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\begin{center}
{\bf Notes on Voltage, Capacitance, Circuits}
\end{center}

\bigskip

\centerline{\bf Voltage}

At the end of the last section we calculated the work done {\bf by} the
electrostatic force when a particle of charge $q$ is moved
directly away from a fixed particle of charge $Q$.  If $r_i$ is
the initial distance and $r_f$ the final distance from the
charge $Q$, then the work was found by integrating $kQq/r^2$ from
$r_i$ to $r_f$:

\begin{equation}
  W_{electrostatic}(r_i \rightarrow r_f) = k{{Qq} \over r_i} -
                        k{{Qq} \over r_f}
\end{equation}

\noindent This result was for the special path that was a straight
line away from the charge $Q$.  We will show in class that one gets
the {\bf same result for any path} that the charge $q$ travels between
$\vec{r}_i$ and $\vec{r}_f$.  this is because the electrostatic force is a central force,
and hence conservative.  The only contribution to the work done by
the electric force is from the component of the path element that is
radially away from the source charge $Q$.  Only the initial and final
distance matter in the net work done by the electrostatic force on
the point charge $q$.

This is a very nice result, and can be generalized to any number of
"point" source particles.  Suppose we have $N$ small "point" particles.
Label them 1, 2, ...., i, ... N, and let them remain fixed in place.  
Label the charge on the i'th particle as $Q_i$.  Since the superposition
principle applies to the electric force, the work done {\bf by} the
electric force as the point charge $q$ follows a path from
$\vec{r}_a$ to $\vec{r}_b$ is given by:

\begin{equation}
  W_{electrostatic}(r_a \rightarrow r_b) = \sum_{i=1}^N k{{Q_iq} \over r_{ia}} -
                        k{{Q_iq} \over r_{ib}}
\end{equation}

\noindent where $r_{ia}$ is the initial distance from the particle to $Q_i$ and
$r_{ib}$ is the final distance from the particle to $Q_i$.  Notice in this
equation the $q$ can be factored out!

\begin{equation}
  W_{electrostatic}(r_a \rightarrow r_b) = q \sum_{i=1}^N ( k{{Q_i} \over r_{ia}} -
                        k{{Q_i} \over r_{ib}})
\end{equation}

\noindent This is possible since the electric force is proportional to the charge
on an object.  From this equation we see that it is useful to consider the
work done by the electrostatic force {\bf per charge}.  We call this quantity
the {\bf electric potential difference} between the points $\vec{r}_a$ and 
$\vec{r}_b$.  It is also refered to as the {\bf voltage difference}, 
$V(\vec{r}_a) - V(\vec{r}_b)$, between the two points.

\begin{equation}
  W_{electrostatic}(r_a \rightarrow r_b) = q(V(\vec{r}_a) - V(\vec{r}_b))
\end{equation}

\noindent Comparing with the equation above, we have

\begin{equation}
   V(\vec{r}_a) = \sum_{i=1}^N k{Q_i \over r_{ia}}
\end{equation}

\noindent where the formula for $V(\vec{r}_b)$ is the same with
$a$ replaced by $b$.  $V(\vec{r}_a)$ is called the electric
potential at the position $\vec{r}_a$.  Usually the subscript $a$
is omitted, and the expression for the electric voltage 
at the position $\vec{r}$ is 

\begin{equation}
   V(\vec{r}) = \sum_{i=1}^N k{Q_i \over r_{i}}
\end{equation}

\noindent where $r_i$ is the distance between the position
$\vec{r}$ and the point charge $Q_i$.  If there is only one
source charge, then $V(\vec{r}) = k Q/r$ where $r$ is the distance
from the position $\vec{r}$ to the charge $Q$.  If there is a
continuous distribution of charge, then the sum becomes an
integral:

\begin{equation}
  V(\vec{r}) = \int k {{\rho (\vec{r}') dV'} \over {|\vec{r} - \vec{r}'|}}
\end{equation}

\noindent In class we will do examples for which we calculate the voltage
and voltage difference for various point and continuous charge distributions.\\

\noindent {\it Comments}\\

\noindent 1.  An important application of voltage difference is that it
is directly related to the work done by the electric force on the charge
$q$.  If $\Delta V \equiv V(\vec{r}_a) - V(\vec{r}_b)$ then $q \Delta V$ is
the work done on the charge $q$ if it travels from point $a$ to point $b$.
The energy the particle acquires is simply $q \Delta V$.  Often this 
energy gain is equal to the particles gain in K.E.\\

\noindent 2.  $V(\vec{r})$ is defined at a point in space, the point at the
position $\vec{r}$.  Nothing needs to be at the point.  This concept is
similiar to the electric field.  The electric field is the force per charge;
{\bf The voltage is the potential energy per charge}.\\

\noindent 3. Although $\vec{r}$ is a vector, a position vector, the
voltage is a {\bf scalar}.  Since voltage is a scalar, it is much
easier to calculate than the electric field.  There is no direction
involved in using the superposition principle.\\

\noindent 4.  Only voltage differences are measurable.  Our expression for 
the voltage due to a point particle of magnitude $Q$, $k Q/r$, takes as
a reference $V(\infty) = 0$.\\

\bigskip

One can always calculate the voltage at a point in space by summing up
the expression above over all the "point" particles in the system.  However,
it is often easier to express the voltage difference in terms of
the electric field.  This can be done as follows.  First start with
the expression for the work done by the electrostatic force for a path
going from $\vec{r}_a$ to $\vec{r}_b$:

\begin{equation}
 W_{electrostatic}(\vec{r}_a \rightarrow \vec{r}_b) = \int_{\vec{r}_a}^{\vec{r}_b} 
                                                   \vec{F} \cdot d \vec{r}
\end{equation}

\noindent where the integral is over the path from $\vec{r}_a$ to $\vec{r}_b$.
Since the electrostatic force is conservative, one obtains the same result
for any path between the two points.  Since $\vec{F} = q \vec{E}$, we have

\begin{equation}
 W_{electrostatic}(\vec{r}_a \rightarrow \vec{r}_b) = q \int \vec{E} \cdot d \vec{r}
\end{equation}

