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\section{Electrostatics}
\subsection{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 coicidence: 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 in the Thales of Miletus (600 BC).

Amber is not unique in this sense. Many other materials, such as glass, rubber, PVC and ebonite 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}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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 charge attract, see Fig.~\ref{unlikech}. 
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{unlikech.ps}}
\caption{\sl the rod are ball have opposite electrical charge. Now 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 ebonite rod that after being rubbed with a cat's skin is 'negative' electrified.

Negative electrification is due to the excess of electrons (compared to the normal number of electrons in a neutral atom).
Positive electrification is due to the deficiency of electrons (compared to the normal number of electrons in a neutral atom).
For example, on rubbing an ebonite rod with fur or cat's skin, the electrons from the cat's skin get transferred to the ebonite rod. The cat's skin now has a deficiency of electrons and so is positively-charged. On the other hand, the ebonite rod has an excess of electrons and hence is negatively-charged.

\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}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{balance.ps}}
\caption{\sl A torsion balance to study electrical forces.}
\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 $Q$. (Note that the role and orde of $Q$ and $q$ is completely arbitrary.)
 The direction of the force $\vec{F}$ is given by $\dakje{r}$, the unit vector pointing from  $Q$ to $q$, see also Fig.\ref{fcoulomb}. 
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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 permitivity'. The electrical permitivity has the numerical value
\begin{equation}
\epsilon_0 = 8.85419 \time 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 \time 10^{9} \rm{\frac{Nm^2}{C^2}}
\end{equation}

Coulomb's law is about the electrical force is 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 'siperpostion 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_i}-\vec{r_0}|^2} \dakje{r_i-r_0}
\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}.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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, $\vec{r_1} - \vec{r_0}$ between charge $Q_1$ and the test charge $q$ is also indicated.
}
\label{superpos}
\end{center}
\end{figure}

Electrical charge in nature comes in discrete amounts; it is quantised. Furthermore electical charge is conserved: the net charge before an interaction is
equal to the charge after an interaction.
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 eletrical currents. R.A. Millikan discoverd 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}
Protons are charged opposite but 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 
confied in other particles (like the proton). Quarks and gluons are
studied in particle collisions at particle laboratories like CERN and
 Fermilab using large accelerators. The detectors to measure the products (i.c. new particles) of these collisions are large, 
typically 10m $\times$ 10m $\times$ 10m. In the Netherlands,
 NIKHEF in Amsterdam, is the main institute where research in this field
is being conducted.


\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_i}-\vec{r_0}|^2} \dakje{r_i-r_0} \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_i}-\vec{r_0}|^2} \dakje{r_i}-\dakje{r_0}
\label{eq:efield} 
\end{equation}
Hence, 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 eletrical field
is a real physics quantity! The best interpretation is, for now, that 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'.  
In case of the eletrical field virtual photons are exchanged between the charges. These virtual particles constitute the field.


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 continuosly 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 and see some
examples of continuous charge distributions.

\subsubsection{Line charge}
A line charge can be described by a charge density $\lambda(\vec{r})$ with has the unit C/m. Instead of the summation over point charges $Q_i$ as in 
equation~\ref{eq:efield} we integrate over the infinitesimal pieces of charge
 $\lambda(\vec{r}) dl$ to find the electrical field:
\begin{equation}
\vec{E}(\vec{r_P}= \frac{1}{4\pi\epsilon_0} \int_{line} 
          \frac{ \lambda(\vec{r}) }{\vec{r}^2} \dakje{r} dl
\label{eq:eline}
\end{equation}
Note that $\vec{r}$ is the connection vector between a piece of charge and the
point $P$ with coordinate $\vec{r_P}$ where the eletrical field is evaluated as illustrated in Fig.~\ref{eline}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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 $dq=\lambda(\vec{r}) dl$: 
\begin{equation}
\vec{E}(\vec{r_P}= \frac{1}{4\pi\epsilon_0} \int_{charge} 
          \frac{ dq }{\vec{r}^2} \dakje{r}
 \approx 
    \sum_{point charges} \frac{1}{4\pi\epsilon_0} \frac{Q_i}{\vec{r_i}^2} \dakje{r_i}
\end{equation}
Perhaps the interpretation of a continuous charge distribution as a collection
of point charges helps you to make the math less abstract.

