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\begin{center}
{\bf Notes on Electrostatics}
\end{center}
These notes are meant for my PHY133 lecture class, but all are free to use them
and I hope they help. The ideas are presented roughly in the order in
which they are taught in my class, and are designed to supplement
the text. I will be writing these notes as I teach the class, so
they will be constantly updated and modified.
\bigskip
\centerline{\bf General features of the Electrostatic interaction}
We will be begin our discussion by doing a simple demonstration in
lecture. We start with a PVC pipe and a paper towel. Initially
there is no force between the two. However, after rubbing the
paper on the PVC pipe, we find they are attracted to each other!
If we rub the paper towel on two similar PVC pipes, we now find that
the two PVC pipes push each other apart (repulsion)! What kind of
force is this, and what property of nature causes the force?
The force is certainly not gravity, and the source is not the
objects mass as it is in Newtonian gravity.
We call this the electrostatic (or electrical) interaction, and we call the
property of nature that causes the force {\bf charge}. It is quite
different than the gravitational force, since the electrostatic force can be
attractive or repulsive (whereas gravity is only attractive). Note that repulsion
occured between the two similar PVC pipes. Whatever charge they had, they
had the same charge. Thus, the experiments would imply that like charges repel.
What about attraction? Initially the PVC pipe and the paper did not
attract or repel. However, after they were rubbed together then
they attracted each other. We know that the PVC pipe acquired
some charge, since two PVC pipes repel each other. The
experiments would suggest that initially the paper and pipe
each had zero {\bf net} charge, and after rubbing, some charge
was transfered from the paper to the pipe. If the paper was
initially neutral (0 net charge), then the paper might have
negative (or the opposite) charge as the pipe. Thus, "opposite"
charges attract.
To understand what is really going on, we need to understand what the
materials are made of. Today, we know that matter is made up of atoms.
The atoms themselves are comprised of a massive nucleus (protons and neutrons)
surrounded by electrons. When two {\bf different} materials are rubbed
together, some electrons can be rubbed off one and onto the other. The
electrons and protons possess charge, as well as mass and other properties.
However, it is the property of charge that produces the electrical force.
Consider the following set of experiments. First we place two electrons a
distance of $r_0$ away from each other and measure the electrical force they exert
on each other. We find that they {\it repel} each other with a force $F_0$. We now
repeat the same experiment except we do it with protons separated by a distance
of $r_0$. We find they also {\it repel} each other with the same force $F_0$. Finally,
we place an electron and a proton a distance $r_0$ apart and measure the force
they exert on each other. We find that they {\it attract} each other with a force of the
same magnitude $F_0$. These experiments tell us that the electron and the proton have the
same magnitude of charge, but are "opposite". We usually represent the amount of charge
on the proton and the electron as $e$.
Protons have been assigned a positive charge ($+e$), and hence electrons have a
negative charge ($-e$). Neutrons have zero net charge. There is no electrostatic
force between a neutron and either an electron or a proton. A neutral atom
has an equal number of protons and electrons.
In the case of the PVC pipe and the paper, electrons were rubbed off of the
paper onto the PVC pipe. The PVC pipe acquired an excess of electrons
(more electrons than protons) and had an overall negative charge. The paper
lost some of its electrons and had an overall positve charge. The extra electrons
will not stay on the pipe forever, but will gradually be pulled off by
moisture in the air. Thus, after a while both the pipe and the paper will be
neutral again. On dry days, the objects can keep their charge for quite a while.
Note: if two identical materials are rubbed together, charge is {\bf not} transfered
from one to the other.
\bigskip
\centerline{\bf Quantitative properties of the Electrostatic Interaction}
The last two quarters we have seen that many aspects of nature can be
understood in "simple" mathematical terms, and we hope that this is true
for the electrostatic interaction. As was done with the force of gravity,
it is best to start with the simplest objects and build up to more
complicated ones. Let's consider the electrical force between two
very small "point" particles that possess charge. Let particle
"1" have an amount of charge equal to $q_1$, and let particle "2"
have an amount of charge equal to $q_2$.
Suppose they are separated by a distance $r$ and are not moving. How
does the electrostatic force depend on $q_1$, $q_2$, and $r$?
In lecture we will demonstrate that if the number of electrons transfered to
the PVC pipes increases, so does the force between the pipes. We will also
show that the force gets stronger as $r$ decreases.\\
\noindent {\bf Units for Charge}\\
In past quarters we have discussed the units of the fundamental
quantities of mass, length, and time, as well as derived quantities such
as velocity, acceleration and force. For the electrostatic force
we have this new quantity: Charge. Physicists believe that charge
is a fundamental quantity, and will have its own unit. However,
there is one interesting property of charge that mass, length and
time do not have: charge is quantized in multiples of a basic charge.
This means that the charge on any particle or object is an integer multiple of a fundamental
charge unit called $e$. For the three particles that make up all of
our stable matter the chart below lists their charges and masses:
\begin{center}
\begin{tabular}{c|c|c}
Particle & Charge & Mass (Kg)\\
\hline
proton & $+e$ & $1.673 \times 10^{-27}$\\
neutron & $0$ & $1.675 \times 10^{-27}$\\
electron & $-e$ & $9.11 \times 10^{-31}$\\
\end{tabular}
\end{center}
\noindent Since all stable matter is made up of protons, neutrons, and electrons,
the net charge $q$ of a material is $(N_p - N_e)e$ where $N_p$ is the number of
protons in the material and $N_e$ is the number of electrons. If there are an
equal number of protons and electrons, the material is neutral. If the
material has an excess of electrons (protons), then the material has a
net negative (positive) charge. Thus, $q = (N_p - N_e) e$ is quantized
in multiples of $e$ since $N_p$ and $N_e$ are integers. This quantization
doesn't only hold for stable mater, but for all subatomic particles that
are directly observed in the laboratory (quarks having a charge that is
a fraction of $e$).
Thus, one really doesn't need a special unit to denote the charge on
an object. One only needs an integer denoting the number of fundamental
charges on the object. Whether $e$ itself has units is a question that
is considered in quantum mechanics. We will take the classical physics
approach and assign a unit to charge, that being the {\bf Coulomb}.
We will return to the discussion of charge units when we talk
about the magnetic force.
In terms of Coulombs, $e \approx 1.6 \times 10^{-19}$ Coulombs.
