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\begin{center}
{\Large {Solution Derivations for Capa \#6\\[20pt]} }
\end{center}
1) Charges are distributed with uniform charge density $\lambda = 4.35\ \mu
C/m$ along a semicircle of radius $R = 26.0\ cm$ centered at the origin of a
coordinate system (as shown in the diagram below.) What is the potential at
the origin?
\begin{quote}
$\lambda $ = Given\\
$r$ = Given\\
The potential difference between two points is defined as
\[
\Delta V=-\int_{A}^{B}E\cdot dl
\]
The first thing to remember is that potential is a scalar. Thus, the
direction of $E$ is irrelevant. The length $dl$ away from the charged
surface is a constant $r$. In my case, it was antiparallel to $E$ (the ring
is positively charged). This means that the dot product is negative. If $E$
points in the same direction as $r$, then the dot product will be positive.
It will also be easier to switch to polar coordinates when integrating.
Thus,
\[
\Delta V=-\int_{0}^{\pi }E\cdot dl=-\int_{0}^{\pi }-rE=\int_{0}^{\pi
}rE=\int_{0}^{\pi }r\frac{k\,dq}{r^{2}}=\int_{0}^{\pi }\frac{k}{r}dq
\]
where the last two steps follow because the electric field at each portion
of the arc can be regarded as that of a point charge. Since $r\,$ and $k$
are constants, they can be pulled out of the integral. Thus,
\[
\Delta V=\frac{k}{r}\int_{0}^{\pi }dq
\]
The limits of integration are in terms of $\theta $, but the differential is
in terms of charge. We must relate the two.
$dq$ can be found from $\lambda $, which is charge per length. Thus, a
charge $dq$ equals a length times $\lambda $. This length is a portion of
the arc over which we are integrating. Thus,
\begin{eqnarray*}
dq & = &\lambda r\,d\theta \\
\Delta V & = &\frac{k}{r}\int_{0}^{\pi }\lambda r\,d\theta =r\frac{k\lambda }{r}
\int_{0}^{\pi }d\theta =k\lambda \left( \pi -0\right) =\pi \lambda k
\end{eqnarray*}
\end{quote}
2) An isolated metal sphere of radius $10.5\ cm$ is at a potential of $5200\
V$. What is the charge on the sphere?
\begin{quote}
$r$ = Given \\
$V$ = Given \\
The potential for a point charge (which essentially what the sphere reduces
to if one is outside the sphere) is
\[
V=\frac{kq}{r}
\]
Thus,
\[
q=\frac{rV}{k}
\]
\end{quote}
3) Determine the energy density of the electric field outside the sphere and
integrate this throughout all space in order to calculate the total energy
in the electric field.
\begin{quote}
The energy density of an electric field is
\[
u_{E}=\frac{1}{2}\varepsilon_{0}E^{2}
\]
which is on page 647 of the textbook. What this question is asking is to
find the energy density associated with the electric field in space and add
it all up. Thus, you would have to start at all points right outside the
sphere's radius, take every point in 3-space, and do a calculation for it.
Might as well get started; it will take a while. Actually there is a
shortcut way. An integral will do this summation for you. But instead of
using the radial vector $\hat{r}$ from the electric field and translating
its coordinates into rectangular coordinates, we can transform our variables
into spherical coordinates and integrate over all of their range. Since
there are three variables in 3-space, we will be performing a triple
integral. Without using much Calculus 3, the proof for this integral is a
little messy. You can look it up in the calculus 3 portion of your calc
book. I am not sure how it is presented, but I have a different calc book
and this is the way it was presented there.
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.28in]{capa06-03-1.png}
\end{center}
\end{figure}
This graphic shows a point in 3-space. It is the point at the upper right.
As you can see, the rectangular coordinates of the point are $P(x,y,z)$. The
spherical representation of the point is also shown. The point is described
as a distance from the origin $\rho $, an angle in the xy-plane $\theta$,
and an angle from the z-axis $\phi$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.33in]{capa06-03-2.png}
\end{center}
\end{figure}
Instead of evaluating the required triple integral over the variables $x,y,z$,
this picture shows how we can do it with the spherical coordinates. The
distance from the point charge is $\rho $. The side of the solid closest to
the origin has length $\rho\,d\phi $. This is just an arc length -- the
radius times the angle. The top of the solid closest to the origin has
length $\rho \sin \phi\,d\theta $. To see this, imagine projecting $\rho $
into the xy-plane. The length will be $\rho \sin \phi $. The length of this
arc is simply the radius $(\rho \sin \phi)$ times the angle $(d\theta)$ or
just $\rho \sin \phi\,d\theta $. The depth of the box is just the change in
the radius from the origin $d\rho $. Instead of integrating over $dx\ast
dy\ast dz$ we can now use the conversion of $d\rho \ast \rho \sin \phi
d\theta \ast \rho d\phi $ or just $\rho ^{2}\sin \phi \ast d\rho \ast
d\theta \ast d\phi $. Keep in mind that the radius from the sphere will be
represented by $\rho $.
