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\begin{document}
\newcommand{\bold}[1]{\mbox{\boldmath $#1$}}
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\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
Spring 2008

\end{bf}

\begin{center}
\begin{large}
Solution to {\o}ving 11
\end{large}
\end{center}

\vspace{0.5cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
$a)$ This system can be viewed as three resistances connected in series:
the two 60 cm long Cu wires and the resistance $R = 20 \; \Omega$.
The resistance of the two Cu wires becomes
$$R_A = \frac{l}{\sigma A} = 
\cdot \frac{1.20{\rm m}}{5.8\cdot10^7\;\Omega^{-1}{\rm m}^{-1}\cdot
2\cdot10^{-6}{\rm m}^2} = 0.01\;\Omega$$
The same current $I$ passes through the whole system. It is
$$I = \frac{V}{R+R_C} = \frac{1.5\;{\rm V}}{20.01\;\Omega} =
0.07496\;{\rm A} \simeq 0.075\;{\rm A}$$
according to Ohm's law. Thus, we obtain the voltage drops
$$V_R = RI = 20\;\Omega \cdot 0.075\;{\rm A} \simeq 1.5\;{\rm V}$$
over the resistance $R$ and
$$V_C = R_C I = 0.01\;\Omega \cdot 0.075\;{\rm A} = 0.00075\;{\rm V}$$
over the two Cu wires together. In conclusion: Negligible voltage drop
in the two Cu wires.

\vspace{0.2cm}
\noindent
$b)$ We found the current $I$ in $a)$ above. The dissipated effect in the
resistance $R$ becomes
$$P = V_R I = 1.5\;{\rm V} \cdot 0.075\;{\rm A} = 0.1125\;{\rm W} \simeq 0.11\;{\rm W}$$

\vspace{0.2cm}
\noindent
$c)$ Here, we must first find the density of free electrons $n$. Next, we
may use $I = j \cdot A = n e v A$ in order to calculate the mean drift velocity $v$.

\noindent
In Cu, we have a mass density 8960 kg pr m$^3$. This corresponds to
8960/0.06354 mol = 141014 mol = 141014 $\cdot 6.02\cdot10^{23}$ atomer = $8.49\cdot10^{28}$
atoms, and hence equally many free electrons, assuming one free electron pr Cu atom.
Mean drift velocity becomes
$$v = \frac{I}{neA} = \frac{0.075}{8.49\cdot10^{28}\cdot1.6\cdot10^{-19}
\cdot2\cdot10^{-6}} = 2.76\cdot10^{-6} \; {\rm m/s} = 2.76 \mu{\rm m/s}$$
Average thermal velocity for the electrons may be estimated by setting the
kinetic energy equal to the thermal energy:
\begin{eqnarray}
\frac{1}{2}mv^2 &=& \frac{3}{2}k_B T \nonumber \\
\Rightarrow v &=& \sqrt{\frac{3k_B T}{m}} \simeq 10^5 \; {\rm m/s} \nonumber
\end{eqnarray}
Here, $k_B = 1.38\cdot10^{-23}$ J/K is Boltzmann's constant. We see that the
mean drift velocity is roughly 11 orders of magnitude smaller than the average
thermal velocity. In other words, it takes several hours for a given electron
to get from one end to the other in our system!


\vspace{1.5cm}
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
$a)$ At first, we should try to realize that what we have here is the following
circuit: [a parallel connection of
$R_1$, $R_2$ and $R_3$] coupled in series with [a parallel connection of $R_4$ and $R_0 = 0$]
in series with [$R_5$]. In other words, the resistance $R_4$ is "cut short", so that no
current passes through $R_4$.
(Alternatively: We have the same value for the potential on each side of $R_4$. Then, no
current can pass through it.)
Thus, the total resistance is
$$R = \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} + R_5$$

\vspace{0.2cm}
\noindent
$b)$ It should be clear that the total current $I$ in the circuit must be the same
as the current $I_5$ passing through $R_5$. Further, it should also be clear that
$I$ must distribute itself on the three currents passing through
$R_1$, $R_2$ and $R_3$: $I = I_1 + I_2 + I_3$. In $a)$ above, we have already
concluded that no current passes through $R_4$: $I_4 = 0$.

\noindent
The voltage drop across the three upper resistances is the same:
$$V' = R_1 I_1 = R_2 I_2 = R_3 I_3$$
The voltage drop across $R_5$ is
$$V'' = R_5 I_5 = R_5 I = R_5 \frac{\cal E}{R}$$
These two together must equal the value of the voltage source:
$${\cal E} = V' + V''$$
Thus
$$V' = {\cal E} - V'' = {\cal E} - R_5 \frac{\cal E}{R} = {\cal E} \left(1 - \frac{R_5}{R} \right)$$
And finally,
$$I_1 = \frac{V'}{R_1}$$
$$I_2 = \frac{V'}{R_2}$$
$$I_3 = \frac{V'}{R_3}$$


\newpage
%\vspace{0.5cm}
\noindent
{\em Exercise 3}

\vspace{0.2cm}
\noindent
The capacitor will be discharged by electrons moving from the negatively charged side,
through the resistor, to the postively charged side. In other words, a positive current
will run through $R$, from the positively charged plate.