The charge on any material object is equal to: $q = (N_p - N_e)
1.6 \times 10^{-19}$ Coulombs, and is consequently quantized.
When macroscopic objects are "charged" large numbers of electrons
are transfered (around $10^{10}$ or more). Because of this, we
are not sensive to the quantization of charge and tend to think
of charge as a continuous quantity. We might say that an object
has a charge of $2$ Coulombs, which might not be an exact multiple
of $e$. As we go through the course, try not to forget that
charge is quantized. Millikan (1909) was the first to demonstrate experimentally that
charge is quantized, and received a Nobel Prize for his discovery.\\
\noindent {\bf Electrostatic Force Law}\\
Coulomb (1736-1806) carried out accurate experiments to determine how the
electrostatic force depends on the distance between charged "point" objects as
well on the magnitudes of their charges. He demonstrated that
the magnitude of the electrostatic force varies as $1/r^2$ where $r$
is the distance between the point objects.
We showed in lecture that the electrostatic force increases when
the objects have acquired more charge. Experiments show that
the force is proportional to the product of the charges $q_1$ and $q_2$, where
$q_i$ is the net charge on object $i$. This
result follows from the superposition principle
for the electrostatic force, and can be understood as follows:
Suppose the magnitude of the force between
two electrons separated a distance $r$ is $F_0$. If we replace one of the
electrons with two electrons glued together, then the force between these
two electrons and the other one will be $2F_0$, since we believe that electrostatic forces
add like vectors. We can continue with this reasoning for any number of extra
electrons. Suppose "point" object 1 has $n_1$
extra electrons, and "point" object $2$ has $n_2$ extra electrons. Each electron
on object 1 will feel a force of $n_2 F_0$. Now, since there are $n_1$ electrons on
object 1, the magnitude of the net force between them will be $n_1 n_2 F_0$:
\begin{equation}
|\vec{F}_{electrostatic}| \propto {{n_1 n_2} \over r^2}
\end{equation}
\noindent It is more convenient to express the electrostatic force law
in terms of the net charge on the "point" objects instead of the numbers
of excess (electron or protons). The net charge on object
1 is $q_1 = (N_e(1) - N_p(1))e$ and on object 2 $q_2 = (N_e(2) - N_p(2))e$,
where $N_e(i)$ ($N_p(i)$) are the number of electrons (protons) on object "i".
In terms of the net charges on the point object we have the usual form
of Coulomb's Law:
\begin{equation}
|\vec{F}_{electrostatic}| \propto {{|q_1| |q_2|} \over r^2}
\end{equation}
\noindent The absolute values are used since the charges can be positive or negative and
the equation is an expression for the magnitude of the electrostatic force. {\bf The
direction of the force is along the line joining the two point particles}. If they are
the same charge, then each particle feels a force directed away from the other one.
If they charges are opposite, then each particle feels a force towards the other. The
magnitudes are equal (Newton's Third Law). The
proportionality sign can be replaced with an equal sign plus constant:
\begin{equation}
|\vec{F}_{electrostatic}| = k {{|q_1| |q_2|} \over r^2}
\end{equation}
\noindent where $k \approx 9 \times 10^9$ $Nm^2/C^2$ and carries the units of
charge. The constant $k$ is also written in terms of the permittivity of
free space $\epsilon_0$: $k=1/(4 \pi \epsilon_0)$, where $\epsilon_0 =
8.854 \times 10^{-12}$ $C^2/(Nm^2)$.
It is sometimes convenient to express Coulomb's electrostatic force law with direction, that is
to use vectors for the approriate quantites. If we define $\vec{F}_{12}$
as the force {\bf on} point particle $2$ {\bf due to} point particle $1$,
then
\begin{equation}
\vec{F}_{12} = k {{q_1 q_2} \over r^2} \hat{r}_{12}
\end{equation}
\noindent where $\hat{r}_{12}$ is a unit vector pointing {\bf from} particle $1$
{\bf to} particle $2$. Note, that for like charges the product $q_1 q_2$ is positive
and particle $2$ feels a force away from particle $1$, $+\hat{r}_{12}$. For unlike charges the
product $q_1 q_2$ is negative the force $1$ feels is toward $2$, $-\hat{r}_{12}$.
\bigskip
\noindent {\bf Comments on Coulomb's Law}\\
\noindent 1. {\bf The formula is only valid for point particles}. How small
is small enough to be considered a "point particle"? If the largest dimension of the
particles is $d$, and the separation of the particles is $r$, then if
$d/r <<1$ the particles can be considered as points and the formula should apply.\\
\noindent 2. {\bf The mass of the particles does not effect the electrical force}.
An objects mass will affect its motion, since mass is a measure of inertia.
Mass and charge are very different. Mass has a dual role: it is a measure of
resistance to change in motion, and it is the source of gravity (Newton).
Charge does not add to the inertia of an object, nor does it affect the
gravitational force.\\
\noindent 3. {\bf Why is the force called electro-static}? It is because the force
does not depend on the velocity of the particles. The Coulomb's law force is the
same if the particles are stationary or moving. We will see (later in the course)
that if the charged particles are moving there is another force that depends on
their speeds: the magnetic force. At first we will just consider the electrostatic
part of the force.\\
\noindent 4. {\bf Why does the magnitude of the force decrease as $1/r^2$}. We have
seen this inverse-square law dependence with the gravitational force law (Newton),
as well as sound and light intensity. For light and sound intensity, the
inverse-square law was a result of the energy being equally distributed
on a spherical surface. Since the area of a sphere is $4 \pi r^2$, the
sound or light intensity decreased as $1/r^2$. For the electostatic force
there is nothing that is spread out evenly over a spherical surface,
however the $1/r^2$ dependence is a result of the geometry of space. A more
advanced understanding of this dependence comes from quantum electrodynamics.
Here we will marvel over the simple inverse-square decrease of the electrical
force and use this nice property to formulate Coulomb's law using geometry
via Gauss' Law to be covered shortly.\\
\noindent 5. Coulomb's Law holds at atomic length scales. It is one of the
more important forces in the atom.\\
\noindent 6. Between particles, the electrostatic force is much larger compared
to the gravitational force. A stationary proton and a stationary electron are attracted to each
other by two forces: gravity and the electrostatic force. They are attracted
via gravity since they both have mass, and they are attracted via the
electrostatic force since they both have charge. However, the electrical forces
within the atom are much much stronger than the gravitational ones, which are
usually neglected in calculations.\\
\noindent 7. It is a bit mysterious that objects with charge can influence each
other even though they are not touching. This so-called "action at a distance"
is not very satisfying. Our present understanding, from quantum mechanics, is
that the interaction can be described locally. The charge is related to the
probability that a photon will be emitted or absorbed locally. If you want to
know more about quantum electrodynamics you should be a physics major.
