the capacitance of this configuration
increases when the distance between the two conductors becomes smaller.

\subsubsection{Spherical Capacitor}
We now consider a capacitor consisting of a spherical surface (thus not a massive sphere, but a shell with
some thickness) of a conducting material with radius $b$. In its center we placed a 
conducting sphere with radius $a$. The conductors carry opposite charge $Q$.
What is the capacitance of this configuration?
To calculate the electrical field we use a spherical Gaussian surface with
 radius $r$ such that $a<r<b$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/csphere.ps}}
\caption{\sl A spherical capacitors. The inner conductor has radius $a$ and the
outer one has radius $b$. The conductors have opposite charge.  Note that the outer conductor is a
apherical surfacd (a shell) an not a massive sphere.
} 
\label{csphere}
\end{center}
\end{figure}
For the electrical field we obtain:
\begin{equation}
\vec{E}=\frac{\sigma a^2}{\epsilon_0 r^2}\dakje{r}
\end{equation}
Now we calculate  the potential difference between the conductors:
\begin{equation}
V=-\int_b^a \vec{E} \dot d\dakje{r} = 
  -\int_b^a \frac{\sigma a^2}{\epsilon_0 r^2} dr =
   \frac{\sigma a^2}{\epsilon_0 r} |_b^a =
   \frac{\sigma a^2}{\epsilon_0} ( \frac{1}{a} - \frac{1}{b} ) 
\end{equation}
for the capacitance follows:
\begin{equation}
C=\frac{Q}{V}= \frac{4\pi a^2 \sigma \epsilon_0}{\sigma a^2(1/a-1/b)}
= 4\pi \epsilon_0 \frac{ab}{b-a}
\end{equation}
Note that the capacitance is  independent on $V$ and/or $Q$ and thus
is indeed a purely geometrical quantity.


\subsection{Knowledge and Skills}
The knowledge and skills  
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You understand the properties of a conductor.
\item
When a conductor is placed in an electrical field you can identify
the surface charge densities and calculate the resulting electrical
field.
\item
You understand the basics of the mirror charge technique
\item
You can explain what capacitors are and deduce the relation:
\begin{equation}
C = \frac{Q}{V} \nonumber
\end{equation}
\item
You can calculate the the capacitance of a parallel plate capacitor.
\item
The energy of a capacitor is given by
\begin{equation}
U=\frac{1}{2} C V^2 \nonumber
\end{equation}
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.



\section{Insulators}
All matter that cannot be classified as a conductor we call an insulator. 
In insulators there are no free electrons to cancel the electric
field inside the 'bulk'. However, by polarization of the molecules and atoms
inside an insulator there is some cancellation of the field. When an atom
polarizes we can discriminate between a positive and negative side. 
For this reason, insulators are historically referred to as 'dielectrics'.
 The properties and behavior of dielectrics in electrical fields are discussed
 in this Section.

\subsection{Polarization of atoms and molecules}
What happens microscopically when we electrically polarize an atom?
The classical picture of an atom is a big positive nucleus surrounded
by tiny electrons orbiting it. For our purpose, that is electrostatics,
we adapt this picture by taking the 'average' of the atom over time. 
Yes, you need some imagination, but this leads to an  atom consisting of
 a small positive nucleus in  the center of a uniform cloud of negative charge.
 Consider the illustration in Fig.~\ref{polatom}. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=9cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/polatom.ps}}
\caption{\sl A view of an electrostatic atom. It consists of a positive 
nucleus surrounded by a uniform cloud of negative charge. The cloud has radius (R). Left: Unpolarized, the nucleus is centered; Right: Polarized, the nucleus is shifted away from the center of the cloud. This results effectively in a
dipole. 
} 
\label{polatom}
\end{center}
\end{figure}
When we apply an external electrical field $E$ as is depicted in the right
 figure, the
negative charge is attracted, while the positive nucleus is repelled.
Consequently, the nucleus is shifted by a distance $d$ with respect to the
center of the cloud. We have assumed that the spherical shape of the cloud is
conserved. The net effect is that we produced an electrical dipole with moment
 $\vec{p}=Q \vec{d}$.