\subsubsection{Surface charge}
A surface charge is described by a charge density $\sigma(\vec{r})$ with has
the unit C/m$^2$. To calculate the electrical field we integrate over
the contribution of the infinitesimal pieces of charge $dq=\sigma(\vec{r}) do$.
Hence,
\begin{equation}
\vec{E}(\vec{r_P}= \frac{1}{4\pi\epsilon_0} \int_{surface} 
          \frac{ \sigma(\vec{r}) }{\vec{r}^2} \dakje{r} do
\label{eq:esurf}
\end{equation}
An illustration of a surface charge is depicted in Fig.~\ref{evol}a.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{evol.ps}}
\caption{\sl a) (left) Illustration of a surface charge. b) (right) illustration of a volume charge.}
\label{evol}
\end{center}
\end{figure}

\subsubsection{Volume charge}
The volume charge density $\rho(\vec{r})$ is the most general continuous
charge density and has the unit C/m$^3$. The contribution of the 
infinitesimal pieces of charge $dq=\rho(\vec{r}) dv$ lead to the following
expression:
\begin{equation}
\vec{E}(\vec{r_P}= \frac{1}{4\pi\epsilon_0} \int_{volume} 
          \frac{ \sigma(\vec{r}) }{\vec{r}^2} \dakje{r} dv
\label{eq:evol}
\end{equation}
An illustration of a volume charge is depicted in Fig.~\ref{evol}b.


\subsubsection{Example of a line charge}
\label{diline}
As an example we will calculate the electrical field in a point $z=P$ of a piece of wire with length $L$ on the $x$ axis. The wire carries an uniform charge
density $\lambda$.  The configuration is shown in Fig.~\ref{exline}.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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}{\sqrt(z_P^2 + x^2) sin(\alpha} \\
dE_z&=&\frac{1}{4\pi\epsilon_0} \frac{\lambda dx}{\sqrt(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          \\
   &=&
        int_{-L/2}^{+L/2} \frac{1}{4\pi\epsilon_0}          
        \frac{\lambda dx}{\sqrt(z_P^2 + x^2} \frac{z_P}{\sqrt(z_P^2 + x^2} \\
   &=&  \frac{\lambda}{4\pi\epsilon_0} 
       \frac{ x}{z_P \sqrt(z_P^2 + x^2}|^{+L/2}_{-L/2}       \\
   &=& 
      \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 it charge $Q$ and drops with the square of the distance.
If we look from very lrge 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, as expected.

\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}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{efieldvec.eps}}
\caption{\sl The electrical field of a point charge.}
\label{efieldvec}
\end{center}
\end{figure}

Tradionally, the electrical field is depicted by field lines as show in Fig.~\ref{efield}. The density of the lines represent the magnitude of the field.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{efield.eps}}
\caption{\sl The electrical field of a point charge.}
\label{efield}
\end{center}
\end{figure}
Note that such drawing are 2 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 on a surface at constant distance from
a point charge drops with  $\frac{1}{r^2}$ (the surface of a sphere is $4 \pi r^2$.

Have another look to the field lines in Fig.~\ref{efield} and imagine a spherical surface around a point charge. The number of filed lines through the surface
is constant, thus indepent on the size, or beter radius, of the sphere.
You find an animation of this phenomen on the webpage.

The number of field lines can be expressed by the electrical flux:
\begin{equation}
\Phi=\int_{surface} \vec{E} \dot d\dakje{o}
\end{equation}
This the flux through a surface. The infinitesimal surface element 
 $d\dakje{o}$ is a vector with magnitude $do$ and its direction
 perpendicular to
the surface. We can write
\begin{equation}
 \vec{E} \dot d \dakje{o} = \vec{E} \dot \vec{e}_n do= 
  |\vec{E}| cos(\phi) do
\end{equation}
 with $e_n$ the normal on the surface and where
 $\phi$ represents the angle between the electrical field and surface.
Figure~\ref{fluxes} shows three examples of the electrical flux through a
surface for different angles between the field and surface.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{fluxes.ps}}
\caption{\sl the electrical flux through a surface $S$ for different angles between the field and surface. The normal $e_n$ is also indicated.}
\label{fluxes}
\end{center}
\end{figure}
With the eletrical 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 \nonumber \\
\end{eqnarray}