\bigskip
\noindent {\bf Superposition Principle}\\
Coulomb's law holds for small "point" objects with charge,
or small point charges. What if the objects are not small?
The electrostatic force behaves like a normal force, and is
a vector. That means that if point object "1" experiences an
electrostatic force from two other point objects, say point
object "2" and from point object "3", the net force on
object "1" is the {\bf vector sum} of the two forces:
\begin{equation}
\vec{F}_1(Net) = \vec{F}_{21} + \vec{F}_{31}
\end{equation}
\noindent Remember that forces are vectors, so the sum on the right
is a sum of {\bf vectors}. We saw this same principle with the
gravitational force, and it also holds for the electrostatic force.
{\bf The net electrostatic force on a point particle is the vector sum
of the electrostatic forces due to all other point particles.}
Stated in a more precise mathematical way:
\begin{equation}
\vec{F}_1(Net) = \sum_{i=2}^\infty k {{q_1 q_i} \over r_{1i}^2} \hat{r}_{i1}
\end{equation}
\noindent where $r_{1i}$ is the distance between point particle "1" and the
i'th point particle, and the sum goes over all particles in the system.
If there is a continuous (solid) distribution of charges, one can break up the
solid into small "points" and sum {\bf or integrate} over the solid. We will do
a number of examples of the force on a point object with charge due to other
point charges as well as continuous (or solid) objects.\\
Coulomb's law plus the superposition principle contains all the "physics" of the
electrostatic force. For the next three weeks we will not be introducing any
new principles of physics, but rather build upon this one important law of nature.
\bigskip
\centerline {\bf Electric Field}
\bigskip
A nice feature of the electrostatic force law is that
{\bf the force that a point particle feels is proportional to
the charge the particle has}. Suppose at a particular point
in space a particle with a charge of $1$ Coulomb feels a force of $10$ Newtons.
If this particle were replaced with another point particle that
had a charge of $2$ Coulombs, the new particle would feel a force of
$20$ Newtons in the same direction as the first. A charge of $3$
Coulombs $\rightarrow$ $30$ Newtons, etc. This proportionality
is evident from the equation above for the net force on particle
"1" with charge $q_1$. The quantity $q_1$ factors out of the sum:
\begin{equation}
\vec{F}_1(Net) = q_1 \sum_{i=2}^\infty k {{q_i} \over r_{1i}^2} \hat{r}_{i1}
\end{equation}
\noindent The force on a particle located at the point where particle "1"
is located is proportional to the amount of charge on the particle. Thus,
it is useful to consider the electrostatic force per charge. We
refer to this quantity as the {\bf electric field}, and it is given the
symbol $\vec{E}$. From the expression above, we see that the
electric field at the point where particle "1" is located is
$\vec{F}_1(Net)/q_1$ which is the sum on the right. If we let
$\vec{r}_1$ be the position vector to point "1", then
\begin{equation}
\vec{E}(\vec{r}_1) = \sum_{i=2}^\infty k {{q_i} \over r_{1i}^2} \hat{r}_{i1}
\end{equation}
\noindent where $i$ sums over all point particles in the system.
In the above equation for $\vec{E}$, $\vec{r}_1$ is the point in space
where particle "1" was located. However, we can define the
electric field at this point without particle "1" even being there.
The electric field can be defined at any point in space, nothing
needs to be there! We usually label the point simply as $\vec{r}$
without the index "1".
$\vec{E}(\vec{r})$ is the force per charge {\bf if}
a charge were to be placed at the point $\vec{r}$.
The electric field is somewhat abstract. It is
similar to the price per weight of a quantity, say bananas.
Let the price of bananas be $50$ cents per pound. This value,
$50$ cents/pound, can exist even if no one ever buys bananas.
However, {\bf if} someone wants to buy bananas, the bananas
are weighed and the price determined. The electric field can
be defined at a point in space even if a charged particle is never
placed there. However, {\bf if} a particle which
has an amount of charge $q$ is placed at the point $r$, {\bf then}
the electrostatic force that the particle will feel is
\begin{equation}
\vec{F}_{electrostatic} = q \vec{E}(\vec{r})
\end{equation}
\noindent In this equation, $q$ is a scalar and can be
positive or negative. The force $\vec{F}$ is a vector,
and thus $\vec{E}(\vec{r})$ is a vector. There is a difference
between the two vectors $\vec{F}$ and $\vec{E}$ in the
equation above. The force vector $\vec{F}$ is the force
on the particle placed at the position $\vec{r}$. If there is
no particle, there is no force vector. The electric field vector,
on the other hand, whether or not a particle is located at
$\vec{r}$. It can be defined at every point in space.
Such a quantity is called a {\bf vector field}. {\it If there is
a vector assigned to every point in space, this collection of
vectors is called a vector field}. Before we calculate the
electric field vectors for different charge distributions,
lets comment on some ideas of the electric field.\\
\noindent {\it Comments on the electric field}\\
\noindent 1. As noted, the electric field is a vector field, it
can be defined (or can exist) at all points in space. It
depends only on the sources of charge, i.e. the charge
distribution.\\
\noindent 2. The direction of the electric field at $\vec{r}$ is
the same as the direction of the force on a positive charge
if it is placed at $\vec{r}$.\\
\noindent 3. The force on a negatively charged object placed at
$\vec{r}$ is in the opposite direction as the direction of $\vec{E}(\vec{r})$.\\
\noindent 4. Is the electric field "real", or is it just a mathematical
quantity defined for convenience? Which vector is more "real" the
force $\vec{F}$ or $\vec{E}$? The way we have defined $\vec{E}$ above,
it is just a mathematical quantity. If you continue in physics, you
will see that light can be described (classically) by electric (and magnetic)
fields. Light seems to be real, so maybe $\vec{E}$ is also. However, experiments
in quantum mechanics are consistent with the electric field being related
to the probability of finding a photon at a particular location. This
interpretation would make one believe that $\vec{E}$ is not "real".
I'll let you decide how you want to think about the electric field.
