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\begin{center}
{\Large {Solution Derivations for Capa \#3\\[20pt]} }
\end{center}
1) Consider two charged parallel planes of infinite extent as shown above.
The magnitudes of the charge densities on the two planes are equal. (For
each statement select T True, F False). WARNING! You have 4 tries only
for this problem.
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.9in]{capa03-01.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) If both plates are negatively charged, the electric field at $a$ points
towards the top of the page.\\
B) If both plates are positively charged, there is no electric field at
$b$\\
C) If the plates are oppositely charged, there is no electric field at
$c$\\
ANSWER:\\
A) False, the field points toward the bottom of the page.\\
B) True, the component of the electric field from each plane cancels the
component from the other.\\
C) True, the electric field vector pointing away from the positive plane is
cancelled by the field vector pointing toward the negative plane.\\
Remember to answer in True or False. For this problem, FTT
\end{quote}
2) Consider a spherical CONDUCTING shell with NO NET CHARGE, with a point
charge, $+Q$, placed at its center. (For each statement select T True, F
False). WARNING! You have 8 tries only for this problem.
\begin{figure}[h]
\begin{center}
\includegraphics[width=2.26in]{capa03-02.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) The electric field at $e$ is zero.\\
B) The electric field at $a$ is zero.\\
C) The inner surface of the shell carries a charge $-Q$.\\
D) The electric field at $c$ is zero.\\
ANSWER:\\
A) False, there is an enclosed charge, namely the point charge at the
center. The overall shell has no $net$ charge, however.\\
B) False, it encloses the point charge.\\
C) True. Charge must reside on a surface. The interior of the shell
cannot have any electric field. Thus, a Gaussian surface through $c$ must
have no field. The only way for there to be no field through $c$ is for
the Gaussian surface to enclose no net charge.\\
D) True, see (C).\\
Again, you must enter True or False for this question. FFTT for this
example.
\end{quote}
3) Consider a sphere of radius $R = 8.5\ m$ where a charge $Q=4.0\ \mu C$
is uniformly distributed through the volume of the sphere. What is the
magnitude of the electric field at a point half way between the center
of the sphere and the surface?
\begin{quote}
$R$ = Given\\
$Q$ = Given\\
The volume charge density of this sphere is
\[\rho =\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^{3}}
\]
This also means that $Q=\rho V$. A point halfway from the center to the
surface would have radius
\[
r=\frac{R}{2}.
\]
To find the volume this smaller sphere has,
\[
V=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi
\left( \frac{R}{2}\right) ^{3}=\frac{4}{3}\pi \frac{R^{3}}{8}=\frac{1}{6}\pi
R^{3}.
\]
The charge enclosed will be proportional to the amount of volume the smaller
sphere has compared to the larger sphere. Thus,
\[
q_{enclosed}=\rho \left( \frac{1}{6}\pi R^{3}\right) =\frac{Q}{\frac{4}{3}
\pi R^{3}}\left( \frac{1}{6}\pi R^{3}\right) =\allowbreak \frac{1}{8}Q.
\]
Now, Gauss's law will tell us the flux through this surface which we can
then relate to the enclosed charge.
\[
\oint E\cdot dA=\oint E\cos \theta\,dA=E\oint dA=4\pi r^{2}E=4\pi \left(
\frac{R}{2}\right) ^{2}E=\pi R^{2}E.
\]
Now,
\[
\oint E\cdot dA=\frac{q_{enclosed}}{\epsilon _{0}}.
\]
So,
\begin{eqnarray*}
\pi R^{2}E &=& \frac{1}{8}\frac{Q}{\epsilon _{0}}\\
E &=& \frac{Q}{8\pi R^{2}\epsilon _0} =\frac{kQ}{2R^{2}}
\end{eqnarray*}
\end{quote}
4) An electric flux of $159\ N\cdot m^2/C$ passes through a flat horizontal
surface that has an area of $0.82\ m^2$. The flux is due to a uniform
electric field. What is the magnitude of the electric field if the field
points $15^{\circ}$ above the horizontal?