\begin{eqnarray*}
\int \int \int u_{E}\rho ^{2}\sin \phi\,d\rho\,d\theta\,d\phi &=&\int \int \int
\frac{1}{2}\varepsilon _{0}E^{2}\rho ^{2}\sin \phi\,d\rho\,d\theta\,d\phi \\
&=&\frac{1}{2}\varepsilon _{0}\int \int \int E^{2}\rho ^{2}\sin \phi\,d\rho\,
d\theta\,d\phi
\end{eqnarray*}
Before going further, it is important to note the limits of integration.
When evaluating multiple integrals, one starts in the center and works
outwards. Thus, the limits on the integral furthest inside correspond
differential furthest inside. The first differential is $d\rho $ which is
the change in radius from the sphere. Its value can be anywhere outside of
the sphere itself to infinity. Thus, $\rho $ ranges from $r$ to $\infty $.
To be technically correct, limits of integration cannot range to infinity.
The reason is how can you evaluate the definite integral if you don't know
what infinity is in order to put it in? The answer is to use a limit.
However, I feel the use of one will complicate things more. Just realize
that when plugging in infinity after integrating that you are not
technically plugging in infinity, but rather evaluating the limit as that
variable approaches infinity. The next differential is $d\theta $. This is
the angle in the xy-plane. It can be anywhere in the xy-plane as the radius
vector can point in any direction. Thus, $\theta $ ranges from $0$ to $2\pi $%
. The last differential is $d\phi $. However, $\phi $ does not range from $0$
to $2\pi $ like $\theta $. Its range is from $0$ to $\pi $. To see this,
imagine if $\phi $ was larger than $\pi $. This would mean that the vector
is now pointing in the other half-space ($y<0$). However, by letting $\theta
$ take on values from $0$ to $2\pi $ this region is already accounted for.
Thus, by revolving the vector around the z-axis, $\phi $ only needs to take
on values to get from one side of the z-axis to the other. It is not
mandatory that it be set up this way: $\phi $ could range from $0$ to $2\pi $
but then $\theta$ could only range from $0$ to $\pi $. The choice is yours.
But now, let's put limits on the integrals.
\[
\frac{1}{2}\varepsilon _{0}\int_{0}^{\pi }\int_{0}^{2\pi }\int_{r}^{\infty
}E^{2}\rho ^{2}\sin \phi\,d\rho\,d\theta\,d\phi
\]
Outside of the sphere, the sphere acts like a point charge. Thus, we can use
the formula for the electric field of a point charge, $E=\frac{kq}{r^{2}}$.
Remember, however, that the radius from the sphere in this problem is $\rho $
while the radius of the sphere itself is $r$. Thus, $E=\frac{kq}{\rho ^{2}}$.
\begin{eqnarray*}
&&\frac{1}{2}\varepsilon _{0}\int_{0}^{\pi }\int_{0}^{2\pi }\int_{r}^{\infty
}\left( \frac{kq}{\rho ^{2}}\right) ^{2}\rho ^{2}\sin \phi\,d\rho\,d\theta\,
d\phi \\
&=&\frac{1}{2}\varepsilon _{0}\int_{0}^{\pi }\int_{0}^{2\pi
}\int_{r}^{\infty }\frac{k^{2}q^{2}}{\rho ^{4}}\rho ^{2}\sin \phi\,d\rho\,
d\theta\,d\phi \\
&=&\frac{1}{2}\varepsilon _{0}\int_{0}^{\pi }\int_{0}^{2\pi
}\int_{r}^{\infty }\frac{k^{2}q^{2}}{\rho ^{2}}\sin \phi\,d\rho\,d\theta\,d\phi
\\
&=&\frac{1}{2}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }\int_{0}^{2\pi
}\int_{r}^{\infty }\frac{1}{\rho ^{2}}\sin \phi\,d\rho\,d\theta\,d\phi
\end{eqnarray*}
Now comes the part of integrating this which is surprisingly simple once all
the constants are removed. Since the integrand is a function of more than
one variable but it is inside a multiple integral, we use what is called
partial integration. This is similar to partial differentiation in that
while integrating with respect to one variable, we hold the rest constant.