\subsection{Gauss' law}
Gauss' law states that the electrical flux through a closed surface, independent of 
the shape, is equal the total charge enclosed by this surface (multiplied by
the factor $\frac{1}{\epsilon_0}$)
Thus:
\begin{equation}
\Phi=\int_{closed surface}\vec{E} \dot d\dakje{o} = 
   \sum_{charge enclosed} \frac{Q_i}{\epsilon_0}
\end{equation}
As we will see this relation allows us to calculate the electrical field in
many 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, $Q=Q_{enclosed}$, and calculate the electrical flux trhough an spherical surface with radius $R$. This configuration
is illustrated in Fig.~\ref{gausspoint}.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{gausspoint.ps}}
\caption{\sl the electrical flux through a surface $S$ for different angles between the field and surface. The normal $e_n$ is also indicated.}
\label{gausspoint}
\end{center}
\end{figure}
The flux is given by:
\begin{equation} 
\Phi = \int_{surface} \vec{E} \dot d\dakje{o}
\label{eq:gp}
\end{equation}
Now we use the argument that both $\vec{E}$ and $\vec{o}$ point in the radial direction, hence
\begin{equation} 
\Phi = \int_{surface} |\vec{E}| do cos(\alpha_{\vec{E},\vec{o}}) = \int_{surface} |\vec{E}| do
\end{equation}
The electrical field at the surface follows from Eq.~\ref{eq:efield}, $|\vec{E}|=E_r(R)=\frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$, independent of the location
 on the surface.
We find
\begin{eqnarray}
\Phi &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \int_{surface} do \\
     &=& \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} 4 \pi R^2 \\
     &=& \frac{Q}{epsilon} 
\end{eqnarray}
in accordance with Gauss' law.

\subsubsection{Intermezzo: a more mathematical approach}
To evaluate (i.c. to get rid off) the dot-product $\vec{E} \dot d\dakje{o}$
 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 had written:
\begin{equation} 
\Phi = \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{equation}
with $\dakje{n}$ the normal vector on the spherical surface. This is a terrible
expression because the physical symmetry of the problem is obscured.

Obviously we need to work in spherical coordinates. Then, formally we obtain:
\begin{eqnarray} 
\Phi &=& \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 \\
     &=& \int_{surface} (E_{\phi} n_{\phi} \dakje{\phi} + E_{\theta} n_{\theta} \dakje{\theta} +  E_r n_r \dakje{r}) do \\
\end{eqnarray}
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 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 $\vec{n}$ on the surface points in the radial direction and has the unit length: $\vec{n}=(0,0,\dakje{r})$. 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 \\
     &=& \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi} \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 sin(\theta)  d\phi d\theta \\
     &=&  \int_{\theta=0}^{\pi} \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 
       (\phi)|^{2\pi}_0  sin(\theta)  d\theta   \\
    &=&   \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 2\pi   (-cos(\theta))|_0^{\phi}    \\
    &=&   \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}  R^2 2\pi    2  \\
    &=&   \frac{Q}{epsilon_0}    \\
\end{eqnarray}
Probably this intermezzo did not increase your confidence in Gauss' law. It showd however the basic steps which many textbooks tend to skip. In the following
we also often will skip such steps; they contains 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 $r^2$
 lead to a cancellation.

\subsection{Validity of Gauss law}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{blop.ps}}
\caption{\sl a) Point charge Q in an closed surface.  b) Point charge Q outside an close surface.}
\label{blop}
\end{center}
\end{figure}
We checked Gauss' law for a spherical surface with a point charge located 
at its centre. 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 surface grows with $r^2$.
Also, we could have put many point charge at different location within the closed surface. Obvioulsy 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 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 many cases to evaluate the electrical field in an elegant way. Below, we desribe a few of these configuration.

\subsubsection{The line charge}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{gline.ps}}
\caption{\sl A line charge (dark grey). A Gaussian cilinder (light grey) with height $h$ and radius $r$ is also drawn. }
\label{gline}
\end{center}
\end{figure}
Figure~\ref{gline} shows a, infinitely long and infitinitely thin, line charge with uniform charge
 density $\lambda$. The optimal
 Gaussian surface is a cilinder. To obtain this insight, first deduce the shape
of the electrical field. It needs some practise 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 is zero. 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 concluse that the eletrical field has only a  radial (cilinder-wise) component. Therefore we try a cilindrical Gaussian surface with arbitrary heigth 
$h$ and radius $r$. The flux through the cilinder is given by:
\begin{equation}
\Phi=\int_{cilinder} \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 field is independ 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   \\
E_{\phi} &=& 0 \\
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 shake it. When the line has a finite thickness of radius $\rho$ this method can be applied as we will see later. 

