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%TCIDATA{Created=Tue Mar 24 18:31:51 1998}
%TCIDATA{LastRevised=Tue Jan 26 18:05:28 1999}
%TCIDATA{Language=American English}

\input{tcilatex}
\begin{document}

\SetTitle{Physics 312 -- Assignment 1}
\SetAuthor{}
\Setdate{}
\TitlePage{}

\textbf{Problem 1 -- Questions C1-C5 on page 620 of \emph{Tipler}}

True or False: If the statement is true, explain why it is true. If it is
false, give a counterexample.

\textbf{(C1)} The electric field of a point charge always points away from
the charge.

False. This is only true for a positive point charge. The electric field of
a negative point charge points toward the charge.

\textbf{(C2)} The charge of the electron is the smallest unit of charge
found.

True. Every particle that has ever been isolated in a laboratory has a
charge equal to an integer times the charge of an electron. (It is true that
quarks carry charges that are smaller than that of an electron, but
individual quarks have never been isolated.)

\textbf{(C3)} Electric field lines never diverge from a point in space.

False. Electric field lines diverge from a positive point charge.

\textbf{(C4)} Electric field lines never cross at a point in space.

True. If electric field lines crossed at a point in space then the electric
field at that point would have two different values at the same time, which
is impossible.

\textbf{(C5)} All molecules have electric dipole moments in the presence of
an external electric field.

True. The external electric field polarizes the molecule, producing a dipole
moment.

\vspace{0.5cm}

\textbf{Problem 2 -- Questions C1-C8 on page 652 of \emph{Tipler}}

True or False: If the statement is true, explain why it is true. If it is
false, give a counterexample.

\textbf{(C1)} If there is no charge in a region of space, the electric field
on a surface surrounding the region must be zero everywhere.

False. The flux through the surface must be zero, but the field itself need
not be. Consider a point charge at the origin and a sphere of radius one
meter located three meters from the origin. The sphere contains no charge,
but the electric field is nonzero everywhere on its surface.

\textbf{(C2)} Gauss's law holds only for symmetric charge distributions.

False. Gauss's law holds for \emph{any} charge distribution, but it is not
usually useful for determining the electric field due to the charge
distribution unless the distribution is symmetric.

\textbf{(C3)} The electric field inside a uniformly charged spherical shell
is zero.

True. Because of symmetry we know that if there were an electric field
inside the shell it would be radial. But there is no charge inside the
shell, so Gauss's law gives $E_{r}=0$. Therefore the entire electric field
must be zero inside the uniformly charged spherical shell.

\textbf{(C4)} In electrostatic equilibrium, the electric field inside a
conductor is zero.

True. Charges can move freely throughout a conductor, so if there were a
nonzero electric field anywhere inside the conductor the charges would
quickly move to cancel it.

\textbf{(C5)} The result that $\mathbf{E}=0$ inside a conductor can be
derived from Gauss's law.

False. To show this result one must consider the physical characteristics of
a conductor.

\textbf{(C6)} If the net charge on a conductor is zero, the charge density
must be zero at every point on the surface of the conductor.

False. The \emph{total} charge on the conductor must be zero, but there can
be regions of positive charge so long as there are regions of negative
charge to balance them.

\textbf{(C7)} The electric field is discontinuous at all points at which the
charge density is discontinuous.

False. If the charge density is discontinuous then the \emph{divergence} of
the electric field must be discontinuous because of Gauss's law, but the
field itself need not be. Consider, for example, the electric field of a
solid sphere of charge (Figure 19-25 on page 643 of \emph{Tipler}). The
charge density is certainly discontinuous at the surface of the sphere, but
the field itself is continuous.

\textbf{(C8)} Half of the electric field at a point just outside the surface
of a conductor is due to the charge on the surface in the immediate vicinity
of that point.

True. See page 647 of \emph{Tipler} and especially Figure 19-32.

\vspace{0.5cm}

\textbf{Problem 3 -- Exercise 17 on page 621 of \emph{Tipler}}

Using Maple or the like, make a field plot and an equipotential plot for
this charge configuration.