To good approximation the dipole moment is proportional to the external
field: $\vec{p}=Q\vec{d}=\alpha \vec{E}$, with proportionally factor $\alpha$,
 called the 'polarizability'.
Table~\ref{alpha} lists some experimentally obtained polarizabilities.
 \begin{table}[thb]
\begin{center}
\begin{tabular}{|l|c|c|} \hline
Atom & Z &  $\alpha$ ($10^{-30}$ m$^3$)   \\ \hline
Helium &  2     &    3 \\ 
Neon &    10  &    5    \\
Argon &   18    &    20 \\
Water vapor& -  & 500 \\ \hline
\end{tabular}
\caption{\sl Several examples of the polarizability for some atoms and water vapor. }
\label{alpha}
\end{center}
\end{table}
As expected, the factor $\alpha$ grows with increasing charge $Z$.
 For water vapor we observe that $\alpha$ is relatively large.
 Such a-typical behavior points in the 
direction of a different physical mechanism. 
Indeed, there is an additional
effect in water vapor and other so called 'polar' molecules. The electrons
in water molecules are attracted by the positive oxygen nucleus, more than
by the hydrogen nuclei. The result is that water (and other polar molecules)
have a built in electrical dipole moment. Figure~\ref{polwater} shows what
happens when these molecules are placed in an electrical field.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=9cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/polwater.ps}}
\caption{\sl 
Left: Water molecules without electrical field. the electrical dipole moment
of the molecules is indicated. Right: Water molecules in an external
electrical field. The dipole moment of the molecules are aligned by
the electrical force.
} 
\label{polwater}
\end{center}
\end{figure}
The dipole moments of the water molecules are aligned by the external 
electrical field which leads to the relatively large polarization 
 factor $\alpha$.

 In this Section we have discussed two different microscopic
effects leading to polarized matter. We have postulated a straightforward
relation ($\vec{p}=\alpha \vec{E}$) between a new phenomenon called
 polarization in the external electrical field. In Electrostatics however,
 we do not care about this microscopic 
behavior so much. We only want to describe the behavior of the fields, averaged
over all atoms, as we will see in the follow Section.

\subsection{Macroscopic Polarization}
What is the electrical field in a dielectric when put in a known 
external field?
 Forget for the moment the atoms and molecules in the dielectric 
and put on your 'abstract glasses'.   
In electrostatic theory, dielectrics consist
 of infinitesimally small electrical dipoles. 
A dielectric can be electrically
polarized by putting it in an external field $\vec{E}_0$ as illustrated by
Fig.~\ref{polmac}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/polmac.ps}}
\caption{\sl 
A dielectric in an external electric field $\vec{E}_0$. The microscopic dipoles
polarize (polarization $\vec{P}$). The resulting total field is given by
the sum of the original external field and that of the polarized dipoles.
} 
\label{polmac}
\end{center}
\end{figure}

Inspired by the microscopic view (see previous Section), the dipole moment
 per unit volume is defined as the polarization $\vec{P}$
which we $assume$ proportional to the electrical field:
\begin{equation}
\label{echip}
\vec{P}=\epsilon_0 \chi_e \vec{E}
\end{equation}
 The factor of proportionality $\chi_e$ is called the 
electrical susceptibility
for historical reasons.
We say that the dielectric is $linear$ when this equation is valid. In general
it is valid for relatively small electrical fields. Note that the word linear
in this context is not related to shape of the dielectric; it describes
 an electrical property.
 In this expression, 
the electrical field $\vec{E}=\vec{E}_{total}$, the total field, which
usually leads to confusion and/or mistakes!
 It is the sum of the external field
and the, yet unknown, field from the polarized dipoles ($\vec{E}_{pol}$).

What is the field of the polarized dipoles? Well, the polarization
leads to net charge separation. When we can quantify this charge, that is, 
determine the charge configuration,  we know how
to calculate the field using standard techniques. So, let's try to
find the charge density inside a dielectric first.

