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\begin{center}
{\Large {Solution Derivations for Capa \#4\\[20pt]} }
\end{center}
1) Consider two separate systems, each with four charges of magnitude $q$
arranged in a square of length $L$ as shown above. Points $a$ and $c$ are in
the center of their squares while points $b$ and $d$ are half way between the
lower two charges. (Give ALL correct answers, i.e., B, AC, BCD...)
\begin{figure}[h]
\begin{center}
\includegraphics[width=4.69in]{capa04-01.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) The electric potential at $d$ is zero.\\
B) The electric potential at $c$ is zero.\\
C) The electric field at $d$ is zero.\\
D) The electric field at $c$ is zero.\\
E) The electric potential at $b$ is zero.\\
F) The electric field at $a$ is zero.\\
G) The electric field at $b$ is zero.\\
H) The electric potential at $a$ is zero.\\
ANSWER:\\
A) False, the potential from the positive charges is greater than the
potential from the negative charges since the positive charges are closer.\\
B) True, it lies on the eqipotential line.\\
C) False, the field from the positive charge cancels out, but there is
still a field from the negative charges.\\
D) False, the diagonal lines from $-q$ to $c$ to $+q$ both point toward $-q$,
resulting in a net field.\\
E) True, the positive and negative charges nearest $b$ cancel out as well as
the pair furthers from it.\\
F) True, see (E)\\
G) False, there is a flux toward the lower, $-q$. The upper two provide a
lesser field that cancels out part of the field moving along the bottom from
right to left.\\
H) True, $a$ is equal distance from each point charge. As such, the
potential sums to zero.\\
Remember that potential is a scalar and electric field is a vector. CAPA
is also looking for the letters of the correct answer, in this case, BEFH
\end{quote}
2) Using the diagram above, find the magnitude of the electric field at point
$d$. DATA: $q=0.750\ \mu C$, $L=0.40\ m$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=2.21in]{capa04-02.png}
\end{center}
\end{figure}
\begin{quote}
$q$ = Given\\
$L$ = Given\\
The magnitude of the electric field is $E=\frac{kq}{r^2}$. Since the
field due to the positive point charges will cancel each other and the
horizontal components of the negative point charges will also cancel, only
the vertical component of each negative point charge contributes to the
field. The field due to the vertical component will be
$E=\frac{kq}{r^2}\cos \theta$. Since there are two negative charges, the
electric field is twice that from one charge,
\[
E=2\frac{kq}{r^2}\cos \theta
\]
$r$ is found from the right triangle shown, and $\cos \theta $ can also be
calculated from the picture. Thus,
\begin{eqnarray*}
E & = & 2\frac{kq}{L^2+\left( \frac{L}{2}\right) ^{2}}\frac{L}{\sqrt{
L^{2}+\left( \frac{L}{2}\right) ^{2}}}\\
& = & \frac{2kqL}{\left( L^{2}+\left( \frac{L}{2}\right) ^{2}\right)
^{\frac{3}{2}}}
\end{eqnarray*}
\end{quote}
3) The figure below shows two points in an E-field: Point {\bf 1} is at
$(X_1, Y_1) = (3,4)$ in $m$, and Point {\bf 2} is at $(X_2, Y_2) = (12, 9)$ in
$m$. The Electric Field is constant, with a magnitude of $80\ V/m$, and is
directed parallel to the +X-axis. The potential at point {\bf 1} is $1000\ V$.
Calculate the potential at point {\bf 2}.
\begin{quote}
$P_1=(x_1,y_1)$ = Given\\
$P_2=(x_2,y_2)$ = Given\\
$E$ = Given\\
$V_1$ = Given\\
First find the distance the point moved parallel to the electric field.
Since the electric field is conservative, it does not matter how you get
from point A to point B. The only thing that matters is the net distance
you move parallel to the field in between. This distance is probably
$d=x_2-x_1$. Since the electric field is constant, the equation for
$\Delta V$ becomes
\[
\Delta V=\int_{A}^{B}dV = -\int_{A}^{B}E\cdot dl = -E\int_{A}^{B}dl = -El
\]
where the last steps follow because the displacement is in the same direction
as the field. Thus, $\Delta V=-Ed$
This is the voltage {\em change} from point 1 to point 2. The potential
at point 2 is $V_1+\Delta V$. This will probably be less than the
potential at point 1.
\end{quote}
4) Calculate the work required to move a negative charge of $Q = -562\ \mu C$
from point {\bf 1} to point {\bf 2}.
\begin{quote}
In a constant electric field, the work is defined to be $W=-qEl$. Since you
just calculated the potential ($El$) in the previous problem, this answer is
just the charge times the field. (Watch the signs).
