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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