Have another look at  Fig.~\ref{polmac}. Inside the the dielectric there
is macroscopically no net charge. The dipoles are aligned in chains of 
positive and negative charge and all the
 '+' are canceled by '$-$'. However, this is not the case at 'the start and 
end of the chains', at the boundary of the dielectric.
 At the left side, all the chains start with '$-$', while at the right side
all chains end with '+'. There is a net charge separation 
 at the surface of the dielectric. Note that the charges themselves
are localized in contrast with  the free charge in a conductor.
 For this reason, the net charge  
at the surfaces of a dielectric is called $bound$ charge. The amount of
bound charge is given by the polarization:
\begin{equation}
\sigma_{bound}=\vec{P} \dot \dakje{n}
\label{eq:sfrp}
\end{equation}

Let's put in a small intermezzo. The above Equation~\ref{eq:sfrp} is sufficient
for linear dielectrics; there cannot be any bound charge inside the
dielectric. However, more generally the relation between polarization
and charge is given by:
\begin{equation}
\rho_{bound}=-\divnodot \dot \vec{P}. 
\label{eq:rfrp}
\end{equation}
It says that there is bound charge inside an dielectric 
when the polarization inside a dielectric is not constant. In such case there
is no (perfect) cancellation between positive and negative bound charge
inside the dielectric in contrast to a linear dielectric. At the surface of any dielectric this relation transforms to Equation~\ref{eq:sfrp}. In this reader,
 like in many textbooks, we further only consider linear dielectrics, unless
stated otherwise. This ends our small intermezzo.

In principle, we can now deal with a dielectric (with known $\chi_e$) 
placed in a known original external field $E_0$:
\begin{itemize}
\item[1]  the (yet unknown) polarization $P$ determines the (bound) charge 
configuration (Equation~\ref{eq:sfrp}),
\item[2]
the charge configuration determines the contribution $E_{pol}$
 to the electrical field, enabling us to calculate $E_{total}$,
\item[3] $E_{total}$ fixes the polarization (Equation~\ref{echip}).
\end{itemize}
To you, it all may look a bit circular, but in the following Sections
 we will discuss some examples. There is one suggestion to consider: memorize the three Equations above (and their meaning)!

\subsubsection{The electrical field in a flat dielectric}
We consider a flat dielectric with a given $\chi_e$ and we place it in
a known uniform electrical field $\vec{E}_0$ in the $z$ direction 
as shown in Fig.~\ref{diflat}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/diflat.ps}}
\caption{\sl 
A piece of dielectric in an external electric field $\vec{E}_0$. 
The microscopic dipoles
polarize (polarization $\vec{P}$). The resulting bound charge at the
surface is also indicated.
} 
\label{diflat}
\end{center}
\end{figure}
The dipoles in the dielectric polarize, leading to a polarization $\vec{P}$.
The bound net 
charge on the surfaces of the dielectric is according to Equation~\ref{eq:sfrp}:\begin{equation}
 \sigma_b=\vec{P}\dot \dakje{n} =\vec{P}
\end{equation}

Now, forget about the dieletric and only use the equivalent configuration.
 Hence, besides $\vec{E}_0$, we have a configuration of two oppositely charged
flat surfaces, which leads to an electrical field:
\begin{equation} 
 \vec{E}_{pol}= \frac{- \sigma_{pol}}{\epsilon_0} \dakje{z}
\end{equation}
 Here we simply used the  formula of a plate capacitor,
 derived in Section~\ref{parplates}
 (but I am sure, you could derive it yourself by now).

In the dielectric, the (total) electrical field is now:
\begin{eqnarray}
\vec{E}  & = & \vec{E}_0 + \vec{E}_{pol} \nonumber \\
         & = & \vec{E}_0 - \sigma_{pol}/\epsilon_0 \dakje{z}\nonumber \\
         & = & \vec{E}_0 - \vec{P}/\epsilon_0 
\end{eqnarray}
Using equation~\ref{echip} we can write
\begin{equation}
\vec{E}  =  \vec{E}_0 -  \chi_e \vec{E} = \frac{1}{(\chi_e + 1 )} \vec{E}_0
\label{eeo}
\end{equation}
That's it!