\smallskip

Figure 18-26 shows the electric-field lines for a system of two point
charges.

\textbf{(a)} What are the relative magnitudes of the charges?

We see that the charge on the left has a total of 32 field lines
intersecting it while the charge on the right has a total of 8 field lines
intersecting it. It follows, then, that the charge on the left is four times
the magnitude of the charge on the right.

\textbf{(b)} What are the signs of the charges?

The field lines diverge from the charge on the left and converge on the
charge on the right. Therefore the charge on the left is positive and the
charge on the right is negative.

\textbf{(c)} In what regions of space is the electric field strong? In what
regions is it weak?

The electric field is strong near the charge on the left, and is especially
strong in the area between the two charges. The electric field is weak far
from the charges.

\smallskip For the plots, we choose units so that the left charge has charge
equal to $+4$ and is located at $\left( -1,0,0\right) $, while the right
charge has charge equal to $-1$ and is located at $\left( 1,0,0\right) $.
Then the potential due to the two charges is (up to a multiplicative
constant) 
\[
V\left( x,y\right) =\frac{4}{\sqrt{\left( x+1\right) ^{2}+y^{2}}}+\frac{-1}{%
\sqrt{\left( x-1\right) ^{2}+y^{2}}}. 
\]
We can then use the Maple functions \texttt{gradplot} and \texttt{contourplot%
} to make plots of the electric field and the equipotentials, respectively.
The is done in the file \texttt{assignment1.mws}, which can also be found on
the solutions webpage.

\vspace{0.5cm}

\textbf{Problem 4 -- Exercise 33 on page 622 of \emph{Tipler}}

\smallskip

An electron with a kinetic energy of $2\cdot 10^{-16}$ $\mathrm{J}$ is
moving to the right along the axis of a cathode ray tube as shown in Figure
18-29. There is an electric field $\mathbf{E}=\left( 2\cdot 10^{4}\text{ }%
\frac{\mathrm{N}}{\mathrm{C}}\right) \mathbf{j}$ in the region between the
deflection plates. Everywhere else, $\mathbf{E}=0$.

\textbf{(a)} How far is the electron from the axis of the tube when it
reaches the end of the plates?

Call the initial kinetic energy of the electron $K_{i}$. Then 
\[
K_{i}=\frac{1}{2}m_{e}v_{i}^{2}, 
\]
where $m_{e}$ is the mass of the electron and $v_{i}$ is the initial speed
of the electron. It follows, then, that the initial velocity of the electron
is 
\[
v_{i}=\sqrt{\frac{2K_{i}}{m_{e}}}\mathbf{i}. 
\]
While between the plates the electron experiences a force $\mathbf{F}=e%
\mathbf{E}=eE\mathbf{j}$, where $E$ is the magnitude of the electric field
between the plates. Because this force does not act in the horizontal
direction the horizontal component of the velocity is constant, and so the
time taken by the electron to pass between the plates is given by 
\[
t_{plates}=\frac{x_{plates}}{v_{i}}=x_{plates}\sqrt{\frac{m_{e}}{2K_{i}}}, 
\]
where $x_{plates}$ is the horizontal extent of the plates. The vertical
component of the velocity when the electron leaves the plates is then given
by the basic kinematics equation $v=v_{0}+at$ as 
\[
v_{vertical}=\left( \frac{F}{m_{e}}\right) t_{plates}=\frac{eE}{m_{e}}%
x_{plates}\sqrt{\frac{m_{e}}{2K_{i}}}=\frac{eE}{\sqrt{2m_{e}K_{i}}}%
x_{plates}. 
\]
The kinematics relation $v_{f}^{2}=v_{0}^{2}+2ax$ yields 
\[
y_{plates}=\frac{v_{vertical}^{2}}{2\left( \frac{F}{m_{e}}\right) }=\frac{%
\frac{e^{2}E^{2}}{2m_{e}K_{i}}x_{plates}^{2}}{2\left( \frac{eE}{m_{e}}%
\right) }=\frac{eE}{4K_{i}}x_{plates}^{2} 
\]
for the distance the electron is from the axis of the tube when it reaches
the end of the plates. Plugging in values, we find that 
\[
y_{plates}=\frac{\left( -1.602\cdot 10^{-19}\text{\textrm{C}}\right) \left(
2\cdot 10^{4}\text{ }\frac{\mathrm{N}}{\mathrm{C}}\right) }{4\left( 2\cdot
10^{-16}\text{ }\mathrm{J}\right) }\left( 4\cdot 10^{-2}\text{\textrm{m}}%
\right) ^{2}=-6.4\cdot 10^{-3}\mathrm{m}. 
\]
Thus when the electron leaves the plates it is $6.4$ $\mathrm{mm}$ below the
axis of the tube.