Thus,
\begin{eqnarray*}
&=&\frac{1}{2}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }\int_{0}^{2\pi
}\left. \left( -\frac{1}{\rho }\sin \phi \right) \right| _{\rho =r}^{\infty
}d\theta\,d\phi \\
&=&\frac{1}{2}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }\int_{0}^{2\pi
}\left( -\frac{1}{\infty }\sin \phi +\frac{1}{r}\sin \phi \right) d\theta\,
d\phi \\
&=&\frac{1}{2}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }\int_{0}^{2\pi }\frac{%
1}{r}\sin \phi\,d\theta\,d\phi
\end{eqnarray*}
Where the last part follows since the first term approaches zero as the
denominator approaches infinity (the technical limit explanation). We can
pull the $r$ outside the integral now since it is a constant for the sphere.
Now, we do the second integral.
\[
=\frac{1}{2r}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }\left.\sin \phi\,\theta
\right|_{\theta=0}^{2\pi }\,d\phi
\]
This is because when integrating with respect to $\theta $, $\sin \phi $ is
treated as a constant. Even though it is a constant, we cannot pull it out
of the integral because we still have to integrate it.
\begin{eqnarray*}
&=&\frac{1}{2r}\varepsilon _{0}k^{2}q^{2}\int_{0}^{\pi }2\pi \sin \phi\,d\phi
\\
&=&\frac{\pi \varepsilon _{0}}{r}k^{2}q^{2}\int_{0}^{\pi }\sin \phi\,d\phi \\
&=&\frac{\pi \varepsilon _{0}}{r}k^{2}q^{2}\left.\left( -\cos \phi \right)
\right|_{\phi=0}^{\pi } \\
&=&\frac{\pi \varepsilon _{0}}{r}k^{2}q^{2}\left( -\left( -1\right) +1\right)
\\
&=&2\frac{\pi \varepsilon _{0}}{r}k^{2}q^{2}\vspace{1pt}
\end{eqnarray*}
I've heard you can also approximate the energy by simply using the equation $%
U=\frac{1}{2}QV$.
\end{quote}
4) Select T-True, F-False, If the first is T and the rest F, enter TFFFF.
\begin{quote}
QUESTION:\\
A) The `drift velocity' of electrons in house-hold electric circuits is
close to the speed of light.\\
B) A negative temperature coefficient of resistivity means that the material
is a superconductor.\\
C) A $5\ m$ length of copper wire has a resistivity of $1.6\ \mu\Omega\cdot cm$.
The wire is cut in half. The resistivity of the wire is halved.\\
D) Current density is a vector.\\
E) The mean free time between collisions in a current carrying conductor is
independent of the applied voltage.\\
ANSWER:\\
A) False, the drift velocity is very slow. See lecture notes 27-7.\\
B) False, a negative temperature coefficient doesn't make sense.\\
C) False, resistivity is a constant for a given material.\\
D) True, see page 672-3.\\
E) True, the equation for drift velocity does not depend on the applied
voltage.\\
Remember to input whether each term is true or false. In this case, FFFTT.
\end{quote}
5) At the Aladdin Synchrotron in Madison, Wisconsin, there is an electron
beam with a current of $140.0\ mA$. The electrons have an average kinetic
energy of $970\ MeV$, and a speed approximately equal to the speed of light.
How many electrons pass a given point in the accelerator in one hour? Give
your answer in the \emph{numberofelectronsperhour} (i.e. if the answer is 50
then enter ``50 1/hr'' or ``50 hr\^-1'' into CAPA).
\begin{quote}
$I$ = Given\\
$KE$ = Given\\
$v$ = Given\\
$t$ = Given\\
$n$ = ?\\
For this problem, all you need from the given information is the current.
The current is given in milliamperes (that is, $10^{-3}$). Current is
defined as charge per second. Thus, to find out the amount of electrons per
second flowing through, we need to divide by the charge of an electron.
\[
\frac{electrons}{second}=I\ast \frac{1\ electron}
{1.6 \times 10^{-19}\ C}
\]
Now, to find the amount going through in one hour, simply multiply by 1 hour
(3600 seconds).