It really doesn't matter what you think, as long as you do.\\
Let's finish the section on electric fields by calculating the electrostatic
field for some interesting charge distributions. After the calculation,
we will comment on the result. We start with the simplest
possibility: the electric field produced by a small "point" particle
that has a net charge of magnitude $Q$. We refer to this object
as a {\bf point charge}.\\
\noindent {\it $\vec{E}$ in the vicinity of a point charge}\\
Suppose there is a "point" particle located at the origin that has
a net charge of $Q$. What is the electric field at all points in
space due to this small charge? Consider a location at the
position $\vec{r}$ from the origin. If a particle of charge
$q$ is placed there, the electrostatic force $\vec{F}$ it feels
is given by Coulomb's law:
\begin{equation}
\vec{F} = k {{Qq} \over r^2} \hat{r}
\end{equation}
\noindent where $\hat{r}$ is a unit vector that points away from the origin.
Since the electric field is the force/charge, $\vec{E} = \vec{F}/q$ or
\begin{equation}
\vec{E}(\vec{r}) = k {{Q} \over r^2} \hat{r}
\end{equation}
\bigskip
\noindent {\it Comments}\\
\noindent 1. We can now use this simple result {\bf and the principle of
superposition} to find the electric field produced by any charge distribution.
It was a real theoretical breakthrough to
come up with a method of finding the electric produced by "complicated"
extended objects. The superposition principle is the key to the
simplification: the electric field from many sources add like vectors.
One needs only to understand the electric field produced by a "point" particle,
then integrate over the solid object to find the complete electric field.
It turns out that the electric field produced by a point particle is simple:
$\sim Q/r^2$. In this way, one can break down a
complicated problem into simpler ones.
In lecture we will first consider a collection of "point charges". Below I
summarize the results for some continuous (or solid) charge
distributions that will be covered in lecture.\\
\noindent {\it The Electric field on the axis a long thin rod}\\
Suppose we have a long thin rod of total charge $Q>0$ and length $l$.
Consider a point on the axis of the rod a distance
$d$ from one end. See the figure.
What is the electric field at the point located a distance $d$
from the end of the rod?
\begin{figure}
\includegraphics[width=14cm]{fig1331a.png}
\end{figure}
First we divide up the rod into small pieces. Lets divide it up
into $N$ equal pieces. Each piece will have a charge $\Delta Q$ and
a length $\Delta x$. Consider the electric field at the point
due to a small piece of rod a distance $x$ from the end.
The magnitude of the electric field, $\Delta E_x$,
from Coulomb's law is:
\begin{equation}
\Delta E_x = k {{\Delta Q} \over {(d+x)^2}}
\end{equation}
\noindent Since we will integrate over $x$, we need to express the
charge of the small piece $\Delta Q$ in terms of $\Delta x$. Since
the whole rod has a net charge $Q$ and length $l$, the charge of the small
piece is $\Delta Q = Q (\Delta x / l)$. Thus, we have
\begin{equation}
\Delta E_x = k {{Q \Delta x} \over {l (d+x)^2}}
\end{equation}
\noindent The final step is to add up the contribution from all
the little pieces of the rod as $\Delta x \rightarrow 0$. This
leads to the integral expression:
\begin{equation}
E_x = \int_0^l k {{Q \; dx} \over {l (d+x)^2}}
\end{equation}
\noindent with the result being:
\begin{equation}
E_x = k {{Q} \over {d(d+l)}}
\end{equation}
\noindent with the direction of the electric field pointing away from the rod
(since $q>0$).\\
\noindent {\it Comments}\\
\noindent 1. One can check the result for the
limiting case of $d>>l$. As $d$ becomes much greater than $l$, the
term $(d+l)$ is very close to $d$. Thus the limit of $E_x$ as
$d \rightarrow \infty$ is $kQ/d^2$ which is the electric field produced
by a point charge as expected.\\
\noindent 2. The ratio $Q/l$ is the charge per length of the rod. This
is a kind of density. You normally think of density as mass/volume.
$Q/l$ is the charge/length, and is called the {\bf linear charge density}.
It is often given the symbol $\lambda \equiv q/l$.\\
\noindent 3. Note that the electric field is not
equal to $kQ/(d+l/2)^2$:
\begin{equation}
E_x \neq k {{Q} \over {(d + l/2)^2}}
\end{equation}
\noindent That is, {\bf in general the electric field is not
the same as if all of the objects charge were located at its center of mass}.
There is one exception to this: the electric field produced by an object with spherical symmetry.
We will consider this case using a clever method discovered by Gauss. But first,
some other extended charge distributions.\\
\bigskip
\noindent {\it The Electric Field at a point on the perpendicular bisector
of a long rod}\\
If time permits we will calculate the electric field at a point that
is equal distant from the ends of the rod. Consider a rod of length
$l$ with a total charge $Q$ distributed uniformly on it. Lets choose
a point "P" that is equal distant from the ends and a distance $y$ from
the center of the rod. Let the rod lie on the x-axis with its
center at $x=0$. As before, we divide the rod up into small
segments, find the electric field at "P", $\Delta \vec{E}$, due to one of the segments,
and add up (integrate) all the segments of the rod (superposition principle).
Consider a segment located at a distance $x$ from the origin having a
charge of $\Delta Q$. The magnitude of $\Delta \vec{E}$ at the point
"P" due to the segment is
\begin{equation}
|\Delta \vec{E}| = k {{\Delta Q} \over {x^2 + y^2}}
\end{equation}
\noindent with the direction away from the segment. After
integrating over the rod, only the component away from the
rod will survive. That is, only $E_y$ will be non-zero.
The component of $\Delta \vec{E}$ in the "y" direction,
$\Delta E_y$ is given by
\begin{eqnarray*}
\Delta E_y & = & |\Delta \vec{E}| cos \theta\\
& = & {{k y \Delta Q} \over {(x^2 + y^2)^{3/2}}}
\end{eqnarray*}
\noindent As in the previous example, the charge $\Delta Q$
can be expressed in terms of $\Delta x$: $\Delta Q = (Q/l)
\Delta x$. $Q/l$ is the charge per length, or the
linear charge density. It is often given the symbol
$\lambda \equiv Q/l$. Substituting $\Delta Q = \lambda \Delta x$,
the integral becomes:
\begin{equation}
E_y = k \lambda y \int_{-l/2}^{+l/2} {{dx} \over {(x^2 + y^2)^{3/2}}}
\end{equation}
\noindent This integral can be solved analytically, with the result
being:
\begin{equation}
E_y = {{2k \lambda} \over {y \sqrt{1 + 4 y^2/l^2}}}
\end{equation}
\noindent {\it Comments}\\
\noindent 1. An interesting case of this result is to let the
length of the rod, $l$, approach infinity. As $l \rightarrow \infty$,
$E_y$ becomes:
\begin{equation}
E_y = {{2k \lambda} \over y}
\end{equation}
\noindent Thus, for the infinite charged rod, the electric field
decreases inversely with the distance from the rod.\\
\noindent 2. We will derive the case of the infinite rod using
Gauss' Law later, and obtain the same result.