\smallskip

\textbf{(b)} At what angle is the electron moving with respect to the axis?

The horizontal component of velocity is constant throughout the journey
through the plates, so when the electron leaves the plates it is 
\[
v_{horizontal}=v_{i}=\sqrt{\frac{2K_{i}}{m_{e}}}. 
\]
We found in part (a) that the vertical component of the velocity when the
electron leaves the plates is 
\[
v_{vertical}=\frac{eE}{\sqrt{2m_{e}K_{i}}}x_{plates}. 
\]
Thus, when the electron leaves the plates it is travelling at an angle 
\[
\theta =\tan ^{-1}\left( \frac{\frac{eE}{\sqrt{2m_{e}K_{i}}}x_{plates}}{%
\sqrt{\frac{2K_{i}}{m_{e}}}}\right) =\tan ^{-1}\left( \frac{eE}{2K_{i}}%
x_{plates}\right) 
\]
with respect to the axis of the tube. Plugging in the values, we find that 
\[
\theta =\tan ^{-1}\left( \frac{\left( -1.602\cdot 10^{-19}\text{\textrm{C}}%
\right) \left( 2\cdot 10^{4}\text{ }\frac{\mathrm{N}}{\mathrm{C}}\right) }{%
2\left( 2\cdot 10^{-16}\text{ }\mathrm{J}\right) }\left( 4\cdot 10^{-2}\text{%
\textrm{m}}\right) \right) =-17.7^{\circ }. 
\]
Thus when the electron leaves the plates it is traveling at an angle $%
17.7^{\circ }$ below the axis of the tube.

\smallskip

\textbf{(c)} How far from the axis will the electron be when it strikes the
florescent screen?

After the electron exits the plates, both components of its velocity will
remain constant because no force acts on the electron. The time between the
electron exiting the plates and striking the florescent screen is then 
\[
t_{screen}=\frac{x_{screen}}{v_{i}}=x_{screen}\sqrt{\frac{m_{e}}{2K_{i}}}, 
\]
where $x_{screen}$ is the distance from the end of the plates to the screen.
Because the vertical component of the velocity is constant during this time,
the electron moves a distance 
\[
y_{screen}=v_{vertical}t_{screen}=\left( \frac{eE}{\sqrt{2m_{e}K_{i}}}%
x_{plates}\right) \left( x_{screen}\sqrt{\frac{m_{e}}{2K_{i}}}\right) =\frac{%
eE}{2K_{i}}x_{plates}x_{screen}. 
\]
Plugging in values, we see that 
\[
y_{screen}=\frac{\left( -1.602\cdot 10^{-19}\text{\textrm{C}}\right) \left(
2\cdot 10^{4}\text{ }\frac{\mathrm{N}}{\mathrm{C}}\right) }{2\left( 2\cdot
10^{-16}\text{ }\mathrm{J}\right) }\left( 4\cdot 10^{-2}\text{\textrm{m}}%
\right) \left( 12\cdot 10^{-2}\text{ }\mathrm{m}\right) =-3.84\cdot 10^{-2}%
\mathrm{m}. 
\]
The total deflection in the horizontal direction is then 
\[
y_{plates}+y_{screen}=-4.48\cdot 10^{-2}\mathrm{m}. 
\]
Therefore the electron strikes the screen at a distance $4.48$ $\mathrm{cm}$
below the axis of the tube.

\end{document}