\noindent However, the left side of the equation is just the difference in voltage
$\Delta V$ times $q$.  Thus, the $q$'s cancel and we are left with

\begin{equation}
 \Delta V  = \int_{\vec{r}_a}^{\vec{r}_b} \vec{E} \cdot d \vec{r}
\end{equation}

\noindent or

\begin{equation}
 V(\vec{r}_a) - V(\vec{r}_b)  = \int_{\vec{r}_a}^{\vec{r}_b} \vec{E} \cdot d \vec{r}
\end{equation}

\noindent This equation is also written as

\begin{equation}
 V(\vec{r}_b) - V(\vec{r}_a)  = - \int_{\vec{r}_a}^{\vec{r}_b} \vec{E} \cdot d \vec{r}
\end{equation}

\noindent We will do a number of examples in which we calculate the difference in
electric potential by integrating $\vec{E} \cdot d \vec{r}$ over a path.  The right
side of the equation is called a line (or path) integral.

If the electric field is known at every point in space, one can use the
above equation to calculate the voltage difference between any two
points.  Thus, from knowledge of the electric field, the
electric potential can be determined.  What about the converse?
If the voltage is known at every point in space, can one determine
the electric field?  The answer is yes.  Since $V$ is obtained by
integrating $\vec{E}$, then you might guess that $\vec{E}$ is related
to the derivative of $V$.  However, $\vec{E}$ is a vector, and $V$ is
a scalar.  We will probably need three derivatives of $V$, one for each
component, to find all three components of $\vec{E}$.  Consider first
$E_x$.

Let's choose a path for the above equation to be on the x-axis.  If we 
choose $b$ to be at $x + \Delta x$ and choose $a$ to be at $x$, then
the above equation for the voltage difference between these two
positions is:

\begin{equation}
 V(x+\Delta x) - V(x) = - E_x \Delta x
\end{equation}

\noindent Dividing by $\Delta x$ and taking the limit as
$\Delta x \rightarrow 0$ we have:

\begin{equation}
  E_x = lim_{\Delta x \rightarrow 0} - {{V(x+\Delta x) - V(x)} \over {\Delta x}}
\end{equation}

\noindent which yields

\begin{equation}
  E_x = - {{\partial V} \over {\partial x}}
\end{equation}

\noindent The partial derivative is used, since $V$ depends on 
$x$, $y$, and $z$.  When one takes the partial derivative with
respect to $x$, $y$ and $z$ are held constant.  The same
procedure can be done with the y- and z- directions:

\begin{eqnarray*}
  E_x & = & - {{\partial V(x,y,z)} \over {\partial x}}\\
  E_y & = & - {{\partial V(x,y,z)} \over {\partial y}}\\
  E_z & = & - {{\partial V(x,y,z)} \over {\partial z}}
\end{eqnarray*}

In summary, we have discussed two ways to find the electric potential at a point,
or electrical potential difference between two points:\\

\noindent 1) use the formula for a point charge source $kQ/r$ and sum up (or integrate) 
over the source charges\\

\noindent 2) First determine $\vec{E}$ and carry out the line integral $\int \vec{E}
\cdot d\vec{r}$.  We will do examples in class, and I present an example of
each method here in the notes.\\

\bigskip

\noindent {\it Electric Potential on the axis of a Charged Ring}\\

Suppose we have a uniformly charged ring of radius $R$ and total charge $Q$.
Lets find the electric potential $V$ on the axis of the ring a distance
$x$ from the center.  For this problem, it is best to "chop" up the ring 
into small pieces, and sum up the contribution from each piece. 
Let the pieces be labeled as 1, 2, 3, ... , i, ...., and $\Delta Q_i$
be the charge on the i'th piece.  The potential at the position $x$ due to
the i'th piece, $\Delta V_i$, is just $k \Delta Q_i/\sqrt{x^2 + R^2}$.  This is true
since each small piece can be treated as a point particle and the distance
from $x$ to the piece is just $\sqrt{x^2 + R^2}$ via Pythagorus Theorum.

\begin{equation}
  \Delta V_i = k {{\Delta Q_i} \over {\sqrt{x^2+R^2}}}
\end{equation}

\noindent Note: {\bf there is no direction to $\Delta V_i$}, potential
is a scalar.  When we add up the contributions of the pieces over the
whole ring, $x$ and $R$ do not change, thus

\begin{equation}
  V = {k \over {\sqrt{x^2+R^2}}} \sum \Delta Q_i
\end{equation}

\noindent The sum over the $\Delta Q_i$ just adds up to $Q$:

\begin{equation}
  V(x,0,0) = k {Q \over {\sqrt{x^2+R^2}}}
\end{equation}

We can differentiate the above expression to find the electric field on the axis:

\begin{eqnarray*}
  E_x & = & - {{\partial V} \over {\partial x}}\\
      & = & {{kQx} \over {(x^2 + R^2)^{3/2}}}
\end{eqnarray*}

\noindent Note: this is the same result we obtained by using Coulomb's Law
for the electric field and integrating over the ring.  Find the electric
potential first and differentiating was easier, and often is.  Sometimes it
is easier to find $\vec{E}$ first and then $V$, and sometimes it is easier
to determine $V$ first and differentiate to obtain $\vec{E}$.\\

\noindent {\it Potential Difference between two parallel plates}\\

Suppose we have two flat identical plates, each with an area $A$.
Suppose the plates are parallel to each other and separated by a distance
$d$.  Lets also assume that $\sqrt{A} >> d$, that is the plates are
very large compared to the distance separating them.  Lets place a total of
$+Q$ amount of charge on one plate and a total of $-Q$ on the other.
We spread the charge uniformly over the surface area $A$.  What is the
electric potential difference (voltage difference) from one plate to the
other?

