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\begin{center}
  {\bf University of Kentucky, Physics 416G \\
    Problem Set \#9 (Rev. A), due Monday, 2011-11-21 }
\end{center}

\vspace{.3 in}

1.~~A spherical capacitor of radius $R$ is made of two hemispherical
conductors separated by a small gap.  The top hemisphere is held at potential
$+V_0$ and the bottom hemisphere at $-V_0$.  Evaluate the first two non-zero
terms in each case.

\quad a) Calculate the potential everywhere inside the sphere.  Note that this
solution is independent of the potential outside the sphere, because of the
boundary condition $V=\pm V_0$ at $r=R$.

\quad b) Calculate the potential everywhere outside the sphere, which is
likewise \mbox{independent of part a).}

\quad c) Calculate the surface charge on both the inside and outside of the
sphere.

\quad d) Integrate the total charge on either hemisphere to calculate the
capacitance.

\quad e) Draw the equipotentials and field lines, demonstrating the dominant
features of the lowest non-zero multipole, both inside and outside the sphere.

\quad f) If the capacitor is filled inside with a dielectric of relative
strength $\epsilon_r$, how are the field and surface charge modified inside?
On the outside?  Calculate the capacitance with the dielectric.

\quad g) [bonus] Sum the terms of all orders to derive the exact formula for
the capacitance.
                               
\hspace{1cm}

2.~~The {\bf Clausius-Mossotti} relation.  Our purpose is to derive the
relation between atomic polarizability $\alpha$ (defined by $\bs p = \alpha
\bs E$ in a linear dielectric) and electric susceptibility $\chi_e$ (defined
by $\bs P = \epsilon_0\chi_e \bs E$ in the same medium).

\quad a) The simplest relation can be derived by ignoring the contribution
from individual dipoles to $\bs E$.  Using the definition of $\bs P$, explain
why $\bs P = N \bs p$, where $N$ is the number density of molecules
(number/volume).  Show that in this case $N \alpha = \epsilon_0\chi_e$.

\quad b) Now correct for the fact a dipole does not feel it's own field.  Thus
the definition of polarizability in a dielectric material must be modified to
$\bs p = \alpha \bs E_0$, where $\bs E_0$ is the macroscopic field due to
everything but $\bs p$.  Calculate $\bs E_0$ at the location of $\bs p$ as a
function of the macroscopic field $\bs E$ and susceptibility $\chi_e$ assuming
that $\bs p$ is at the center of a bubble of radius $R$ (the size of the
molecule) inside the dielectric material.  Hint: see example 4.2 and problem
4.16.

\quad c) Using the modified equation from part a), $N \alpha \bs E_0 =
\epsilon_0\chi_e \bs E$, and part b), derive the relation $N \alpha = \epsilon_0
\chi_e \frac{3}{3+\chi_e} = 3\epsilon_0 \frac{\epsilon_r-1}{\epsilon_r+2}$.
Show that this reduces to the na\"ive relation in a) as $\chi_e\to0$.

\quad d) Use part a) to compare the $\alpha$ and $\chi_e$ for elements which
appear in both tables 4.1 and 4.2.  Calculate the fractional correction term
$\frac{3}{3+\chi_e}$ derived in part c).

\newpage

3. The {\bf Langevin} formula.  In this problem we calculate the electric
susceptibility $\chi_e$ of a polar material, composed of molecules with
permanent dipole moment $p$.  While the dipole has a tendency to align along
the total field $\bs E$, thermal agitations prevent perfect alignment.  Ludwig
Boltzman showed that the number of atoms occupying a state $i$ of energy $U_i$
is proportional to the Boltzman factor $w=e^{-U_i/kT}$, where $k$ is the
Boltzman constant and $T$ the temperature of the system.  Thus colder systems
are more likely to settle into the lowest energy state because the probability
of being in an excited state is exponentially small.  In our case, the state
of a dipole in an electric field is given by its rotational direction $i
\equiv (\theta,\phi)$ with respect to the field $\bs E=E\bh z$.

a) Calculate the potential energy $U(\theta)$ of the dipole $\bs p$ in the
field $\bs E$.

b) By considering the solid angle of a ring of constant energy (ie. constant
$\theta$), show that the number of states $dn$ of energy between $U(\theta)$
and $U(\theta+d\theta)$ is proportional to $\sin\theta d\theta$.  The ratio
$g=\frac{dn}{d\theta}$ is called the {\it degeneracy of states}, the {\it
  density of states}, or the {\it phase space}.

c) Calculate the weighted average energy $\langle\, U\rangle \equiv \int dn\,
w(\theta) U(\theta) / \int dn\, w(\theta)$ of a dipole $p$ in the field $E$,
weighted by the Boltzman factor (distribution).

d) Calculate the weighted average $\langle p_z\rangle$ of $p_z=p\cos\theta$ as
a function of the electric field $E$ and temperature $T$, also weighted by the
Boltzman distribution.  Eplain why $P=N\bs \langle p_z\rangle$ and use it to
show that $P=Np[\coth(pE/kT)-(kT/pE)]$.  Graph this as a function of $E$.

e) In the limit where $pE \ll kT$, determine the linear relationship between
$E$ and $P$ to determine the susceptibility $\chi_e$.  Calculate $\chi_e$ for
liquid and vapor water, using $p=3.8\times 10^{-9}~e\,$cm.

f) [bonus] As in \#2, correct for the fact a dipole does not feel it's own
field.  How much does $\chi_e$ improve for liquid water?

\vspace{1cm}
Also Griffiths chapter 4, problems 
1, 
4, 
5, 
6, 
7, 
8, 
9, 
10, 
12, 
13, 
and 14.

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