%%FIGURE
\vspace{0.8cm}
\centerline{\epsfxsize=2.4in \epsfbox{los11_utladning.eps}}
\vspace{0.5cm}
%%END FIGURE

\noindent
In the figure, $I$ has been drawn with direction towards the positive plate, despite
the conclusion above, that positive $I$ runs in the opposite direction. This is done, since
I prefer to keep the relation
$$I = + \frac{dQ}{dt}$$
(and not with a minus sign). Kirchhoff's voltage rule yields
$$-RI - \frac{Q}{C} = 0$$
i.e.
$$\frac{dQ}{Q} = -\frac{dt}{RC}$$
Integration on both sides gives
$$Q = k e^{-t/RC}$$
and the initial condition $Q(0) = Q_0$ determines the integration constant $k$:
$$Q_0 = k e^0 = k$$
so that
$$Q(t) = Q_0 e^{-t/RC}$$
The current then becomes
$$I(t) = \frac{dQ}{dt} = -\frac{Q_0}{RC} e^{-t/RC}$$
As expected with a minus sign, so that positive current $I$ runs counterclockwise in
the figure above.


\newpage
%\vspace{0.7cm}
\noindent
{\em Exercise 4}

\vspace{0.5cm}
\noindent
$a)$ Kirchhoff's voltage rule (K2) gives
$$\ems = \frac{Q}{C}$$
Also, we have
$$I = \frac{dQ}{dt}$$
Hence:
$$Q(t) = V_0 C \cos\omega t$$
and
$$I(t) = \dot{Q} = -V_0 \omega C \sin\omega t$$
Sketch of $\ems(t)$ (in the figure, $V(t)$, solid line) and $I(t)$ between
$t=0$ and $t=T=2\pi/\omega$:

%%FIGURE
\vspace{1.0cm}
\centerline{\epsfxsize=5.0in \epsfbox{EC-skisse.eps}}
\vspace{1.0cm}
%%END FIGURE

\noindent
We see that the applied voltage $\ems(t)$ and the resulting current $I(t)$ in
the circuit are not at maximum, zero etc at the same instants. We say that we
have a {\em phase difference} between $\ems(t)$ and $I(t)$. For this simple
circuit, with only a single capacitor, we see that $I$ is zero when $\ems$ is
at a maximum value, and the other way around. Hence, the two have a phase
difference of $\pi/2$. This is most clearly seen if we write $I$ on the form
$$I(t) = I_0 \cos(\omega t - \alpha)$$
We have
$$-\sin\omega t = \cos(\omega t + \pi/2)$$
Hence, we see that $I_0 = V_0 \omega C$ and $\alpha = -\pi/2$. The impedance of
this simple circuit, i.e., a single capacitor with capacitance $C$, becomes
$$Z_C(\omega) = \frac{V_0}{I_0} = \frac{1}{\omega C}$$
Notice that the impedance increases when the angular frequency becomes smaller.
In the limit $\omega \rightarrow 0$ the impedance $Z_C$ approaches infinity, which
is quite reasonable: No direct current through a (ideal) capacitor. In the opposite
limit, $\omega \rightarrow \infty$, $Z_C$ approaches zero, which is also reasonable:
The charge on a given plate oscillates between $+V_0 C$ and $-V_0 C$,
no matter what the frequency is. When the frequency is increased, the current in
the circuit will increase linearly with the frequency. (Later, we will see how
the magnetic field stops the current from increasing beyond limits.)


\vspace{0.5cm}
\noindent
$b)$ Kirchhoff's voltage rule (K2) gives
$$\ems = \frac{Q}{C} = R I_R$$
whereas Kirchhoff's current rule (K1) gives
$$I = I_C + I_R$$
Also, we have
$$I_C = \frac{dQ}{dt}$$
Hence:
\begin{eqnarray}
I_R(t) &=& \frac{V_0}{R} \, \cos\omega t \nonumber \\
Q(t) &=& V_0 C \, \cos\omega t \nonumber \\
I_C(t) &=& -\omega C V_0 \, \sin\omega t = \omega C V_0 \, \cos(\omega t +\pi/2) \nonumber
\end{eqnarray}
Total current delivered by the voltage source becomes
$$I(t) = \frac{V_0}{R} \, \cos\omega t - \omega C V_0 \, \sin\omega t$$
We want to have $I(t)$ on the form
$$I(t) = I_0 \, \cos(\omega t - \alpha)$$
with amplitude $I_0 = V_0/Z$, where $Z$ is the impedance of $R$ and $C$ coupled in parallel,
while $\alpha$ becomes the phase difference between $\ems(t)$ and $I(t)$. We have
$$\cos(\omega t - \alpha) = \cos\omega t \, \cos\alpha + \sin\omega t \, \sin\alpha$$
Hence, upon direct comparison:
\begin{eqnarray}
\frac{\cos\alpha}{Z} &=& \frac{1}{R} \nonumber \\
\frac{\sin\alpha}{Z} &=& -\omega C \nonumber
\end{eqnarray}
These two equations, with the two unknowns $Z$ and $\alpha$, are easily solved,
and we find
\begin{eqnarray}
Z &=& \frac{R}{\sqrt{1 + (\omega R C)^2}} \nonumber \\
I_0 &=& \frac{V_0}{R} \, \sqrt{1 + (\omega R C)^2} \nonumber \\
\alpha &=& -\arctan(\omega RC) \nonumber
\end{eqnarray}
In the limit $\omega \rightarrow 0$, we should recover well known results from the
DC examples in the lectures, and indeed, we do: $Z \rightarrow R$ and $\alpha \rightarrow 0$
so that $I_0 \rightarrow V_0/R$. All the current goes through the resistor $R$, and the
capacitor $C$ now represents an open circuit where no direct current can run.