\bigskip
\noindent {\it The electric field on the axis of a ring}\\
Suppose we have a ring of total charge $Q>0$ and radius $a$. Consider a
point on the axis of the ring a distance $x$ from the
center (See the figure). What is the
electric field at the point at $x$ due to the charged ring?
\begin{figure}
\includegraphics[width=14cm]{fig1331b.png}
\end{figure}
First we divide up the ring into small "point" pieces of charge $\Delta Q$.
Next, we find the magnitude of the electric field, $|\Delta \vec{E}|$,
at the point due to the small piece of the ring:
\begin{equation}
|\Delta \vec{E}| = k {{\Delta Q} \over {x^2 + a^2}}
\end{equation}
\noindent The direction of the electric field $\Delta \vec{E}$ points away from
the edge of the ring where $\Delta Q$ is. Now we need to sum up the forces due to all
the little pieces of the ring. This involves an integration: $\int \Delta \vec{E}$.
However, for the ring this is simple. Upon integrating around the ring, the only component
that survives is the one along the axis: $|\Delta \vec{E}| cos \theta$ or
\begin{equation}
\Delta E_x = k {{\Delta Q} \over {x^2 + a^2}} cos(\theta)
\end{equation}
\noindent towards the ring. Since $x$, $a$ and $\theta$ are the same for all the pieces of the
ring, the integral is simple:
\begin{equation}
E_x = k {1 \over {x^2 + a^2}} cos(\theta) \int \Delta Q
\end{equation}
\noindent The integral over $\Delta Q$ is just $Q$, so
\begin{equation}
E_x = k {{Q} \over {x^2 + a^2}} cos(\theta)
\end{equation}
\noindent away from the ring. Since $cos\theta = x/\sqrt{x^2 + a^2}$, we have
\begin{equation}
E_x = k {{Qx} \over {(x^2 + a^2)^{3/2}}}
\end{equation}
\noindent with the direction away from the ring. \\
\noindent {\it Comments}\\
\noindent 1. We will derive this result later by a different
method: solving for the electric potential and differentiating
it. The result will be the same.
\bigskip
\noindent {\it Electric field on the axis of a circular disk}\\
As a final example of integrating over an extended charge distribution,
we will consider calculating the electric field on the axis of
a thin circular disk that is uniformly charged. Suppose we have a
very thin disk of radius $a$ that has a total charge of $Q$ uniformly
distributed on its surface. We are interested in finding the
electric field at a point on the axis a distance $x$ from the center.
We can divide the disk into a series of concentric rings. We just derived
an expression for the electric field on the axis of a ring, so we can just
add up the contributions of the concentric rings that make up the
circular disk. Let $s$ be the radius of a concentric ring with
thickness $\Delta s$, and let the ring have a charge of $\Delta Q$.
The magnitude of the electric field on the axis due to this ring is
\begin{equation}
E_x = k {{\Delta Q x} \over {(x^2 + s^2)^{3/2}}}
\end{equation}
\noindent from our example of the ring above. To find the charge of
the ring, we use: $(\Delta Q/Q) = (2 \pi s \Delta s)/(\pi a^2)$.
Substituting above, we have:
\begin{equation}
E_x = { {2 \pi k Q x s \Delta s} \over {\pi a^2 (x^2 + s^2)^{3/2}} }
\end{equation}
\noindent Integrating the rings over the whole disk we have:
\begin{equation}
E_x = {{2 \pi k Q x} \over {\pi a^2}} \int_0^a {{s \; ds} \over {(x^2 + s^2)^{3/2}}}
\end{equation}
\noindent Solving the integral yields:
\begin{equation}
E_x = {{Q/(\pi a^2)} \over {2 \epsilon_0}} (1 - {{x} \over {\sqrt{x^2 + a^2}}})
\end{equation}
\noindent The ratio $Q/(\pi a^2)$ is the charge per area of the disk, and is also
a type of charge density. In this case it is an area density. The symbol
$\sigma$ is often used to denote charge/area, {\bf the surface charge density}
$\sigma \equiv \Delta Q / \Delta A$. With
this definition, we have
\begin{equation}
E_x = {{\sigma} \over {2 \epsilon_0}} (1 - {{x} \over {\sqrt{x^2 + a^2}}})
\end{equation}
\noindent We looked at this example for two reasons. One, to introduce the
idea of an area charge density ($\sigma$). The second is to show what happens
in the case of an infinitly large disk.\\
\noindent {\it Comments}\\
\noindent 1. In this case of the infinitly large disk, $a \rightarrow \infty$.
We then have
\begin{equation}
E_x = {\sigma \over {2 \epsilon_0}}
\end{equation}
\noindent for the magnitude of the electric field a distance $x$ from an
infinitly large thin disk. This is a very interesting result, it means
that the electric field is constant near an infinitely large
plane of charge.\\
\noindent 2. The above result tells us how to make a constant electric
field: uniformly distribute charge on a very large plate.\\
\noindent 3. We will derive the case of the infinite plane of charge
again using Gauss' Law, and obtain the same result.\\
\bigskip
\centerline{\bf Flux in General}
\bigskip
An important mathematical quantity for the formulation of the
laws of electrodynamics is flux. We will first define what flux
is, then see how we can recast Coulomb's Law and the superposition
principle for electrostatics in terms of the flux of the electric
field.
In order to define flux, one needs two things:\\
\noindent 1. {\it A vector field}: A vector field is a "field of
vectors". That is, at every point in space there is a vector
defined. We have seen one so far, the electric field $\vec{E}$.