\subsubsection{A flat surface charge}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{gplate.ps}}
\caption{\sl A flat surface charge (dark-grey). A Gaussian cubical surface is also indicated. }
\label{gplate}
\end{center}
\end{figure}
Figure~\ref{gplate} shows a, infinitely large and infitinitely thin, surface charge with uniform charge density $\sigma$..
What are the components of the electrical field?
\begin{itemize}
\item
 Imagine the plate constitutes of small point charges $d\sigma$. Every point charge produces a 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 pointing away from the sufrace charge.
\item
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 remais 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 electical field does not depend
on $x$ or $z$ and thus can only be dependend 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 $\vec{E}= \frac{\sigma}{2 \epsilon_0} \dakje{y}$. 
Note that the field is constant, but opposite for positive and negative $y$ values respectively.
 


\subsubsection{A massive spherical charge}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{gsphere.ps}}
\caption{\sl A spherical volume with radius $R$ carrying a uniform charge density $\rho$ (dark-grey). 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 $R$.
What are the components of the electrical field? Suppose there are non-radial (spherical-wise) components. Rotate the configuration around its centre such that the non-radial components change direction. Realize that the physical
configuation is invariant which allows no non-radial components. The same argument can be used to deduce that the radial compont of the field only depend on $r$. Hence, $\vec{E}=E_r(r) \dakje{r}$.
We did not specify whether we discussed the field inside or outside the sphere. Well, it doesn't matter. Both inside and outside the sphere we can use the
above arguments.

For electrical flux (inside or outside) 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}
Outside the spherical charge, $r>R$, the enclosed charge is $\int_{Volume} \rho dV= \rho \frac{4}{3}\pi R^3$:
\begin{equation}
\Phi=E_r(r) 4\pi r^2 = \frac{Q_{enclosed}}{\epsilon_0} = \rho \frac{4}{3}\pi R^3\end{equation}
Thus
\begin{equation}
\vec(E)= \frac{\rho R^3}{3 \epsilon_0 r^2} \dakje{r} \,\,\,\,\, r>R
\end{equation}
Inside the spherical charge, $r<R$, the Gaussian spherical surface encloses
 a charge $\rho \frac{4}{3}\pi r^3$.
The leads to an electrical field of
\begin{equation}
\vec(E)= \frac{\rho r}{3 \epsilon_0} \dakje{r} \,\,\,\,\, r<R
\end{equation}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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 a graph of size of the eletrical field as function of $r$. Starting form the centre, the field grows linaerly with $r$ till the radius $R$ of the sperical charge is reached. Then it drops with $1/r^2$, similar to
the field of a point charge in the origin.

\section{The divergence of the electrical field}
The divergence of the electrical field is defined as
\begin{equation}
\div \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}

In this Section we will introduce this quantity in a natural way and describe its link with Gauss' law.

\section{Flux and divergence}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{box.ps}}
\caption{\sl A infinitesimal box with volume= $dxdydz$ in an electrical field $\vec{E}$. 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 inifitesimally small box, which is placed in an electrical field $\vec{E}$. We work in cartesian coordinates and calculate the electrical flux trhough the box. The flux $\int_{cover} \vec{E} \dot d\dakje{o}$ through each of the covers a to f  of the box is given by:
\begin{eqnarray}
\rm{a:} -E_x(x,y,z) dydz
\rm{b:}  E_x(x+dx,y,z) dydz
\rm{c:} -E_y(x,y,z) dxdz
\rm{d:}  E_y(x,y+dy,z) dxdz
\rm{e:} -E_z(x,y,z) dxdy
\rm{f:}  E_z(x,y,z+dz) dxdy
\end{eqnarray}
Note the relative minus sign for oppsite covers which comes from the oppsite
direction of the normal vectors on these covers.
Now we add all contribions to obtain the flux through the box:
\begin{eqnarray}
\int_{box} \vec{E} \dot d\dakje{o} =
 & & [E_x(x+dx,y,z) - E_x(x,y,z)] dydz \label{c1} \\
 & & [E_y(x,y+dy,z) - E_y(x,y,z)] dxdz \label{c2} \\
 & & [E_z(x,y,z+dz) - E_z(x,y,z)] dxdy \label{c3} \\
\end{eqnarray}
Remember the rule for elementary calculus that $df=f(x+dx)-f(x)$. The obove equation can be rewritten as:
\begin{equation}
\int_{box} \vec{E} \dot d\dakje{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 mathematically equivalent to muliplying with unity.
We find
\begin{eqnarray}
\int_{box} \vec{E} \dot d\dakje{o} &=& \frac{dE_x}{dx} dxdydz + 
                       \frac{dE_y}{dy} dxdydz + \frac{dE_z}{dz} dxdydz \\
		       &=& [ \frac{dE_x}{dx} +
                       \frac{dE_y}{dy} + \frac{dE_z}{dz} ] dxdydz \\
		       &=&  [ \div \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 the boxes together to make you 'any' volume
 as illustrated in Fig.~\ref{boxes}.
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{boxes.ps}}
\caption{\sl Any volume consists of an ensemble of infinitesimal boxes.
\label{boxes}
\end{center}
\end{figure}
 We then obtain
\begin{equation}
\int_{surface} \vec{E} \dot \dakje{o} = \int_{volume} [ \div \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's law and Gauss's theorem}
Gauss law for the electrical flux:
\begin{equation}
\int_{surface} \vec{E} \dot d\dakje{o} = \frac{1}{\epsislon_0} Q_{enclosed}
   = \frac{1}{\epsislon_0} \int_{volume} \rho dV
\end{equation}
can be combined with Gauss theorem:
\begin{equation}
\int_{surface} \vec{E} \dot \dakje{o} = \int_{volume} [ \div \dot \vec{E} ] dV
\end{equation}
This leads to the following expression:
\begin{equation}
\int_{surface} \vec{E} \dot \dakje{o} = \int_{volume} [ \div \dot \vec{E} ] dV
 = \frac{1}{\epsislon_0} \int_{volume} \rho dV
\end{equation}
which implies
\begin{equation}
 \div \dot \vec{E} = \frac{\rho}{\epsislon_0} 
\end{equation}
This relation is Gauss' law in differential form.