\end{quote}
6) Assume that copper has one electron per atom to carry charge. Given that
the mass density of copper is $8.9 \times 10^3\ kg/m^3$ and that its molecular
weight is $63.5\ g/mol$, calculate the drift speed of the electrons in a
copper wire that carries $1.20\ A$ and has a circular cross section
$0.80\ mm $ in radius.
\begin{quote}
$d$ = density = Given\\
$w$ = weight = Given\\
$I$ = Given\\
$r$ = Given\\
$v_{d}$ = ?\\
From the book (p.671) we get the equation
\[
I=nqAv_{d}
\]
Solving for $v_{d}$,
\[
v_{d}=\frac{I}{nqA}
\]
We must now calculate the number of electrons present. Using the density and
the molecular weight, we can find the number of atoms per volume. To see
this, consider density $/$ molecular weight = $\frac{kg}{m^3}/
\frac{kg}{mol} = \frac{kg}{m^3} \ast \frac{mol}{kg} = \frac{mol}{m^3}$.
Thus, $n=\frac{d}{w}$. However, this tells us the number of moles of atoms
per unit volume. We want the number of atoms (and ultimately the number of
electrons). We must multiply by Avogadro's Number which is
$6.02 \times 10^{23}$ atoms/mol. So,
\[
n=\frac{d}{w}\left( 6.02 \times 10^{23}\right)
\]
This gives us number of atoms per unit volume. Since each atom contributes
an average of one electron, this is also the number of electrons per unit
volume. The charge of an electron, $q$, is $1.6 \times 10^{-19}\ C$. The
cross sectional area is just $\pi r^{2}$, but remember to convert the radius
to meters. Since we know $I$ and $q$ and have calculated $n$ and $A$, we can
now find $v_d$.
\end{quote}
7) A number 16 copper wire has a diameter of $1.291\ mm$. Calculate the
resistance of a $31\ m$ long piece of that wire. Use
$\rho = 1.72 \times 10^{-8}\ \Omega \cdot m$ for the resistivity of copper.
Use ``Ohm'' as your units, i.e. if the answer is 5 ohms, enter ``5 Ohm.''
\begin{quote}
$d$ = Given\\
$l$ = Given\\
$\rho $ = Given\\
Resistance is given by the equation (from lecture notes 27-4)
\[
R=\frac{\rho l}{A}
\]
$A$ is easily calculated and the others are given. Note that you are given a
diameter in mm.
\end{quote}
8) For safety, the National Electrical Code limits the allowable amount of
current which such a wire may carry. When used in indoor wiring, the limit
is $6\ A$ for rubber insulated wire of that size. How much power would be
dissipated in the wire of the above problem when carrying the maximum
allowable current?
\begin{quote}
$I$ = Given\\
Power has many forms. The one of interest here is
\[
P=I^{2}R
\]
Using the value from the previous problem, simply plug in the numbers. Power
has units of watts.
\end{quote}
9) What would be the voltage between the ends of the wire in the last
problem?
\begin{quote}
The equation needed is
\[
V=IR
\]
\end{quote}
10) The graph represents the Voltage-Current characteristic for an unknown
resistor. Determine the value of the resistor from the graph. Use ``Ohm'' as
your units, i.e. if the answer is 5 ohms, enter ``5 Ohm.''
\begin{quote}
Get out the vernier calibers... You need to find the slope of that line.
Note that the vertical axis is given times $10^{3}$.
\end{quote}
11) A typical household circuit is capable of carrying $15.0\ A$ of current
at $120\ V$ before the circuit breaker will trip. How many $1500\ W$ hair
dryers can run off one such circuit? (Give your answer as a unitless
integer.)
\begin{quote}
$I$ = Given\\
$V$ = Given\\
$P_{hair}$ = Given (amount of power each hair dryer uses)\\
For this problem, it will be convenient to use the following equation for
power
\[
P=IV
\]
This is the total power that the circuit can handle. Divide this number by
the power required by each hair dryer to find the amount of hair dryers you
can run.