We will encounter another one this quarter, the magnetic field
$\vec{B}$. Other vector fields in physics include the
velocity vector in a fluid and the graviational field to name
a few.\\
\noindent 2. {\it A surface}. The surface can be a flat surface,
a curved surface, or a closed surface.\\
We will define flux as a scalar quantity. The vector field is a
vector. So to make a scalar out of a vector, we will need to
use the scalar (or dot) product with another vector. We will
do this by associating a vector with the surface. This will be
possible if the surface is flat, or very small such that it can
be considered flat. Consider a flat surface. {\bf We define the area
vector associated with the flat surface as a vector whose
direction is perpendicular to the surface and whose magnitude
equals the area of the surface}. Last quarter, we also
defined the area vector associated with a surface the same way.
We will first define flux for a constant vector field and a flat surface.
Suppose there exists a constant vector field $\vec{V}$. This means that
at every point in space there is a vector $\vec{V}$, the same vector
at all points in space. Also suppose there is a flat surface that has
an area $A$. The area vector $\vec{A}$ associated with this surface
is perpendicular to the surface. The flux, usually labeled as $\Phi$,
is defined as:
\begin{equation}
\Phi \equiv \vec{V} \cdot \vec{A}
\end{equation}
\noindent Remember this definition holds for a constant vector field
and a flat surface. Flux is a scalar. The dot product is maximized
when the two vector fields are parallel to each other. Thus, the
flux will be maximized when the vector field is perpendicular to the
surface. The dot product is zero when the two vector fields
are perpendicular to each other. Thus, the flux will be zero when
the surface is "parallel" to the vector field. Qualitatively,
one can think of the flux as how much of the vector field passes
through the surface. If we let $\theta$ be the angle between
the area vector and the vector field,
\begin{equation}
\Phi \equiv |\vec{V}| |\vec{A}| cos(\theta)
\end{equation}
\noindent for a constant vector field and a flat surface. How do we generalize the
definition of flux to include a vector field that can vary from point
to point in space and to a surface that is not necessarily flat? We
can do this by dividing up the surface $S$ into very small surfaces,
$\Delta S_i$ where $i$ is the i'th little surface piece. If the
$\Delta S_i$ are small enough, they will be very close to being a
flat surface. The smaller they are, the closer they will be to being
flat. Also, if the surface is small enough, the vector field will be
approximately constant on it. Thus, if $\vec{V}_i$ is the vector
at the center of the i'th surface, we can define the flux of the
vector field $\vec{V}(\vec{r})$ over the surface $S$ as
\begin{equation}
\Phi_V = \lim_{\Delta S_i \rightarrow 0} \sum_i \vec{V}_i \cdot \Delta \vec{A}_i
\end{equation}
\noindent where $\Delta \vec{A}_i$ is the area vector associated with the
i'th surface. Thus, the limiting process allows us to extend our definition
of flux for a flat surface to a curved one. As the small surfaces
$\Delta S_i$ get smaller and smaller, the surfaces approach flat ones and
the vector field approaches a constant one over the surface. The number of
surfaces goes to infinity and the sum becomes an integral. This kind of
integral is called a {\bf surface integral} and is written as
\begin{eqnarray*}
\Phi_V & = & \lim_{\Delta S_i \rightarrow 0} \sum_i \vec{V}_i \cdot \Delta \vec{A}_i\\
& = & \int \! \! \int_S \vec{V}(\vec{r}) \cdot d \vec{A}
\end{eqnarray*}
\noindent The two integral signs denote that it is a two dimensional integral over
the surface $S$. Often $\vec{V}(\vec{r})$ is written simply as $\vec{V}$. However,
one must remember that $\vec{V}$ is a vector field and in general can depend on the
location of the surface. The subscript $V$ on $\Phi$ reminds us that the flux
is for the vector field $\vec{V}$. Remember, when discussing flux, one needs
to specify the {\bf vector field} and the {\bf surface}. The above integral
can be complicated. In this course, we will only consider surface integrals
that are easy to compute. We will usually choose our surfaces such that
the vector field is perpendicular to the surface, and has a simple
dependence over the surface.
You might be wondering why we are introducing the concept of flux. So
far we have considered electrostatics, and haven't needed to use
flux in our description. However, when we investigate changing electric
(and magnetic) fields, flux will be useful for describing the physics.
In the next section we will restate Coulomb's law using flux, via Gauss' Law. Although
Gauss' Law does not add any new "physics", it helps in solving some problems
involving conductors and symmetric charge distributions. It also is a good
introduction to the concept of flux.
\bigskip
\centerline{\bf Electric Flux}
\bigskip
Electric flux, $\Phi_E$, is the flux for which the vector field is the
electric field. The surface must be specified. For electrostatics,
the most interesting surface to calculate electric flux is a
closed surface. A {\bf closed surface} is a surface that is closed,
encloses a volume. For a closed surface there will be an inside
and an outside. We choose the direction of the area vector to
be {\bf positive} if it points {\bf outward from the surface}.
Suppose there is a "point" particle located at the origin that
has a net charge of $Q>0$. Consider the electric field produced
by this charge as the "vector field". For a surface for calculating
$\Phi_E$ let's choose a closed spherical surface that has a radius $r$ whose
center is at the origin. To denote a closed surface, we write a circle
in the integral signs:
\begin{equation}
\Phi_E = \oint \! \! \! \oint \vec{E} \cdot d \vec{A}
\end{equation}
\noindent Now let's calculate the right side for a spherical surface
for the electric field produced by a point charge at the center.
Since the electric field is radial from $Q$, $\vec{E}$ is perpendicular
to the surface at all points on the surface. Also, since all points on
the surface are the same distance from the charge, the magnitude of the
electric field on the surface is constant. So the integral is very
simple. It is just the magnitude of the electric field times the surface
area:
\begin{equation}
\Phi_E = |\vec{E}| 4 \pi r^2
\end{equation}
\noindent Since $\vec{E}$ is produced from a point charge, we have from
Coulomb's Law, $|\vec{E}| = k Q/r^2$:
\begin{equation}
\Phi_E = {{kQ} \over r^2} 4 \pi r^2
\end{equation}
\noindent The $r^2$ cancel, and we are left with
\begin{equation}
\Phi_E = 4 \pi k Q
\end{equation}
\noindent Usually one substitutes for $k$: $k= 1/(4 \pi \epsilon_0$),
\begin{equation}
\Phi_E = {Q \over \epsilon_0}
\end{equation}
\noindent Note that in this expession, the radius of the spherical
surface does not enter. That is, no matter how big or small $r$ is
the electric flux through the spherical surface is $Q/\epsilon_0$.