In this case it is easiest to first determine the electric field between
the plates and carry out $\int \vec{E} \cdot d \vec{r}$.  We already determined
the electric field from an infinite plate using Gauss' Law as well as using
Coulomb's Law and integrating over the plate.  Since $\sqrt{A} >>> d$, we
can approximate the plates as infinite.  Previously we found the electric
field between the parallel plates by adding up the electric field from each
plate: $|\vec{E}| = Q/(2 A \epsilon_0) + Q/(2 A \epsilon_0)$.  So the
net electric field between the plates is $|\vec{E}| = Q/(A \epsilon_0)$ and
is {\bf constant} between the plates.  Since the electric force is conservative,
we can take calculate the potential difference using any path from one plate to
the other.  Lets choose a path straight across, and since $\vec{E}$ is constant, the
line integral is very easy.

\begin{eqnarray*}
  V & = & \int \vec{E} \cdot d \vec{r} \\
    & = & E d\\
  V & = & {{Qd} \over {A \epsilon_0}}
\end{eqnarray*}

\noindent The line integral is simple for the following reasons:
1) since $\vec{E}$ and $d \vec{r}$ are in the same direction, the dot
product is just the produce of the magnitude of $\vec{E}$ and $d \vec{r}$.
2) Since $\vec{E}$ is constant, the line integral is just $|\vec{E}|$ times
the distance between the plates.

Remember that the voltage difference gives us the electrical potential energy
change {\bf per charge} of a particle.  That is, if a particle of charge $q$ 
travels through a potential difference $V$, the work done by the electrical
force is $W = q V$.  {\bf If the electrical force is the only force that
does work on the particle}, then the gain in K.E. is $q V$ so

\begin{equation}
  qV = {{mv_f^2} \over 2} - {{mv_i^2} \over 2}
\end{equation}

\noindent If the particle starts from rest, it attains a speed
$v = \sqrt{2 q V/m}$.  If an electron travels through a voltage
difference of one volt, the work done on it by the electrical force
is $1.6 \times 10^{-19}$C $(1)$Volt = $1.6 \times 10^{-19}$ Joules.
This unit of energy is called an electron-volt or eV.  $1$eV =
$1.6 \times 10^{-19}$ Joules.

\bigskip

\centerline{\bf Capacitance}

\bigskip

Capacitance is a property of {\bf two conductors}.  Suppose we
have two conductors (metals) that are initially uncharged.
Suppose we then put a total charge of $+Q$ on one and a total charge of
$-Q$ on the other.  This can be accomplished by transfering electrons
from one to the other.  If we wait for the charges (electrons) to come
to static equilibrium, the electric fields within each conductor will
be zero.  This means that the whole conductor is at the same electrical
potential.  At points outside the conductors there will be electric 
fields, and $\int \vec{E} \cdot d \vec{r}$ from one conductor to the
other will be the potential difference $V$ between the conductors.

Now, there is a nice property about the potential difference.  If we
were to transfer twice as much charge, the electric fields everywhere
would double and so would the potential difference $V$.  If three times
as much charge is transfered, the electric fields would triple and so
would $V$.  That is, $V$ is proportional to $\vec{E}$ since 
$V = \int \vec{E} \cdot d \vec{r}$, and $\vec{E}$ is proportional to $Q$
from Coulomb's law.  Thus the voltage between two conductors with
equal and opposite charge is proportional to the magnitude of the
charge on them:

\begin{equation}
   V \propto Q
\end{equation}

\noindent This can also be written as

\begin{equation}
   Q \propto V
\end{equation}

\noindent Replacing the proportional sign with an equal sign and a constant
we have:

\begin{equation}
   Q = C V
\end{equation}

\noindent where the constant $C$ is called the capacitance of the two
conductors.  A unit of capacitance is Coulomb/Volt.  The capacitance
depends only on the geometry of the two conductors.  

Lets calculate the capacitance of two parallel plate conductors.  Let each
plate have an area $A$ and let the plates be separated by a distance $d$.\\

\noindent Recipe for calculating capacitance:\\

\noindent a) first place a charge of $+Q$ on one plate and a charge of $-Q$ on the other\\
\noindent b) determine $\vec{E}$ for points between the conductors\\
\noindent c) calculate $V = \int \vec{E} \cdot d \vec{r}$\\ 
\noindent d) divide $V$ by $Q$ to find $C$.\\

\noindent {\it Parallel Plate Capacitor}\\

For the parallel plate conductors, we already did this and found $V = Qd/(A \epsilon_0)$.
Thus, the capacitance for the parallel plates is $C = A \epsilon_0 / d$.  Note that $C$
only depends on the geometry of the conductors and not on any charge or voltage they might have.
{\bf If} the conductors have a potential difference of $V$, {\bf then} they each
acquire a charge of $CV$.\\

\noindent {\it Coaxial Cylinderical Capacitor}\\

Consider two cylindrical shells that are coaxial.  Each shell is a conductor.
Let them both have a length $l$, and let one have a radius $a$ and the other
a radius $b > a$.  Lets also assume that $l >> (b-a)$, so we can
treat the capacitor as being infinitely long.  Now, let's find the 
capacitance of these two conductors.

First, suppose there is a total charge of $+Q$ on the inner most conductor,
and a charge of $-Q$ on the outer one.  Next, we need to find the electric
field in the space between the conductors.  Since $l >> (b-a)$ we can
make the approxmation that the conductors are infinitely long.  In that
case the electric field between the conductors must point away from
the axis of the conductors.  We used Gauss' Law previously for the
case of an infinitely long rod, and the same analysis will apply here.
We can choose as a mathematical surface a "can" that is co-axial with
the cylinders of length $d$ and radius $r$: $a<r<b$.  The flux through the ends is
zero.  Applying Gauss' Law for the "can" surface gives:

\begin{equation}
    2 \pi r d \; E = {{Q (d/l)} \over \epsilon_0}
\end{equation}

\noindent Canceling the length of the can $d$, we obtain the magnitude
of the electric field between the conductors to be:

\begin{equation}
    E = {{Q} \over {2 \pi \epsilon_0 l r}}
\end{equation}

\noindent where the direction is radial.  Next we need to calculate the
voltage difference between the conductors: $\Delta V = \int \vec{E} \cdot
d \vec{r}$.  If we take as a path for the integral a path radially away
from the axis then $\vec{E}$ and $d \vec{r}$ are parallel, the line 
integral becomes:

\begin{equation}
 \int \vec{E} \cdot d \vec{r} = \int_a^b {{Q} \over {2 \pi \epsilon_0 l r}} dr
\end{equation}