\end{quote}
12) In the rush to get ready for lecture, a physics professor leaves the
hairdryer described in the previous problem running and does not turn it off
until he gets home $6.4\ hr$ later. How much will this add, in dollars, to
his next electric bill (assume electricity costs $\$0.078$ per $kWhr$)? %$
Enter the number of dollars without units; i.e. if the answer is 50 cents,
enter ``0.50.''
\begin{quote}
$t$ = Given (hours)\\
$c$ = Given (price of electricity)\\
$P_{hair}$ = Given (previous problem)\\
In order to find the cost of running the hair dryer, we must multiple the
amount of power used by the hair dryer. (dollar/kWhr * kWhr = dollar) In
order to find the power used by the hair dryer, we can calculate the amount
of kilowatt hours used. Thus the power used is
\[
P=P_{hair}\ast t
\]
and the cost is cost = $cP$
\end{quote}
13) A $12\ V$ lead-acid car battery, engineered for `up to 500 or more
charge/discharge cycles,' has a rating of $95.0\ A \cdot hr$. (It sells for
$\$126.00$.) Calculate the total amount of charge that moves through the
battery before it needs to be recharged. %$
\begin{quote}
$V$ = Given\\
rating = Given\\
The rating means that the battery can deliver the specified amount of amps
for one hour (or half as many for twice as long, etc.). Since current is
charge/time, the amount of charge is the current times the time. But that is
basically given to you in the rating. However, since current is defined as
charge per second, you must convert the hour to seconds and then multiply.\\
charge = rating * time
\end{quote}
14) What is the total electrical energy that the battery can deliver before
it needs to be recharged?
\begin{quote}
Power is defined to be energy divided by time. So, energy would then be
power times time. Power can be calculated using the previously given
information.
\begin{eqnarray*}
P & = & IV \\
E & = & P\ast time
\end{eqnarray*}
The current and time are related. You know the rating, so you can come up
with any combination of current and time that will yield that rating. Thus,
the full amperage for one hour, half for twice as long, etc.
\end{quote}
15) An underground telephone cable, consisting of a pair of wires, has
suffered a short somewhere along its length (at point P in the Figure). The
telephone cable is $5.00\ km$ long, and in order to determine where the
short is, a technician first measures the resistance between terminals $AB$;
then he measures the resistance across the terminals $CD$. The first
measurement yields $25.00\ \Omega$; the second $100.0\ \Omega$. Where is the
short? Give your answer as a distance from point $C$.
\begin{quote}
This problem is very general and difficult to explain.\newline
$d_{tot}$ = Given\newline
$R_{1}$ =\ Given (resistance across $AB$)\newline
$R_{2}$ = Given (resistance across $CD$)\newline
We know that resistance is defined as
\[
R=\frac{\rho l}{A}
\]
Let $d_{1}$ be the distance to the short from the terminals $AB$. Let $d_{2}$
be the distance to the short from the terminals $CD$. Since the total wire
length is given, we know that
\begin{equation}
d_{1}+d_{2}=d_{tot} \label{eqn:1}
\end{equation}
Next, set up the equation for the resistance at each point
\begin{equation}
R_{1}=\frac{\rho d_{1}}{A} \label{eqn:2}
\end{equation}
\begin{equation}
R_{2}=\frac{\rho d_{2}}{A} \label{eqn:3}
\end{equation}
Since $d_{1}$ and $d_{2}$ are the unknowns (technically $\rho $ and $A$ are
constants), we can solve for one of them and plug it into the other.
\begin{equation}
d_{2}=\frac{R_{2}A}{\rho } \label{eqn:4}
\end{equation}
From (\ref{eqn:1}),
\[
d_{1}=d_{tot}-d_{2}
\]
Thus,
\[
d_{1}=d_{tot}-\frac{R_{2}A}{\rho }
\]
Plugging back into (\ref{eqn:2}),
\[
R_{1}=\frac{\rho \left( d_{tot}-\frac{R_{2}A}{\rho }\right) }{A}
\]
\[
AR_{1}=\rho d_{tot}-R_{2}A
\]
\[
AR_{1}+R_{2}A=\rho d_{tot}
\]
\[
A\left( R_{1}+R_{2}\right) =\rho d_{tot}
\]
\[
A=\frac{\rho d_{tot}}{\left( R_{1}+R_{2}\right) }
\]
We can now plug this back into equation (\ref{eqn:4}) and solve for the
required distance from terminal $CD$.
\[
d_{2}=\frac{R_{2}\frac{\rho d_{tot}}{\left( R_{1}+R_{2}\right) }}{\rho }
\]
\[
d_{2}=R_{2}\frac{d_{tot}}{\left( R_{1}+R_{2}\right) }
\]
Remember to convert the total distance to meters.
\end{quote}
\end{document}