You might think that this simple result would only apply to the
flux through a closed spherical surface with a charge $Q$ at
the center. However, we will show what Gauss did, that {\bf even if
the surface is not a spherical one the result is the same as
long as the charge is within the surface}. This is because of the
inverse square fall-off of the electric field with distance from
the point source. In lecture we will show that if one considers
wedges eminating from the charge that the flux through any surface
intersecting the wedge will be the same. Thus, the above equation
is valid for any surface that encloses the point charge:
\begin{equation}
\Phi_E = {Q \over \epsilon_0}
\end{equation}
\noindent {\bf for any closed surface that encloses the point charge}.
If there is more than one charge within the surface then we can use
the principle of superposition. Since the electric fields add like
vectors, the electric flux through a closed surface will just be the
sum of all the charge within the surface divided by $\epsilon_0$:
\begin{equation}
\Phi_E = {Q_{Net \; inside \; closed \; surface} \over \epsilon_0}
\end{equation}
\noindent This is known as {\bf Gauss' Law}. One often writes
explicitly the integral expression for the flux:
\begin{equation}
\oint \! \! \oint \vec{E} \cdot d \vec{A} = {Q_{Net \; inside \; closed \; surface} \over \epsilon_0}
\end{equation}
\noindent Gauss' Law can also be stated in words: The electrical flux
(flux for the electric vector field) for any closed surface is equal
to the net charge enclosed by the surface. Before we do examples
of Gauss' Law for conductors and symmetric charge distributions we
mention a few comments.\\
\noindent {\it Comments on Gauss' Law}:
\noindent 1. As previously stated, Gauss's Law is just another way of
expressing Coulomb's Law and the superposition principle. This is
because Gauss's law is only true if the vector field decreases as
$1/r^2$ from a point source.\\
\noindent 2. Gauss's Law is true for any closed surface. The right
side of the equation is easy to evaluate, since one just adds up
all the charge within the surface. The left side of the equation
can be very difficult to evaluate, since one must do a surface
integral. However, if one can choose a surface for which the
electric field is simple then the left side can be determined.
In the applications for Gauss' Law, we will seek out simple
situations for which the left side can be evaluated.\\
\noindent 3. The surface one chooses to apply the equation above
is just a mathematical surface. There doesn't need to be anything
on it. The closed surface that one chooses for evaluating the
electric flux is often called a "Gaussian Surface".\\
\noindent 4. Gauss' Law relates the electric field on a closed
surface with the charge inside the surface. It is a geometrical
approach for describing the electrostatic force.
\bigskip
\centerline{\bf Examples using Gauss' Law}
\bigskip
Gauss's Law relates the surface integral of the electric field for a
closed surface (electric flux) to the charge within the surface:
\begin{equation}
\oint \! \! \oint \vec{E} \cdot d \vec{A} = {Q_{Net \; inside \; closed \; surface} \over \epsilon_0}
\end{equation}
\noindent Gauss's law is true for {\bf any closed surface}, however the flux
integral on the left is often very difficult to evaluate. In this class we
will apply the above equation to situations for which the left side is
"not too difficult" to determine. The trick is to pick a situation for
which there exists a surface for which the electric field is simple.
This will be true for certain symmetric charge distributions as well
as fields within a conductor.
\bigskip
\noindent {\it Infinitely Long (and infinitely thin) rod of charge}\\
Consider the electric field produced by an infinitely long (thin)
rod. Assume that the rod is non-conducting and is uniformly
charged. We can describe the amount of charge in terms of
the {\bf linear charge density} $\lambda$. Since the rod is
infinitely long it has an infinite amount of charge. However
the charge/length is a finite number.\\
What properties does the electric field have?\\
\noindent a) Since the rod is infinitely long, the electric field
must point radially away from the rod. It cannot have any component
along the direction of the rod, since each direction is equivalent.
It cannot have any component circulating the rod since there is no
preference for clockwise versus counter-clockwise.\\
\noindent b) The magnitude of the electric field can only depend on
the radial distance from the rod.\\
\noindent Because of these properties, if we choose a surface that
is a "can" whose axis coincides with the rod, then the electric
flux integral is easy to evaluate. Let the "can" have a radius
$r$ and a height $L$. We need to calculate the electric flux
over the whole closed surface. There will be three parts:
one over the sides of the "can", one over the right end and one
over the left end:
\begin{equation}
\oint \! \! \oint \vec{E} \cdot d \vec{A} = \int (sides) + \int (right)
+ \int (left)
\end{equation}
\noindent The flux integral over the ends of the "can" equal zero. This is
because the electric field is parallel to the ends of the "can". The
flux integral over the side of the "can" just equals the area of the side
times $E(r)$ since $\vec{E}$ is perpendicular to the surface:
\begin{equation}
\oint \! \! \oint \vec{E} \cdot d \vec{A} = 2 \pi r L E(r) + 0 + 0
\end{equation}
\noindent The amount of charge within the surface is just $\lambda L$,
so Gauss' Law yields:
\begin{equation}
2 \pi r L E(r) = {{\lambda L} \over \epsilon_0}
\end{equation}
\noindent Solving for $E(r)$ we have:
\begin{equation}
E(r) = {{ \lambda} \over {2 \pi \epsilon_0 r}} = {{2 k \lambda} \over r}
\end{equation}
\bigskip
\noindent {\it Thin shell which is uniformly charged}\\
Consider the electric field produced by a thin shell of radius
$R$ that has a total charge of magnitude $Q$ distributed uniformly
on its surface.
What properties does the electric field have?\\
\noindent a) Since the charge distribution is spherically
symmetric, the direction of the electric field will be in the
radial direction. That is, it must point away from or towards
the center of the shell.\\
\noindent b) Since all directions are equivalent, the
magnitude of the electric field can only depend on $r$, where
$r$ is the distance to the center of the shell.
That is, at the position $\vec{r}$: $\vec{E}(\vec{r}) = E(r) \hat{r}$. Here
$E(r)$ is the magnitude of the vector field a distance
$r$ away from the center.\\
Thus, if we choose a spherical surface of radius $r$ centered
at the shell, then the flux integral is easy to evaluate. Since
$\vec{E}$ is perpendicular to the surface with a constant magnitude:
\begin{equation}
\oint \! \! \oint \vec{E} \cdot d \vec{A} = E(r) 4 \pi r^2
\end{equation}
\noindent If $r > R$, Gauss' Law gives
\begin{equation}
E(r) 4 \pi r^2 = {Q \over \epsilon_0}
\end{equation}
\noindent Since all the charge is within the surface.