\noindent Carrying out the integral gives:

\begin{equation}
  V = {Q \over {2 \pi \epsilon_0 l}} ln(b/a)
\end{equation}

\noindent where $V$ is the potential difference between the conductors.  
Notice that $V$ and $Q$ are proportional to each other.  If $V$ is
doubled, so must the charge on the conductors also double.  The capacitance
is the ratio of $Q/V$, which gives:

\begin{equation}
  C = {{2 \pi \epsilon_0 l} \over {ln(b/a)}}
\end{equation}

\noindent {\it Comments}\\

\noindent 1.  In both cases the capacitance of the pair of conductors only
depends on the geometry of the conductors.  Not on how much charge or
voltage they might have.\\

\noindent 2.  In both cases the capacitance equals a distance times 
$\epsilon_0$.  In metric units $\epsilon_0 = 8.85 \times 10^{-12}$
C$^2$/(N m$^2$).  These units are equal to $F/m$ (Farads per meter).
Since $\epsilon_0$ is so small, it is difficult to build a 
capacitor with a large value for $C$ in Farads using small materials.
One needs to use a dialectric material between the conductors.\\

\centerline{\bf Capacitors in Circuits}

\bigskip

Capacitors are important in electrical applications, and we now want
to consider how they behave in simple circuits.  We will first
consider their properties when they are placed in parallel and
in series.  

Circuit elements are connected to each other by wires which are
themselves conductors.  There is very little energy loss through
the conducting wires, so a good rule for {\bf steady state circuits} is the
following: {\it Any two points in a circuit that are connected by
wires only are at the same electrical potential}.  By steady
state circuit we mean one in which the currents are not changing
in time and there are no changing magnetic fields through the
circuit.  In the next section we will consider the effects of
changing magnetic fields.  However, we will first start with
circuits for which these effects are small and can be ignored.\\

\noindent {\it Capacitors connected in Parallel}\\

Consider two capacitors whose right sides are connected together by a wire and whose
left sides are connected together by another wire.  Call one "1" and the other "2".
Let the capacitance of "1" be $C_1$, and the capacitance of "2" be $C_2$.  
Since the left sides are connected together by a conducting wire they are at the same
electrical potential.  Since the right sides are connected together by a 
wire they are also at the same electrical potential.  Thus, $V_1 = V_2 \equiv V$.
If the electrical voltage across each of the capacitors is $V$, then there will
be a charge of $Q_1 = C_1 V$ on the plates of capacitor "1".  There will be 
a charge of $Q_2 = C_2 V$ on the plates of capacitor "2".  The net charge
on both plates is just the sum $Q_{net} = Q_1 + Q_2$ which is 
$Q_{net} = V (C_1 + C_2)$.   Since the capacitance is $Q_{net}/V$, we
see that the effective (or equivalant) capacitance of two capacitors in parallel is 
$C_1 + C_2$:

\begin{equation}
  C_{equiv} = C_1 + C_2
\end{equation}

\noindent for capacitors in parallel.\\

\noindent {\it Capacitors connected in Series}\\

Consider two parallel plate capacitors called "1" and "2".  Let the right side
of "1" be connected to the left side of "2".  The left side of
"1" and the right side of "2" are open.  Suppose the capacitors
are initially uncharged.  Now transfer an amount of charge
$+Q$ from the right side of "2" to the left side of "1".
The left side of "1" will have a charge of $+Q$ and the right
side of "2" will have a charge of $-Q$.  What about the other
sides of the capacitors?

Since the left side of "2" is a conductor, the electric field
inside must be zero.  For this to happen, a charge of $+Q$
must come to the left side of "2".  This will leave a charge
of $-Q$ on the right side of "1".  Thus, when equilibrium
is reached, there will be a charge of $+Q$ on the left and
a charge of $-Q$ on the right sides {\it of each capacitor}.
If two capacitors in series are initially uncharged, when
charged {\bf the same amount of charge will be on the plates
of each capacitor}.

Let $V_1$ be the voltage difference between the plates of "1"
and $V_2$ the voltage difference between the plates of "2".
If we evaluate $\int \vec{E} \cdot d \vec{r}$ from the left
side of "1" to the right side of "2" we will obtain
$V_1 + V_2$.  That is, the potential difference across
both capacitors, $V$, will equal the sum of the potential
differences across each capacitor.

\begin{equation}
  V = V_1 + V_2
\end{equation}

\noindent where $V_1$ is the potential difference across "1" and
$V_2$ the potential difference across "2".  However, $V_i = Q/C_i$,
so

\begin{equation}
  V = {Q \over C_1} + {Q \over C_2}
\end{equation}

\noindent The equivalent (or effective) capacitance of the two capacitors in
series is $C_{equiv} = Q/V$ or

\begin{equation}
  C_{equiv} = {1 \over {(1/C_1) + (1/C_2)}}
\end{equation}

\noindent This equation is usually written as:

\begin{equation}
  {1 \over C_{equiv}} = {1 \over C_1} + {1 \over C_2}
\end{equation}

  The main ideas in deriving these equations for combining capacitors is
that 1) for capacitors in parallel the voltage is the same across both
capacitors; and 2) for capacitors in series the charge on each capacitor
is the same (assuming they started out uncharged).

\bigskip

\centerline{\bf Current} 

\bigskip

Consider a parallel plate capacitor with capacitance $C$.  Suppose
we put a charge $Q$ on one plate and a charge $-Q$ on the other plate.
The charges will stay on the plates for a long time (ideally forever).
If we now connect a wire (conductor) from one plate to the other,
what will happen?  Now we have one big conductor, and charge will flow
until both sides are neutral.  Actually electrons will flow from the
negative side to the positive.  We will do a demonstration in lecture
that will demonstrate the charge (electrons) flowing.  The flow of
charge is called {\bf electric current}.  