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.64in]{capa03-04.png}
\end{center}
\end{figure}
\begin{quote}
$\phi $ = Given\\
$A$ = Given\\
$\theta $ = Given\\
The electric flux, you will remember, is also given by $E\cdot A$. This
dot product is also equal to $EA\cos \theta $. However, $\theta $ in this
case is the angle between the surface normal vector and the electric field
vector. The angle you are given is the angle between the electric field
and the surface. So, the angle between the normal vector and the field
vector is $\alpha =\left( 90-\theta \right) $.
\begin{eqnarray*}
\phi &=& EA\cos \alpha\\
E &=& \frac{\phi }{A\cos \alpha }
\end{eqnarray*}
\end{quote}
5) A cubic cardboard box of side $a = 0.340\ m$ is placed so that its edges
are parallel to the coordinate axes, as shown in the figure. There is NO
net electric charge inside the box, but the space in and around the box is
filled with a nonuniform electric field of the following form:
$\stackrel{\rightarrow}{\mathbf{E}}(x,y,z) = Kz\,\mathbf{j} + Ky\,\mathbf{k}$,
where $K = 3.40\ N/C \cdot m$ is a constant.
What is the electric flux through the top face of the box? (The top face
of the box is the face where $z=a$. Remember that we define positive
flux pointing out of the box.)
\begin{figure}[h]
\begin{center}
\includegraphics[width=2.59in]{capa03-05.png}
\end{center}
\end{figure}
\begin{quote}
$a$ = Given\\
$\mathbf{E}=Kz\,\mathbf{j}+Ky\,\mathbf{k}$\\
$K$ = Given
In this case, we have to use a double integral to evaluate $\mathbf{E}$ over
the entire surface. But, $z$ is a constant across the top face. Thus, we
don't have to worry about it in the integral. Since $\mathbf{E}$ also does
not depend on $x$, we can think of the top face of the box simply as a square
in the $xy$-plane. The equation we have is then
\[
\phi =\int E\cdot dA=\int_{x=0}^{a}\int_{y=0}^{a}E\cos \theta\,dy\,dx
\]
The $E$ is the magnitude of the vector $\mathbf{E}$. To figure out what the
magnitude of $\mathbf{E}$ and $\cos \theta$ really are, we have to think about
the vector equation given for $\mathbf{E}$. This says that at any given point
$\left(x,y,z\right)$, the vector points $Kz$ units in the $y$ direction and
$Ky$ units in the $z$ direction. To find the resultant vector, we use the
Pythagorean theorem. Thus,
\[
E=\sqrt{\left( Kz\right) ^{2}+\left( Ky\right) ^{2}}.
\]
To find the angle $\theta$, we can construct a triangle. From this triangle,
we can read off $\cos \theta $. Again, $\theta $ is the angle between the
normal vector and the electric field vector. In this case, the normal vector
will be pointing straight up (in the $+z$ direction). Using alternate
interior angles, we can see that
\[
\cos \theta =\frac{Ky}{\sqrt{\left( Kz\right)^{2}+\left( Ky\right) ^{2}}}.
\]
Plugging this back into the integral,
\begin{eqnarray*}
\phi &=& \int_{x=0}^{a}\int_{y=0}^{a}\frac{Ky}{\sqrt{\left(
Kz\right) ^{2}+\left( Ky\right) ^{2}}}\sqrt{\left( Kz\right) ^{2}+\left(
Ky\right) ^{2}}\,dy\,dx \\
&=& \int_{x=0}^{a}\int_{y=0}^{a}Ky\,dy\,dx\\
&=& K\int_{y=0}^{a}\left( \left.yx\,dy\right|_{x=0}^{a}\right)\\
&=& Ka\int_{y=0}^{a}y\,dy\\
&=& Ka\left( \left.\frac{y^{2}}{2}\right|_{y=0}^{a}\right)\\
&=& \frac{1}{2}Ka^{3}
\end{eqnarray*}
\end{quote}
6) What is the total electric flux through the 5 other faces of the box?
(Again, outward flux is positive.)
\begin{quote}
We can think of the box as a Gaussian surface. Since there is no net charge
inside the box, there will be no net electric flux thorough the surface.