The susceptibility $\chi_e$ for glass and plastic like are of order 10. For water the susceptibility is about 80. 

\subsubsection{Plate capacitor with dielectric}
Consider a plate capacitor with a dielectric in between its plates as
illustrated in Fig.~\ref{plateconddiel}. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{/user/z37/college/em/dictaat/electrostatica/figures/plateconddiel.ps}}
\caption{\sl 
A plate capacitor with dielectric in between the plates.
} 
\label{plateconddiel}
\end{center}
\end{figure}
The plates have surface $A$ and
are separated by a distance $d$. The dielectric has an electrical
susceptibility $\chi_e$. Given the free charge $Q_{free}$
on the plates, what is the electrical field in the dielectric and what is
the capacitance of this capacitor?

That looks like a tough question, but in fact it isn't. Just cut the
problem in relevant pieces and you will see that you already can solve all
pieces one by one.

Starting point, the first piece,  is an $empty$ capacitor. Thus the same configuration  as already discussed in Section~\ref{parplates}.
We repeat our results (you should be able to derive these results by now):
\begin{eqnarray}
\vec{E}_{empty} &=& \frac{Q_{free}}{A \epsilon_0} \dakje{z} =  \frac{\sigma_{free}}{\epsilon_0} \dakje{z} \nonumber \\  
 V_{empty} &=& -\int_-^+ \vec{E}_{empty} \dot dz = \frac{d}{ A \epsilon_0}  Q_{free} \nonumber \\   
 C_{empty} &=& \frac{Q_{free}}{V_{empty}} =\frac{\epsilon_0 A}{d} 
\end{eqnarray}

The second step is to place the dielectric between the plates.
The field inside the empty capacitor is relevant external field for the
dielectric. In fact we have placed a dielectric in an external field just like
we have already seen in the previous Section and
 according to equation~\ref{eeo} the field becomes:
\begin{equation}
\vec{E}= \frac{1}{(\chi_e + 1 )} \vec{E}_{empty}= \frac{1}{(\chi_e + 1 )}  
 \frac{\sigma_{free}}{\epsilon_0} \dakje{z} 
\end{equation}

So-far, so good and we go to the next piece.
We can calculate the potential between the plates by integration of the
electrical field (just like we have done for empty capacitors):
\begin{equation}
 V = -\int_-^+ \vec{E} \dot dz =   \frac{1}{(\chi_e + 1 )}  \frac{d}{ A \epsilon_0}  Q_{free} =  \frac{1}{(\chi_e + 1 )} V_{empty}
\end{equation}
for the capacity follows:
\begin{equation}
  C = \frac{Q_{free}}{V} =   (\chi_e + 1 ) \frac{\epsilon_0 A}{d} =   (\chi_e + 1 ) C_{empty} 
\end{equation}
In textbooks the following definition is often used
\begin{equation}
    (\chi_e + 1 ) \epsilon_0 = \epsilon 
\end{equation}
With this definition most expression related to dielectrics become similar
as the expression in vacuum after substituting $\epsilon$ for $\epsilon_0$.

 Note that the capacitance between the empty and 'filled' capacitor
is just the factor $  (\chi_e + 1 )$. Thus, for plastic fillings the 
capacity grows with a factor of order ten.