In many applications the electrons flow through wires.  If the charge (or
electrons) flow in a "thin" wire, we can count the number of
coulombs that pass a point in the wire per unit time, and can describe
the current in terms of Coulombs/sec.  A Coulomb/sec is defined as an
{\bf Ampere} or simply an Amp.  The current in a wire is often given
the symbol $I$.

If the charge is not flowing in a wire, then one needs a more general
expression.  The expression needs to specify the direction of the
charge flow as well as the amount of charge flowing.  Borrowing the
concept from fluid flow, a useful definition would be the charge
density, $\rho$ times the velocity vector.  A good name for this
quantity is {\bf current density} and is given the symbol $\vec{J}$:

\begin{equation}
  \vec{J} \equiv \rho \vec{v}_d
\end{equation}

\noindent where $\rho$ is the charge density and $\vec{v}_d$ is the
{\bf drift velocity vector} of the charge.  Note that the current density is
defined {\bf at a point}.  A wire is not needed.  However, if the
current is flowing in a wire, then $\vec{J}$ will point in the
direction of the wire.  The current $I$ in a wire is related to
$|\vec{J}|$ in the wire.  The amount of charge $\Delta Q$ that flows past
a point in a time $\Delta t$ is just $\rho v_d \Delta t \; A$, where $v_d$ is the
drift speed of the charge and $A$ is the area of the wire:

\begin{equation}
  \Delta Q = \rho v_d A \Delta t
\end{equation}

\noindent Dividing by $\Delta t$ and A:

\begin{equation}
 {{\Delta Q / \Delta t} \over A} = \rho v_d 
\end{equation}

\noindent Taking the limit as $\Delta t \rightarrow 0$, we have

\begin{equation}
 {I \over A} = \rho v_d = |\vec{J}| 
\end{equation}

\noindent Thus for a wire, the magnitude of the current density is the current/area.

Another way of expressing the current density vector is to use the number of
"charge carriers" per volume, $n_q$.  In terms of $n_q$ the charge density is
$\rho = n_q q$ where $q$ is the charge on each "charge carrier".  With this
notation, 

\begin{equation}
   \vec{J} = n_q q \vec{v}_d
\end{equation}

\noindent If electrons are the charge carriers,
then $n_q$ is the number of electrons per volume, $n_e$, and the charge on
each electron is $e$:  $\vec{J} = n_e e \vec{v}_d$.\\

\centerline{\bf Resistance}

\bigskip

When we connected the plates of the capacitor with a wire, the
charge flowed from one side to the other and the capacitor
became neutral again.  We "discharged" the capacitor.  If we discharge
the capacitor with a more "resistive" wire it takes
longer for the capacitor to discharge.  The current is less,
there is more {\bf electrical resistance} to the flow of charge.

How can we quantify electrical resistance?  Resistance is a property
of the medium (or material) that the charge is flowing in.  For
a material to have low resistance, charges must move more easy than
a material with high resistance.  What makes the charges move?
An electrical force, i.e. an electric field.  So one can quantify
the resistivity of a material by how fast the charges "drift" for a
particular electric field.  For a given electric field $\vec{E}$
a material with a large resistivity will have a small current
density.  We can define the resistivity of a material as
the ratio of $|\vec{E}|/|\vec{J}|$.  The symbol for resistivity
is usually $\rho$, however this is often used to denote density,
so we will use $\rho_e$ for electrical resistivity.  This is
often written as:

\begin{equation}
  \vec{J} = {1 \over \rho_e} \vec{E}
\end{equation}

\noindent Another useful quantity is the conductivity of a 
material.  The lower the resistance, the larger the conductivity:
conductivity = 1/resistance.  Conductivity
is usually given the symbol $\sigma$, but since
we also have been using $\sigma$ as surface charge density,
we will call conductivity $\sigma_e$:

\begin{equation}
   \sigma_e = {1 \over \rho_e}
\end{equation}

\noindent With this definition, 

\begin{equation}
  \vec{J} = \sigma_e \vec{E}
\end{equation}

\noindent {\bf Both $\rho_e$ and $\sigma_e$ are intrinsic properties of a material}.
In the table below we list values of the resistivity for some common conductors
at room temperature:

\begin{center}
\begin{tabular}{c|c|c|c|c|c}
element: & Cu & Al & Fe & Ag & Au\\
\hline
$\rho_e \times 10^{-8}$ ($\Omega /m$)  & $1.7$ & $2.82$ & $10$ & $1.59$ & $2.44$\\ 
\end{tabular}
\end{center}

The above microscopic property can be used to determine the electrical properties
of a large (macroscopic) piece of the material.  Suppose the material is a long
cylinder, like a wire.  Let the cross sectional area be $A$ and the length be
$l$.  If the voltage difference between the ends of the material is $V$,
then the average electric field inside is $E = V/l$.  If the electric
field is $V/l$, then the current density is $J = \sigma_e V/l$.  
Since the current $I$ in the conductor (wire) is $J A$, we have

\begin{equation}
  I = {{\sigma_e A} \over l} V
\end{equation}

\noindent This equation is usually written with $V$ on the left side of the
equation and in terms of $\rho_e$ instead of $\sigma_e$:

\begin{equation}
  V = {{\rho_e l} \over A} I
\end{equation}

\noindent The quantity $\rho_e l/A$ is called the resistance of the object
and given the symbol $R$:

\begin{equation}
  V = R I
\end{equation}

\noindent The resistance of the object is proportional to its length and 
inversely proportional to its area:  $R = \rho_e l/A$.

The above equation allows one to measure the resistance of an object: establish
a voltage difference $V$ across the object and measure the current $I$ that flows.
The resistance is $R = V/I$.  The question we can ask is: Does $R$ depend on
the amount of current $I$ that is flowing through the material?  In general, the
answer to this question is YES.  Often, when more current flows through
a material, the material gets hot and the resistance increases.  However, for
some materials the effects of temperature are small enough that the
resistance is approximately constant over a wide range of currents.
These kinds of materials are called {\bf Ohmic Materials}.