\end{quote}
5) What is the electric field at the point $x = 6.5\ m$? Positive E-fields
point to the right.
\begin{quote}
This problem only requires you to find the slope at the point given. However,
from equation 25-10, the electric field is defined to be the negative of the
slope of the potential graph. Contrary to the expected 1\% margin of error
you are given on the other problems, you always have to get graphs exactly.
At least it seems that way.
\end{quote}
6) The lines show the equipotential contours in the plane of three point
charges, $Q_1$, $Q_2$, and $Q_3$. The values of the potentials are in
{\em kilovolts} as indicated for the +5, 0, and -5 $kV$ contours. The
positions of the charges are indicated by the dots. (Give ALL correct
answers, i.e., B, AC, BCD...)
\begin{figure}[h]
\begin{center}
\includegraphics[width=4.52in]{capa04-06.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) Charge $Q_3$ has the largest magnitude of all.\\
B) Charge $Q_2$ is the largest negative charge.\\
C) $Q_1$ is a negative charge.\\
D) The electric field at $i$ is stronger than at $j$.\\
E) The electric field at $k$ is zero.\\
F) The force on a proton at $g$ points to the bottom of the page.\\
ANSWER:\\
A) True, as shown on a concept test, $Q_3$ has the largest magnitude
potential at the farthest distance from itself making it the largest charge.\\
B) False, the first eqipotential ring around $Q_1$ has greater radius.\\
C) True, negative potentials surround negative charge.\\
D) True, $i$ is closer to a charge than is $j$ so it has a larger potential.\\
E) False, the potential is not the field. The field describes the change
in potential and although the potential is instantaneously zero at $k$, the
field is not. (Sum the vectors from the point charges for proof).\\
F) False, a proton will want to move away from the positive charge near the
bottom of the page; thus the force would be toward the top.\\
\end{quote}
7) Calculate the work required to move a charge of $-0.39 \times 10^{-12}\ C$
from `i' to `b'.
\begin{quote}
$q$ = Given\\
Two points given\\
Work is defined as $-W=qEl$. But $El$ is just the negative of the change in
voltage. That is,
\begin{eqnarray*}
-W & = & -q\Delta V\\
W & = & q\Delta V
\end{eqnarray*}
Where $\Delta V$ is the potential difference between the two potential lines
each point lies on.
\end{quote}
8) Calculate the magnitude of the electric field at `k'.
\begin{quote}
The electric field is given as
\[
E=-\frac{dV}{dl}
\]
which can be approximated as
\[
E=-\frac{\Delta V}{\Delta l}
\]
Thus, find the potential difference between the two potential lines nearest
to the point requested and measure the distance between them. Only measure
the distance perpendicular to both (that is, along a line that
perpendicularly intersects each nearby curve). The letter should be
somewhere near where that line will be. Remember to convert the distance
to meters, and plug in the equation.
\end{quote}
9) Calculate the magnitude of the force on a charge of
$9.60 \times 10^{-19}\ C$ at `g'.
\begin{quote}
$q$ = Given\\
Force is equal to
\[
F=qE
\]
Although it would be nice to think we could use the electric field
calculated in part (8), we can't. Thus, you must get the ruler back out
and calculate the field through this point in the same way as the last
point. Then multiply by the charge.
\end{quote}
10) Calculate the magnitude of $Q_3$. The magnitudes of the three charges
are in the exact rations of 1 to 2 to 3.
\begin{quote}
To find the magnitude of $Q_3$ you can pick a point on the graph
(preferably on one of the equipotential lines) and calculate the potential at
that point. Then, we can solve for $Q$. I chose a point on the $+5\ kV$
equipotential. Thus, the sum of potentials from each other charge had to
add up to $5000\ V$ on that line. That is,
\[
V_1+V_2+V_3=5000
\]
Since the magnitude of the charges are in the ratio of 1:2:3 (for the point
charges 2, 1, and 3, respectively)
\begin{eqnarray*}
Q_1 & = & -2Q\\
Q_2 & = & -Q\\
Q_3 & = & 3Q
\end{eqnarray*}
Thus,
\begin{eqnarray*}
-\frac{2kQ}{r_{1}}-\frac{kQ}{r_{2}}+\frac{3kQ}{r_{3}} & = & 5000\\
kQ\left( -\frac{2}{r_{1}}-\frac{1}{r_{2}}+\frac{3}{r_{3}}\right) & = & 5000\\
Q & = & \frac{5000}{k\left( -\frac{2}{r_{1}}-\frac{1}{r_{2}}+\frac{3}{r_{3}}
\right) }
\end{eqnarray*}
When measuring $r_1$, $r_2$, and $r_3$, remember to measure them along
the perpendicular line connecting the two equipotential lines. Since it
takes no work to move along an eqipotential, the only distance that matters
is that along the perpendicular line. Also remember that the question
asked for the charge on $Q_3$ which has charge $3Q$.