Solving for $E(r)$ we have
\begin{equation}
E(r) = {Q \over {4 \pi \epsilon_0 r^2}} = k {Q \over r^2}
\end{equation}
\noindent If $r < R$ then there is no charge within the surface
and we have
\begin{equation}
E(r) 4 \pi r^2 = 0
\end{equation}
\noindent Or,
\begin{equation}
E(r) = 0
\end{equation}
\noindent for $r R$, Gauss' Law gives
\begin{equation}
E(r) 4 \pi r^2 = {Q \over \epsilon_0}
\end{equation}
\noindent Since all the charge is within the surface.
Solving for $E(r)$ we have
\begin{equation}
E(r) = {Q \over {4 \pi \epsilon_0 r^2}} = k {Q \over r^2}
\end{equation}
\noindent This is the same result as for the spherical shell.\\
Inside the sphere, $r0$, and
the other with a charge of $q>0$. Suppose they are initially
separated by a distance $r_i$. Let the particle with charge
$Q$ be held fixed, and let the other particle be "pushed" directly
away from $Q$ by the electrostatic force to a larger distance $r_f$. How much
work is done by the electrostatic force ($W_{electrostatic}$) in pushing the
particle from $r_i$ to $r_f$?
\begin{equation}
W_{electrostatic}(r_i \rightarrow r_f) = \int_{r_i}^{r_f} \vec{F} \cdot d\vec{r}
\end{equation}
\noindent where $\vec{F}$ is the electrostatic force between the two
point particles. From Coulomb's Law, we have
\begin{equation}
W_{electrostatic}(r_i \rightarrow r_f) = \int_{r_i}^{r_f} k {{Qq} \over r^2} dr
\end{equation}
\noindent Since $q$ moves directly away from $Q$, the direction of $d\vec{r}$ is
in the same direction as the force. This is essentially a one dimensional problem.
Carrying out the integration, we have
\begin{equation}
W_{electrostatic}(r_i \rightarrow r_f) = k{{Qq} \over r_i} - k{{Qq} \over r_f}
\end{equation}
\noindent Note: that this is the work done by the electrostatic force if the
path that $q$ moves is directly away from $Q$. What if $q$ is moved in a circular
direction around $Q$ (i.e. with a constant radius $r$)? In this case the
direction of $\vec{F}$ is {\bf perpendicular} to the direction of the
path. The work done by the electrostatic force will be zero in this case.
Any path connecting $\vec{r}_i$ and $\vec{r}_f$ can be divided up into small
line segments. For each small line segment, the work done by the electrostatic
force will be the force times the change in radial distance, since any circular
path direction does not contribute to the work. Thus, the result of the
above equation will be true for {\bf any path} from $\vec{r}_i$ to $\vec{r}_f$.
Since the work done is path independent, it is a conservative force.
This was the same result we found with the gravitational force, since it was
also a central force and decreases as $1/r^2$. Actually, any central force
will be conservative.
If $q$ has a speed $v_i$ at $\vec{r}_i$, and a speed $v_f$ at $\vec{r}_f$,
then from work-energy theorum:
\begin{eqnarray*}
Net \; Work & = & \Delta K.E.\\
k{{Qq} \over r_i} - k{{Qq} \over r_f} & = & {{mv_f^2} \over 2} -
{{mv_i^2} \over 2}
\end{eqnarray*}
\noindent Moving the initial parameters to the left and the final parameters
to the right, we have:
\begin{equation}
{{mv_i^2} \over 2} + k{{Qq} \over r_i} = {{mv_f^2} \over 2} + k{{Qq} \over r_f}
\end{equation}
\noindent As $q$ moves in the presence of $Q$ (which is fixed), the expression
$mv^2/2 + kQq/r$ remains constant. The above combination of terms is conserved.
We have been calling this combination mechanical energy, with the first term
equal to the kinetic energy, and the second term is defined as the potential energy.
The electrical potential energy $U$ of two point charges is:
\begin{equation}
U = k {{Qq} \over r}
\end{equation}
\noindent {\it Comments}:\\
\noindent 1. Potential energy is a scalar.\\
\noindent 2. Potential energy falls off as $1/r$. The force decreases
as $1/r^2$, but since potential energy involves the integral of force,
one power of $r$ is reduced in the denominator.\\
\noindent 3. The reference for zero potential is at $r = \infty$.\\
\noindent 4. The expression is also true if $Q$ and/or $q$ are
negative. If they have opposite signs, the system has negative potential
energy. It takes work to separate them infinitely far apart.\\
The above expession is also called the electrostatic potential energy
of the two particle system. If there are three point particles in the
system, the total potential energy is found by adding the contributions
from each pair of particles:
\begin{equation}
U = k{{q_1 q_2} \over r_{12}} + k{{q_1 q_3} \over r_{13}} + k{{q_2 q_3} \over r_{23}}
\end{equation}
\noindent where $r_{ij}$ is the distance between particle "i" and particle "j".
One can add up the potential energies from each pair because electric
forces add like vectors (superposition principle). For example, start with
$q_1$ fixed in space. The first term is the energy needed to bring particle
"2" a distance $r_{12}$ from particle "1". Now keep particle "1" and "2" fixed
and bring in particle "3". Since the force on "3" is the vector sum of the
forces due to "1" plus "2" one has the last two terms in the sum. This idea
can be extended to any number of point particles.
\bigskip
\centerline{\bf Summary}
\bigskip
This ends our introduction to electrostatics. All the "physics" is
described by: Coulomb's Law for point particles plus the superposition
principle. Since the force on a particle is proportional to the charge
on it, it was convenient to define the "Force per charge" which we
called the electric field. We discussed a different way to state
Coulomb's Law: Gauss' Law. We found the electric field for different
charge distributions consisting of "point charges" as well as
continuous charge distributions. Finally, we used the concept of
work to determine the electrostatic potential energy of a collection
of point particles. In the next section we will continue to
extend the physics of Coulomb's law to define voltage, determine
how charge behaves on conductors (capacitors), as well as simple electrical
circuits.
\end{document}