For Ohmic materials, $V = RI$ where $R$ is constant over a wide range of currents.
In this course, we will consider currents through Ohmic materials, for which
$R$ is constant.

\bigskip

\centerline{\bf Capacitor Discharge}

\bigskip

If we connect the plates of a charged capacitor with a wire of 
fixed resistance $R$, we can determine how the charge on the capacitor 
changes in time.  Suppose we have a capacitor of capacitance $C$,
with an initial charge $Q_0$.  At $t=0$ we connect the plates with
a wire of resistance $R$.  What is the charge on the capacitor as
a function of time $Q(t)$?  We can't write an algebraic equation
for $Q(t)$, but we can write an equation that describes how $Q(t)$
changes in time.  This can be done as follows:

\noindent The voltage across the capacitor and resistor, $V$, is
related to the current $I$ through the resistor as:

\begin{equation}
  V = RI
\end{equation}

\noindent We now express $V$ and $I$ in terms of the charge $Q$ on the
capacitor:  $V = Q/C$  and the current $I$ is the rate at which the 
charge is leaving the capacitor: $I = - dQ/dt$.  The minus sign is
because if $Q$ is decreasing the current is positive.  Substituting
into the equation above and re-arrainging terms gives

\begin{equation}
  R {{dQ} \over {dt}} = -{Q \over C} 
\end{equation}

\noindent This is a differential equation for $Q$.  It is an equation
relating the rate at which $Q$ changes to how much charge $Q$ is on the
plates.  The more charge, the faster it leaves the plates.  It is
often the case in physics that equations pertaining to the changes
in quantities take on a simple form.  This is why calculus is so
important in physics.   Capacitor discharge is such a case.

We will discuss the solution in class.  We will show that $Q(t)$
is of exponential form:

\begin{equation}
  Q(t) = Q_0 e^{-t/(RC)}
\end{equation}

\noindent The quantity $RC$ is called the time constant and is the time
it takes for the charge to decrease to $1/e$ of its initial value.  Since
the voltage across the capacitor is proportional to $Q$, $V = Q/C$,
the voltage across the capacitor also decreases exponentially in time:

\begin{equation}
  V(t) = V_0 e^{-t/(RC)}
\end{equation}

\noindent where $V_0$ is the voltage at time $t=0$.  Experiments on
discharging capacitors verify the exponential decay, indicating that
the resistance used does not depend on the current through it.

\bigskip

\centerline{\bf Circuits in General}

\bigskip

Capacitors are nice for storing charge as well as other applications,
but after the charge has left the plates no more current can flow.
If one wants to sustain the current then a battery or other form 
of power supply must be used.  In the case of a battery, a chemical
reaction keeps charge available for the circuit.  The potential
across the battery is determined by the chemical reaction, and
therefore as charge flows from one side of the battery to the other 
side the potential remains constant.  {\bf An ideal battery keeps a
constant electrical potential difference across its terminals}, while 
"supplying" as much charge as necessary to flow in the circuit.  The 
value of the potential difference is determined by the chemical elements
used in the battery.

There are only two basic principles needed to analyze {\bf steady state circuits}.
These are called Kirchoff's Law's:\\

\noindent 1.  {\bf The sum of the currents going into a junction (of wires)
equals the sum of the currents leaving that junction}.  Another way is
to say that the charge flowing into equals the charge flowing out of any
junction.  This is essentially a statement that charge is conserved.\\

\noindent 2. {\bf The sum of the electrical potential differences around
any closed path is zero}.  Another way is to say this is that the sum of the
voltages around a closed path is zero.  In terms of the electric
field: $\oint \vec{E} \cdot d \vec{r} = 0$.  This is essentially a statement
that the electrostatic force is a conservative force.  We need to emphasize that
this is only true for circuits that don't have changing magnetic fields.
We will discuss the corrections to this "law" for changing currents and/or
magnetic fields in the next section.\\

With these two "laws" we can determine the current through any circuit element as 
well as the voltage differences between any two points in a circuit once we know
the properties of the circuit elements.  We will do examples in lecture with circuits 
made up of just batteries and resistors, as well as some that contain capacitors.

In some cases, we can find the effective resistance for certain combinations
of resistors.\\

\noindent {\it Resistors connected in Parallel}\\

Resistors in parallel have their left sides connected together and their
right sides connected together.  The current that comes into the 
parallel set-up will split up and divide through each resistor.
Suppose we have two resistors in parallel with resistance $R_1$ and
$R_2$.  Label the current through $R_1$ as $I_1$ and the current through
$R_2$ as $I_2$.  If the current that comes into the combination is
$I$, then $I = I_1 + I_2$.  

Since the right sides are connected together as well as the left sides,
the potential difference $V$ is the same across each resistor.  Thus,
$I_1 = V/R_1$ and $I_2 = V/R_2$.  If the voltage across the resistors
is $V$ and the total current $I$, then the effective resistance of
the two resistors is just $R_{equiv} = V/I$, or $I = V/R_{equiv}$.
Starting with

\begin{equation}
  I = I_1 + I_2
\end{equation}

\noindent and substituting for the current the voltages divided by
the resistances:

\begin{equation}
  {V \over R_{equiv}} = {V \over R_1} + {V \over R_2}
\end{equation}

\noindent dividing out the $V$'s:

\begin{equation}
  {1 \over R_{equiv}} = {1 \over R_1} + {1 \over R_2}
\end{equation}

\noindent If there are more than $2$ resistors in parallel the
method can be extended:

\begin{equation}
  {1 \over R_{equiv}} = {1 \over R_1} + {1 \over R_2} \cdots
\end{equation}

\noindent For resistors in parallel, each resistor has the potential
difference across it.  Resistors in parallel reduces the resistance.
The equivalent resistance for resistors connected in parallel is 
less than the resistance of any one resistor.