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

Outside the charge sphere $\rho=0$, the eletrical 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}
\div \dot  \frac{\vec{r}}{r^3}& =& \div \dot [ \frac{1}{(x^2+y^2+z^2)^{\frac{3}{3}}} (x,y,z) ] = \div \dot  
  ( \frac{x}{(x^2+y^2+z^2)^{\frac{3}{3}}} ,    \frac{y}{(x^2+y^2+z^2)^{\frac{3}{3}}} ,    \frac{z}{(x^2+y^2+z^2)^{\frac{3}{3}}} ) \\
     & =&      \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}}}
     & =&    [ \frac{\partial x}{\partial x}] 
        \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}} +
   x [ \frac{\partial}{\partial x} 
           \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}}] +
  \partial_y .... + \partial_z .....
      & =&  [1  \frac{ 1}{(x^2+y^2+z^2)^{\frac{3}{2}}} +
   x [-\frac{3}{2} 2x    \frac{ 1}{(x^2+y^2+z^2)^{\frac{5}{2}}}] +
  \partial_y .... + \partial_z ..... \\
  &= &  \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}}} =0
\end{eqnarray}
Not convinced? Try it yourself!
 
\section{The loop integral of the electrical field}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{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 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}
\end{equation}
The field is pointing purely radially. No matter waht the exact path is following from $a$ to $b$, the dot-product filters out the radial component ($\dakje{r} \dot d\vec{dl}=dr$. 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, indepent of the path we followed, the integral of a closed path of the
electric field is zero. Using the superpoistion 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 hte second electrical field
 equation in integral form and has no historical name.
 The differential form $\rot{\vec{E}}=0$ we will derive later.

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

\subsection{Work in a gravitational field}
To refresh your memory we first consider work and potential energy
in a graviational field. The work, $W_{person}$ when you lift an object with mass $m$ from
the ground to height $h$ is
\begin{equation}
W=\int_0^h \vec{F} \dot d\vec{l}
\end{equation}
with the force,  $|\vec{F}|=m g$, the mass times the gravitional constant
pointing down.
Hence
\begin{equation}
W_{person}=-mgh
\end{equation}
The potential energy of the object equals $U=-W_{person}=mgh$.
From the view of the gravitational field the work perfomed is 
 $W_{field}=-W_{person}$. 

\subsection{Potential energy in an electric field}
\begin{figure}
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{qpath.ps}}
\caption{\sl A test charge $q$ in the field of a source charge $Q$ is brought in fromm infinity to  $P$.}
\label{path}
\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 potential energy of the configuration is defined as
the work of the field needed to bring the test charge $q$ from infitiy to $P$.
\begin{equation}
U=W_{field}=\int_{\inf}^P Q \vec{E} \dot d\vec{l} = 

Q \int_{r_a}^{r_b} E_r  dr =
 \frac{Qq}{4\pi \epsilon_0} \int_{r_a}^{r_b} \frac{1}{r^2} dr = \frac{Qq}{4\pi \epsilon_0} \frac{-1}{r}|_{r_a}^{r_b} =   \frac{Qq}{4\pi \epsilon_0} (\frac{1}{r_a} -  \frac{1}{r_b})
\end{equation}

 


\end{document}