\vspace{0.2cm}
\noindent
With the given numerical values we have
$$\omega RC = 2\pi \cdot 10^6 \cdot 10 \cdot 16 \cdot 10^{-9} = 1.0$$
so that
\begin{eqnarray}
I_0 &=& \frac{1.0}{10} \cdot \sqrt{2} = 0.14 \; {\rm A} \nonumber \\
\alpha &=& - \arctan 1.0 = -45^\circ \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Sketch of $\ems(t)$, $I(t)$, $I_R(t)$ and $I_C(t)$:

%%FIGURE
\vspace{1.0cm}
\centerline{\epsfxsize=5.5in \epsfbox{los12_skisse_1.eps}}
\vspace{1.0cm}
%%END FIGURE

\newpage
\noindent
Notice that $I_R$ oscillates in phase with $\ems$, whereas $I_C$ has
a phase difference of $\pi/2$ with respect to $\ems$. The phase difference
between $\ems$ and the total current $I = I_R + I_C$ becomes, with the given
numerical values, somewhere in between, i.e., 45 degrees.


%\newpage
\vspace{1.5cm}
\noindent
Sketch of $\alpha$, $I_0$ and $Z$ (with $\alpha$ in radians and
$\omega RC$ between 0.016 and 160 along the horizontal axis):

%%FIGURE
\vspace{1.0cm}
\centerline{\epsfxsize=5.0in \epsfbox{los12_skisse_2.eps}}
\vspace{1.0cm}
%%END FIGURE

\noindent
Notice that the phase difference $\alpha$ between the applied voltage $\ems$
and the total current $I$ is approximately zero for small frequencies. In that
case, most of the current goes through the resistor ($I \simeq I_R$), and the circuit
behaves roughly as it would have done without the capacitor present. For high
frequencies, $\alpha$ is approximately -90 degrees. In this case, most of the current
goes in the ''branch'' with the capacitor ($I \simeq I_C$), and the circuit behaves
roughly as it would have done without the resistor present.

\noindent
Hence, for low frequencies, the current amplitude $I_0$ is appoximately constant,
i.e., independent of the frequency, and for high frequencies, $I_0$ increases
linearly with the frequency.

\noindent
We see that the transition between ''low'' and ''high'' frequencies happens
where $\omega \simeq 1/RC = 1/\tau$, where $\tau$ is the time constant for
such an $RC$ circuit.



\newpage
%\vspace{1.0cm}
\noindent
{\em Exercise 5}

\vspace{0.2cm}
\noindent
1. The full region between $r=a$ and $r=b$ may be viewed as many resistances
$dR$ connected in series, where each resistor is a thin spherical shell with
radius $r$ and thickness $dr$:
$$dR = \frac{\rho\;dr}{4\pi r^2}$$
The total resistance is found by summing up all these individual resistances, i.e.,
by integrating from $r=a$ to $r=b$:
\begin{eqnarray}
R &=& \int dR \nonumber \\
&=& \int_a^b \frac{\rho\;dr}{4\pi r^2} \nonumber \\
&=& \frac{\rho}{4\pi} |_a^b \left(-\frac{1}{r}\right) \nonumber \\
&=& \frac{\rho}{4\pi}\left(\frac{1}{a} - \frac{1}{b} \right) \nonumber
\end{eqnarray}


\vspace{0.2cm}
\noindent
2. The given expression for the current $I$ shows that we may here use
Gauss' law for the electric field to determine $I$:
$$I = \frac{1}{\rho} \cdot \frac{Q}{\varepsilon_0}$$
We must assume that the charge entering into the inner conductor immediately
distributes itself over the spherical surface at $r=a$ before is starts on its
way through the material between $r=a$ and $r=b$.

\noindent
The potential difference between the inner and outer conducting shell is easily
determined, since we know the electric field $\bold{E}$:
\begin{eqnarray}
\Delta V &=& V_a - V_b \nonumber \\
&=& - \int_b^a E(r)\; dr \nonumber \\
&=& \frac{Q}{4\pi\varepsilon_0} |_b^a \frac{1}{r} \nonumber \\
&=& \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) \nonumber
\end{eqnarray}
From these expressions, it follows that the resistance is
$$R = \frac{\Delta V}{I} = \frac{\rho}{4\pi}
\left( \frac{1}{a} - \frac{1}{b} \right)$$


\end{document}