\bigskip

\noindent {\it Resisters connected in Series}\\

Resistors connected in series are connected like train cars,
one after the other.  In this case, the current through each
resister is the same: $I$.  The voltage difference across
the series of resistors is the sum of the voltage differences
across each resistor.  If $V_i$ is the voltage across the
i'th resistor, then

\begin{equation}
  V = V_1 + V_2 + V_3 \cdots
\end{equation}

\noindent where $V$ is the voltage across the series of resistors.
If the equivalent resistance of the series is $R_{equiv}$, then

\begin{equation}
  R_{equiv} I = R_1 I + R_2 I + R_3 I \cdots 
\end{equation}

\noindent Since the current $I$ is the same through each resistor, we have

\begin{equation}
  R_{equiv} = R_1 + R_2 + R_3 \cdots
\end{equation}

\noindent For resistors connected in series the current is the same through each one.
Placing resistors in series increases the resistance in the circuit.

\bigskip

\noindent{\bf Energy considerations in Electric Circuits}

\bigskip

One of the main ways that energy is transfered to our homes is via
electrical circuits, thus the study of energy transfered in electrical circuits is of
upmost importance.  We consider first the energy transfered by a battery,
then we consider resistive elements.\\

\noindent{\it Energy transfer from a Battery}\\

Consider the flow of charge from one side to the other in a battery.  The battery
supplys a certain amount of charge at a specific potential difference $V$.  Suppose
after a time interval $\Delta t$ an amount of charge $\Delta Q$ flows from one
side of the battery to the other side.  When an amount of charge $\Delta Q$ moves 
through a potential difference of $V$, the energy it acquires is $\Delta Q V$.  
Thus, in a time interval $\Delta t$ an amount of energy $\Delta Q V$ is 
supplied by the battery.  The energy per unit time is therefore,
$\Delta Q V/ \Delta t$.  The energy per unit time is the power.
So the power $P$ of the battery is

\begin{equation}
  P = {{\Delta Q V} \over {\Delta t}}
\end{equation}

\noindent $\Delta Q/ \Delta t$ is the current flowing out of and into
the batter, so

\begin{equation}
  P = I V
\end{equation}

\noindent  Current times voltage is the rate at which a battery is
transfering energy to the circuit.  The energy comes from the 
stored chemical energy in the battery.

The capacity of a battery is often given in units of Amp-hours.
For example, D cell batteries have a capacity of around
$10$ amp-hours.  This means that the battery can supply 
a current of 1 amp for $10$ hours, or $1/2$ amp for $20$
hours, etc.  Since an amp is a Coulomb/sec, $10$ amp-hours
is 10 Coulombs/sec times 3600 sec = $36000$ Coulombs.
The voltage across a D cell is $1.5$ Volts, so the initial
energy stored in a typical D cell is $(36000)(1.5) = 54000$
Joules.  In general, if $AH$ is the capacity of a battery
in units of amp-hours, then the stored energy is just
$(AH) \times 3600 \times V$ where $V$ is the voltage of the
battery.  The table below gives approximate values for some
common batteries.

\begin{center}
\begin{tabular}{c|c|c|c}
 Type  &  Internal Resistance ($\Omega$)  & Voltage ($V$)  & Capacity(Amp-hours)\\
\hline
 AAA  &  $0.6$  & $1.5$ &  $0.6$ \\
 AA  &  $0.4$  & $1.5$ & $1.4$ \\
 C  &  $0.2$  & $1.5$ & $4.5$ \\
 D  &  $0.1$  & $1.5$ & $10$ \\
 Alkaline & $2$ & $9$ & $0.5$\\
\end{tabular}
\end{center}

\noindent {\it Energy transfer in a Resistor}\\

Consider a resistor of resistance R with a steady current I
flowing through it.  As with the battery, the energy 
transfered to an amount of charge $\Delta Q$ flowing through the resistor is
equal to the potential difference $V$ times $\Delta Q$.
If the charge is flowing at a steady rate, then the speed 
with which it exits equals the speed at which it entered.  Thus,
the charge does not gain any kinetic energy.  Where does
the energy go?  If the charge were subject to an electric
field without any material, then it would accelerate and
speed up.  However, the material slows it down and keeps
the charge flowing at a constant speed.  Thus, the
energy given to the flowing charge is transfered to the
material of the resistor.  The resistor gains internal
energy and its temperature increases.

As with the battery, in a time $\Delta t$ an amount of
energy equal to $\Delta Q V$ is transfered to the resistor.
Therefore, the energy per unit time transfered to the
resistor is $\Delta Q V/ \Delta t$.  Since $\Delta Q / \Delta t$
is the current flowing through the resistor, we have

\begin{equation}
  Energy \; Transfer \; rate = VI
\end{equation}

\noindent  The energy transfered to the resistor makes the resistor's
temperature increase until it transfers energy to the environment in
terms of internal energy transfer (heat), light, or other forms of
energy.  The above equation can be written in terms of the resistance
and current:

\begin{equation}
  Energy \; Transfer \; rate = I^2R
\end{equation}

\noindent or in terms of $R$ and $V$:

\begin{equation}
  Energy \; Transfer \; rate = {V^2 \over R}
\end{equation}

\noindent Energy loss in wires is an important consideration in power
lines.  To minimize the energy loss, or $I^2R$ loss, the current should
be a minimum.  Since power is $VI$, this means that to minimize the
energy transfered to the environment (i.e. loss) one should use a
LARGE voltage and a small current in the transmission lines.  This
way $V \times I$ can be large, but $I^2 R$ small.

We close this section with some interesting applications of electrostatics
and circuits.

\bigskip

\centerline{\bf Cost of Electrical Energy}

\bigskip

Each month we get a bill for the energy we have used in our
homes that was transfered through wires to our houses.  The
electric company monitors the amount of energy we use.  The
unit that they use is the kilowatt-hour.  This might not seem
like an energy unit, but it is.  A kilowatt is a rate of
energy use: $1 \; kW = 1000 \; Joules/sec$.  An hour ($3600$ sec)
is a measure of time, and a rate times a time is an amount:

\begin{eqnarray*}
   1 \; kW-hr & = & 1000 {{Joules} \over sec} \times 3600 \; sec\\
           & = & 3,600,000 \; Joules
\end{eqnarray*}

\noindent This seems like a lot of energy, but a Joule is a rather
small amount of energy on a human scale.  Your body uses about $100$ Joules/sec
to keep itself going.  So a Kw-hr is a fair amount of energy.  The cost for
a Kw-hr of energy varies from place to place, but is around $14$ cents per
Kw-hr.  That is a pretty small price for $3,600,000$ Joules.  Think how much
you pay for the $54,000$ Joules that are contained in a D-cell battery.