Thus, if one face has electric flux given in (5), then the total electric
flux by the other 5 sides must cancel this. Thus, the answer is simply
\[
\phi =-\frac{1}{2}Ka^{3}
\]
\end{quote}
7) A small conducting spherical shell with inner radius $a$ and outer radius
$b$ is concentric with a larger conducting spherical shell with inner radius
$c$ and outer radius $d$. The inner shell has a total charge $-2q$ and the
outer shell has a total charge $+4q$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=2in]{capa03-07.png}
\end{center}
\end{figure}
\begin{quote}
QUESTION:\\
A) The total charge on the inner surface of the small shell is zero.\\
B)\ The radial component of the electric field in the region $rd$ is given
by $+2q/\left( 4\pi \epsilon _{0}r^{2}\right) $.\\
F) The total charge on the inner surface of the large shell is zero.\\
G) The total charge on the outer surface of the small shell is $-2q$.\\
ANSWER:\\
A) True, all the charges will move to the outside to get as far apart as
possible.\\
B) False, the electric field is zero inside the shell.\\
C) True, the other $+2q$ will be on the inner surface, so a Gaussian
surface in the outer shell has no enclosed charge.\\
D) False. There is no electric field inside of the shell.\\
E) True, shell acts like a point charge (see lecture notes, 24-6).\\
F) False, $+2q$ will be on the inside. See (C).\\
G) True, all the charges will move to the outside.\\
Remember to give the letters of the true answers, in this case, ACEG
\end{quote}
8) A thin {\em infinite} nonconducting sheet with uniform surface charge
density $\sigma = +8.0\ \mu C/m^2$ lies in the {\em y-z} plane. A charge
$Q = +5.5\ \mu C$ is located on the $x$ axis at a distance $x = 36\ cm$
from the sheet as shown.
Find the magnitude of the electric field at a point $P$ with coordinates
$x=y=36\ cm$, $z=0$.
\begin{quote}
$\sigma $ = Given\\
$Q$ = Given\\
$x$ = Given
The electric field at any given point is the vector sum of fields from each
charge present. The electric field from an infinite plane of charge (Lecture
notes, 24-6) is
\[
E_{plane}=\frac{\sigma }{2\epsilon _{0}}.
\]
The field from a point charge is
\[
E_{point}=\frac{q}{4\pi \epsilon_{0}r^{2}}.
\]
Remember to convert the distance to meters and the charges to Coulombs.
The magnitude is then
\[
E=\sqrt{E_{plane}^{2}+E_{point}^{2}}
\]
\end{quote}
9) A test charge of $2.90\ \mu C$ is placed $6.30\ cm$ away from a very
large flat uniformly charged nonconducting surface. The force on the charge
is $230\ N$. The charge is now moved to $3.50\ cm$ away from the surface.
What is the force on the test charge now?
\begin{quote}
$q$ = Given\\
$r_1$ = Given\\
$F_1$ = Given\\
$r_2$ = Given\\
$F_2$ = ?\\
The electric field from an infinite plane is constant everywhere. Thus,
$E=\frac{F}{q}$ is constant. Since the field is not changing and the test
charge is not changing, the force cannot change either.
\end{quote}
10) Three identical metal spheres, x, y, and z, are placed at the corners
of an equilateral triangle, as shown below. Sphere z has a constant
positive charge. Which vector could represent the force on z, when x is
positive and y is negative, with the magnitude of the charge on y less than
that on x? WARNING! You have 4 tries only for this problem.
\begin{quote}
Cannot generalize further. A simple problem, though. Draw the vectors
from each charge and vector add them.
\end{quote}
11) A cork ball of mass $7.30\ g$ is placed between two very large
horizontal planes of charge. The bottom plane has a uniform charge
density of $+0.94\ \mu C/m^2$, whereas the upper plane has a uniform
charge density of $-0.94\ \mu C/m^2$. The cork ball, which carries an
unknown charge, is placed between the planes and is observed to float
motionlessly. What is the charge on the ball?
\begin{quote}
$m$ = Given\\
$\sigma _1$ = Given\\
$\sigma _2=-\sigma _1$ = Given\\
Since the charge density of the two planes are equal and opposite, the
electric field is just twice that of one of the fields. Since the positive
one is on the bottom, the field vector points straight up. The relationship
between the electric field and the electric force is given by $F=qE$. For
this problem, sum all the forces in the $y$-direction, set them equal to
$0$, and solve for the unknown. The electric field for an infinite (or very
large) plane is
\[
E=\frac{\sigma }{2\epsilon_{0}}.