\end{quote}
11) In the picture below, the 3 charges $Q_1$, $Q_2$, and $Q_3$ are located at
positions $(-a,0)$, $(a,0)$, and $(0,-d)$ respectively. (The origin is the
point halfway between $Q_1$ and $Q_2$.)
\begin{figure}[h]
\begin{center}
\includegraphics[width=1.73in]{capa04-11.png}
\end{center}
\end{figure}
\begin{quote}
Consider the special case where $Q_1$, $Q_3>0$, and $Q_2=-Q_1$. Assume that
the zero of potential is at infinity, as is normal for point charges. Which
of the following statements are true? (Choose all correct answers e.g. ABD,
CDFG).\\
QUESTION:\\
A) The electric field at the origin points in the positive y direction, away
from $Q_3$.\\
B) The force on $Q_3$ due to the other two charges is zero.\\
C) The external work done to bring these charges to this configuration (from
infinity) was positive.\\
D) The electric potential at any point along the y-axis is positive.\\
E) The electric potential at the origin equals $Q_3/\left( 4\pi \epsilon
_0 d\right) $.\\
F) The work required to move $Q_3$ from its present position to the origin
is zero.\\
G) If $Q_3$ is released from rest, it will initially accelerate to the
right.\\
ANSWER:\\
A) False, it points into the first quadrant since the field vectors from
$Q_1$ and $Q_2$ both point to the right. The field from $Q_3$ points
up, but the resultant vector is not along the y-axis.\\
B) False, the force points to the right. The vertical components of the
force cancel, but their horizontal components add for the resultant force.\\
C) False, it takes no work to bring the first charge in ($Q_1$). It
takes negative work to bring $Q_2$ in since it is a negative charge being
attracted to $Q_1$. Since this configuration is a dipole, it takes very
little work to bring $Q_3$ in perpendicular to the field (might be zero).
Thus, the resulting work is negative.\\
D) True, the potential from $Q_1$ and $Q_2$ cancel each other out
leaving only the positive potential from $Q_3$.\\
E) True, see (D)\\
F) True, the work is $W=-q\Delta V$. Since the potential along the y-axis
due to $Q_1$ and $Q_2$ is zero (it is only positive in (D) because of
$Q_3$), it takes no work to move $Q_3$ along it.\\
G) True, see (B)\\
Remember to provide the letters of the correct answers, in this case, DEFG
\end{quote}
12) In the previous problem, let $Q_1 = 2.10\ \mu C$, $Q_2 = -3.20\ \mu C$,
and $Q_3 = 3.70\ \mu C$. (Note that $Q_1$ and $Q_2$ are different now.) The
distances are $a=1.20\ cm$ and $d=2.80\ cm$. Calculate the total
electrostatic potential energy of the charge configuration.
\begin{quote}
$Q_1$ = Given\\
$Q_2$ = Given\\
$Q_3$ = Given\\
$a$ = Given\\
$d$ = Given\\
To find the ``electrostatic potential energy'' of the configuration, simply
calculate the amount of energy required to bring the charges to their
present position from infinity. It takes zero work to bring in the first
charge ($Q_1$). To bring in $Q_2$ a distance $2a$ from $Q_1$, from
equation 25-1,
\begin{eqnarray*}
\Delta U_{A\rightarrow B} & = & -\int_{A}^{B}F\cdot dl=-\int_{\infty }^{2a}
\frac{kQ_{1}Q_{2}}{r^{2}}dr\\
& = & kQ_{1}Q_{2}\int_{2a}^{\infty }\frac{dr}{r^{2}}\\
& = & -\left. \frac{kQ_{1}Q_{2}}{r}\right|_{2a}^{\infty }\\
& = & -\left( \frac{kQ_{1}Q_{2}}{\infty }-\frac{kQ_{1}Q_{2}}{2a}\right)\\
& = & -\left( 0-\frac{kQ_{1}Q_{2}}{2a}\right)\\
& = & \frac{kQ_{1}Q_{2}}{2a}
\end{eqnarray*}
To bring in the charge $Q_3$ from infinity, we have to calculate the
potential energy from both $Q_1$ and $Q_2$ that are already in place. Thus,
\[
\Delta U=\frac{kQ_{1}Q_{3}}{r_{1}}+\frac{kQ_{2}Q_{3}}{r_{2}}
\]
$r$ can be calculated from the triangle,
\[
r=\sqrt{a^{2}+d^{2}}
\]
Thus, the total electrostatic potential energy is
\[
\Delta U=\frac{kQ_{1}Q_{2}}{2a}+\frac{kQ_{1}Q_{3}}{\sqrt{a^{2}+d^{2}}}+
\frac{kQ_{2}Q_{3}}{\sqrt{a^{2}+d^{2}}}
\]
\end{quote}
\end{document}