To calculate how much it costs to use an appliance, just multiply the 
Kw-hr's times the number of hours and then multiply by $14$ cents.

\bigskip

\centerline{\bf Energy Stored in a Capacitor}

\bigskip

When a capacitor is charged with an amount of charge
$+Q$ on one plate and $-Q$ on the other plate it has
a certain amount of electrical potential energy.  When the
capacitor discharges, say through a light bulb, one sees
the transfer of this energy.  The question we want to consider
is how much potential energy does a capacitor of capacitance
$C$ have when it is charged?

You might guess that the potential energy stored in a charged
capacitor is just $VQ$, where $V$ is the potential across the
capacitor and $Q$ is the charge on its plates.  However, this
is not correct.  As the capacitor is "charging up", initially
it is easy to transfer charge from one side to the other.
It gets more difficult to transfer charge when there is more charge
on the plates.  

We can solve the problem in a number of ways.  A simple way is to
let the charged capacitor discharge through a resistor of resistance
$R$ and add up the energy transfered to the resistor.  When a 
capacitor discharges through a resistor, the charge on the
plates decreases as $Q_0 e^{-t/(RC)}$.  The current is 
$-dQ/dt$:

\begin{equation}
  I = {{Q_0} \over {RC}} e^{-t/(RC)}
\end{equation}

\noindent Since the energy transfer rate to the resistor is $I^2R$, the total energy
transfer is:

\begin{equation}
  Total \; Energy = \int_0^\infty I^2 R dt
\end{equation}

\noindent Substituting in for the current we have:

\begin{equation}
  Total \; Energy = \int_0^\infty {Q^2 \over {RC^2}} e^{-2t/(RC)} dt
\end{equation}

\noindent Carrying out the integral gives

\begin{equation}
  Total \; Energy = {Q^2 \over {2C}}
\end{equation}

\noindent This energy, which was transfered to the resistor, was the amount of
energy that was stored in the capacitor.  One can express this energy
in terms of $Q$ and $C$ as above, or in terms of $Q$ and $V$ by substituting
$V = Q/C$:

\begin{eqnarray*}
  Energy \; in \; Capacitor & = & {Q^2 \over {2C}}\\
                    & = & {{QV} \over 2}\\
                    & = & {{CV^2} \over 2}
\end{eqnarray*}

Another interesting way to express the energy stored in a capacitor is
in terms of the electric field between the plates.  For a parallel plate
capacitor the magnitude $E$ of the electric field is $E=Q/(A \epsilon_0)$.
Writing $Q$ as $Q=E A \epsilon_0$ and $C=A\epsilon_0 /d$ where $d$ is the
spacing between the plates, one has:

\begin{equation}
  Energy \; in \; Capacitor = {{(A \epsilon_0 E)^2} \over {2 (A\epsilon_0)/d)}}
\end{equation}

\noindent Simplifying, we have

\begin{equation}
  Energy \; in \; Capacitor = {{\epsilon_0 E^2} \over 2} \; Ad
\end{equation}

\noindent The quantity $Ad$ is the volume inside the capacitor.  Thus,
the quantity $\epsilon_0 E^2/2$ is an energy density which depends only
on the electric field.  It can shown (using vector calculus) that in general
the expression $\epsilon_0 E^2/2$ can be interpreted as the energy density
of the electrostatic field.  It is often given the symbol $U_E$:

\begin{equation}
   U_E = {{\epsilon_0 E^2} \over 2}
\end{equation}

\noindent is the energy density of the electric field.  In our upper division
course it is shown that this term is part of an energy conservation equation
for electromagnetic fields and charges.

\bigskip

\centerline{\bf Microscopic View of Electric Current}

\bigskip

We should consider what is happening at the microscopic level
in a wire when current is flowing.  If there is an electric field
present in the wire, the free electrons will feel a force and
accelerate.  If the magnitude of the electric field is $E$, the
acceleration of an electron will be $a = eE/m$ where $m$ is the
mass of an electron.  The electron will keep accelerating until
it is stopped by an atom (or another electron) in the conductor.
If we denote $\tau$ as $1/2$ of the average time it takes the electron to
travel before it is stopped, the average drift velocity is
$v_d \approx e E \tau /m$.  If there are $n$ free electrons per volume,
then the current density is $J = n e v_d$, or

\begin{equation}
   \vec{J} = {{n e^2 \tau} \over m} \vec{E}
\end{equation}

\noindent We see that this simple model predicts that $\vec{J}$ is
proportional to $\vec{E}$.  If $\tau$ doesn't depend on the amount of
current that is flowing (or temperature), then the material will be
Ohmic.

We can get a rough idea of how fast the charge is flowing in a wire.
Suppose we have a wire that has a current of one amp and whose diameter 
is $2 \; mm$.  Since $I =  J A$, and $J = n e v_d$, we have

\begin{equation}
   I = n e v_d A
\end{equation}

\noindent where $n$ is the number of conduction electrons per $m^3$ and $A$ is the
area.  Solving for the drift velocity $v_d$,

\begin{equation}
  v_d = {I \over {n e \pi r^2}}
\end{equation}

\noindent For copper, $n \approx 10^{28}$ electrons/$m^3$.  Using $r = 0.001 \; m$ we
obtain $v_d \approx 0.2 \; mm/sec$.

This ($0.2 \; mm/sec$) is very slow.  Why does the light go on just as the switch is 
closed?  This is because the electric field in the wire is established very fast,
roughly at the speed of light.  As soon as this happens electrons all throughout
the circuit move simultaneously.  The free electrons in the light start moving
as soon as the switch is closed.  Although the drift speed is slow, the current
can be large because there are a lot of electrons moving.

\end{document}