\]
Thus,
\[
F=q\left( 2E\right) =mg
\]
since the field is twice that generated by one plane.
\begin{eqnarray*}
q\frac{2\sigma }{2\epsilon _{0}} & = & mg\\
q & = & \frac{\epsilon _{0}mg}{\sigma }
\end{eqnarray*}
\end{quote}
12) The diagram below shows a dipole centered at the origin and along the
$x$-axis. Determine an expression for the total electric field at a point
A $(r_0 = 3.8L, 0)$ in terms of $q$ and $L$. Calculate the magnitude of the
total electric field at A, when $q=8.15 \times 10^{-7}\ C$ and $L = 44.2\ cm$.
\begin{quote}
$r_0$ = Given\\
$L$ = Given\\
$q$ = Given\\
To solve this problem, we can calculate the electric field at each end of
the dipole and then vector add the fields together. The electric field
experienced by a point charge is
\[
E=\frac{F}{q}=\frac{kq}{r^2}.
\]
The field from the negative charge is
\[
E_1=\frac{-kq}{\left( L+r_0\right) ^2}
\]
The field from the positive charge is
\[
E_2=\frac{kq}{\left( r_0-L\right) ^2}
\]
The total electric field is then
\[
E_1+E_2=\frac{kq}{\left( r_0-L\right) ^2}-\frac{kq}{\left(
L+r_0\right) ^2}
\]
\end{quote}
13) Consider the situation as described in the diagram, and the following
statements. If the statement is true, answer `T', if it is false, answer
`F', and if the answer cannot be determined from the information provided,
answer `C'. For example, if `A' and `E' are true and there is not enough
information to answer `C' and the rest are false, then answer `TFCFTF'.
\begin{quote}
QUESTION:\\
A) The direction of the net electric field at A is $\leftarrow$\\
B) The magnitude of the net electric field at point A is less than the
magnitude of the net electric field at the origin.\\
C) The direction of the net electric field at the origin is $\leftarrow$\\
D) The direction of the net electric field at a point on the negative y-axis
is $\downarrow$\\
E) The direction of the net electric field at the point $(-r_0,0)$ is
$\rightarrow$\\
F) The direction of the net electric field at a point on the positive y-axis
is $\nwarrow $\\
ANSWER:\\
A) False, it is to the right caused by the proximity to the positive charge.\\
B) True, both charges contribute to the electric field in the same direction.\\
C) True. The direction from (B) is to the left.\\
D) False, it will be to the left.\\
E) True, it is opposite that calculated in (12).\\
F) False, it will be to the left, as in part (D)\\
\end{quote}
14) A point charge $+q$ is at the origin. A spherical Gaussian surface
centered at the origin encloses $+q$. So does a cubical surface centered
at the origin and with edges parallel to the axes. Select T-True, F-False,
If the first is T and the rest F, enter TFFFF.
\begin{quote}
QUESTION:\\
A) The Electric Flux through the spherical surface is lesser than that
through the cubical surface.\\
B) The area vector and the E-Field vector point in the same direction for
all points on the cubical surface.\\
C) By symmetry, the E-Field at all points on the cubical surface is zero.\\
D) The flux through the spherical Gaussian Surface is independent of its
radius.\\
E) Now q is moved from the origin but is sill within both the surfaces. The
total flux through both surfaces is changed.\\
ANSWER:\\
A) False, the same amount of field lines exit both surfaces.\\
B) False, the E-Field vector will point radially from the enclosed point
charge where the area vector will point parallel to the axes.\\
C) False. Same argument could be made for the sphere, but the E-Field is
not zero there either.\\
D) True, since the field lines begin on the charge and travel to infinity,
there is no dependence on the radius of the surface.\\
E) False, the flux is only the amount of field lines travelling through the
surfaces.\\
\end{quote}
15) The diagram below depicts a section of an infinitely long cylinder of
radius $R$ that carries a uniform (volume) charge density $\rho$. Develop
an expression for the electric field anywhere inside the cylinder. To
check your result, what is the magnitude of the electric field at
$r = 7.70\ cm$, where $R = 9.20\ cm$ and $\rho = 1.05\ \mu C/m^3$? (Note:
$r$ is measured radially from the center axis of the cylinder.)