\subsection{The electrical Displacement}
To describe electrical fields inside dielectrics, we had to thrown in
 several new 'electrostatic objects' like the polarization $P$ and
the bound charge, $\rho_b$. The question may arise:
is Gauss' Law valid inside a dielectric or should it be adapted?
The answer is that Gauss' Law is always valid, but we should be careful.
Consider Gauss's Law once more:
\begin{equation}
\epsilon_0 \divnodot \dot \vec{E} = \rho
\end{equation}
Realize that $\vec{E}$ in Gauss' Law always is the $total$ electrical field and
 $\rho$ is the $total$ charge density:
\begin{equation}
\epsilon_0 \divnodot \dot \vec{E} = \rho_{free} + \rho_b
\end{equation}
where $\rho_{free}$ is the charge induced on an conductor surface or the charge needed to create an external electrical field. (Keep in mind that we often discussed examples where we put a dielectric in an external field. Well, the external field comes not for free: it is the effect of free charge somewhere.)
Using Equation~\ref{eq:rfrp} we can write:
\begin{equation}
\epsilon_0 \divnodot \dot \vec{E} = \rho_{free} - \divnodot \dot \vec{P}
\end{equation}
or equivalently,
\begin{equation}
 \divnodot \dot (\epsilon_0 \vec{E} + \vec{P}) = \rho_{free}
\end{equation}
Now define the dielectric displacement $\vec{D}=\epsilon_0 \vec{E} + \vec{P}$.
It is just a matter of definition, but $\vec{D}$  has an important feature:
\begin{equation}
\divnodot \dot \vec{D} = \rho_{free}
\end{equation}
It is Gauss' Law for 'field' $\vec{D}$ that only depends on the presence
of free charge. Also the integral form of Gauss' Law is valid:
\begin{equation}
\int_{closed-surface}\vec{D} \dot d\vec{o} = 
   \sum_{free-charges-enclosed} Q_i
\end{equation}
Combining the formula's in this Section you can derive that $\vec{D}$
is also proportional to the (total) field $\vec{E}$:
\begin{equation}
 \vec{D}=\epsilon \vec{E} + \vec{P} =  
         \epsilon \vec{E} + \epsilon \chi \vec{E} = 
       (\chi_e + 1 ) \epsilon_0 \vec{E} = \epsilon \vec{E}
\end{equation}
This mathematical approach completes the theory on electrical fields in matter.


\subsection{Knowledge and Skills}
The knowledge and skills  
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You can explain how a dielectric is polarized in 
an electrical field, resulting in a net bound charge at its surface.
\item
The relation between the polarization and the electrical field is
\begin{equation}
\vec{P}=\epsilon_0 \chi_e \vec{E} \nonumber
\end{equation}
\item
With this relation you can calculate the electrical field in
a parallel plate capacitor filled with a dielectric:
\begin{equation}
\vec{E}= \frac{1}{(\chi_e + 1 )} \vec{E}_{empty}= \frac{1}{(\chi_e + 1 )}  
 \frac{\sigma_{free}}{\epsilon_0} \dakje{z} \nonumber
\end{equation}
You can also show that the capacitance is:
\begin{equation}
  C =  (\chi_e + 1 ) C_{empty} \nonumber 
\end{equation}
The capacitance of the filled capacitor is larger than that of the empty
capacitor. 
\end{itemize}
In addition, make the corresponding exercises of this
section, which you can find in the Appendix.


\section{The icons of Electrostatics}
Throughout the previous Sections we deduced, described, discussed and
derived the aspects of electrical fields in vacuum and matter.
The resulting field equations are listed in Table~\ref{eq:electro}. 
Never forget that  we started with Coulomb's empirical Law,
  based on experiment and constructed the theory on electrostatics around it. 
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|} \hline
Comment              & Integral      &   Differential     \\ \hline
Gauss & $\int_{surface} \vec{E}
                        \dot d\vec{o}=
                        \int_{volume} \frac{\rho}{\epsilon_0} dv$ & $\divnodot \dot \vec{E}=\frac{\rho}{\epsilon_0}$  \\
  &  $\int_{closed-line}\vec{E} \dot d\vec{l} =0$ & 
                              $ \rot{\vec{E}}=\vec{0} $ \\ \hline
\end{tabular}
\caption{\sl
 The complete set of  field equations for electrostatic theory.
}
\label{eq:electro}
\end{center}
\end{table}