\begin{quote}
$r$ = Given\\
$R$ = Given\\
$\rho $ = Given\\
For this problem, it is important to choose the appropriate Gaussian
surface. Important things to remember are that the Gaussian surface should
reflect the original object's shape, the electric field must be constant,
and $E\cdot dA$ must be constant. Since there is a cylinder of infinite
length involved, we can choose our Gaussian surface to be a smaller cylinder
of radius $r$. Gauss's law tells us
\[
\oint E\cdot dA=\frac{q_{enclosed}}{\epsilon _{0}}.
\]
To find the enclosed charge, we can use the given volume charge density, $%
\rho $. \ This tells us charge/volume. So, we have to multiply $\rho $ by
a volume to get the corresponding charge. The volume of the enclosed
cylinder is $\pi r^{2}l$. Thus, $q_{enclosed}=\rho \pi r^{2}l$. \ Now we
can solve this.
\[
\oint E\cdot dA=\frac{\rho \pi r^{2}l}{\epsilon _{0}}
\]
Since the electric field is constant, we can pull it outside of the integral.
\[
E\oint dA=\frac{\rho \pi r^{2}l}{\epsilon _{0}}
\]
The integral of $dA$ is just the surface area of the Gaussian surface, in
this case a cylinder.
\begin{eqnarray*}
E2\pi rl & = & \frac{\rho \pi r^{2}l}{\epsilon _{0}}\\
E & = & \frac{\rho \pi r^{2}l}{2\pi rl\epsilon _{0}}\\
E & = & \frac{\rho r}{2\epsilon _{0}}
\end{eqnarray*}
\end{quote}
16) A thin plastic rod is bent so that it has a shape of a semicircle of
radius $R = 8.20\ m$. An amount of charge $Q = 10.00\ C$ is distributed
along the rod. What is the magnitude of the electric field at the center
of the circle (point $P$)?
\begin{quote}
$R$ = Given\\
$Q$ = Given\\
This problem reverts back to using direct integration. The length of the
semicircle is just $\pi r$ (half of the circumference of a circle). Thus,
the charge per unit length is
\[
\lambda =\frac{Q}{\pi r}.
\]
Coulomb's law for an electric field is
\[
E=\frac{kq}{r^{2}}
\]
If we integrate along the length of the semicircle, each little piece of the
semicircle will have a charge $dQ$ and an associated field $dE$. However,
we are only interested in the vertical component of the field. If we take
the left side of the semicircle to be $\theta =0$ and the right side to be
$\theta =\pi $, then the vertical component is $\sin \theta $. We now have
\[
dE=\frac{k\,dQ}{r^{2}}\sin \theta
\]
We must now relate $dQ$ to $\theta $ since that is the variable we will
integrate with respect to. We know that $\lambda $ is the charge per
length. So if we multiply it by a small length corresponding to $d\theta$,
then we should be able to finish. The formula for arc length is $r\theta
$. Thus, $dQ=\lambda r\,d\theta $. But $\lambda =\frac{Q}{\pi r}$, so
$dQ=\frac{Q\,d\theta }{\pi }$
\begin{eqnarray*}
dE & = & \frac{kQ}{\pi r^{2}}\sin \theta\,d\theta \\
E & = & \int_{0}^{\pi }\frac{kQ}{\pi r^{2}}\sin \theta\,d\theta \\
E & = & \frac{kQ}{\pi r^{2}}\int_{0}^{\pi }\sin \theta\,d\theta \\
E & = & \frac{kQ}{\pi r^{2}}\left( -\cos \theta |_{0}^{\pi }\right) =\frac{kQ}{
\pi r^{2}}\left( 1-\left( -1\right) \right) =2\frac{kQ}{\pi r^{2}}
\end{eqnarray*}
\end{quote}
\end{document}