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The curl free nature of the electric field in electrostatics implies existence of a potential,\(\phi(\vec(r))\), from which the electric field can be derived as \(\vec{E}=-\nabla \phi\). The potential at a point is just the work done in moving a unit point charge from infinity to its current position.

A Carnot engine is made to operate as a refrigerator, operating at $0^o$ C and discharging heat at $20^o$C. Find the minimum amount of work done by the refrigerator in converting one kilogram of water at $0^o$ into ice. ( Assume latent heat to be $L\,=\, 3.35\times 10^5J/Kg.$) 

In this lecture we continue our discussion of canonical partition function. An expression for internal energy is obtained in terms of the partition function. The parameter β is identified with 1/kT by noting that the parameter β in the partition function depends only on the heat bath and not on the system in equilibrium with the heat bath. The partition function is used to compute the energy of an ideal gas and comparing the same with the expression given by the equipartition theorem. The expression for other thermodynamic functions, entropy, free energy, enthalpy and variance of energy are derived in terms of the canonical partition function. Comparing the average energy with its variance gives a criterion so that the average energy may approximately represent the state of the system.

In this section examples are given to show that the flux rule is not always applicable.

Consider a 2 dimensional phase space ( $q,p$) with a rectangular region defined by four corners as shown.

If the region ABCD is the phase space region at time time t = 0 , find the region $A'B'C'D'$ at time t given the Hamiltonian is
$$ H\,=\,\frac{p^2}{2m}\,-\, m a q $$
and explicitly verify that the area is constant. Take the coordinates of A,B,C and D as $(q_A,p_A)\,,\,(q_B,p_A)\,,\,(q_B,p_C)$ and $(q_A,p_C)$ respectively

 Electrodynamics                                               March 11, 2019

                                      Quiz-IV

In case of a charge distribution having symmetry about \(z\)-axis,

\(Q_{xy}=Q_{yz}=Q_{zx}=0\) and \(Q_{yy}=Q_{xx}\) Also trace
\(Q_{xx}+Q_{yy}+Q_{zz}\) is always zero. Thus the quadrupole moment tensor can
be specified by a single number
\(Q= 2(Q_{zz}-Q_{xx})\). Calculate \(Q\) for six equal charges placed on corners
of a regular hexagon of sides \(a\).

The definition of quadrupole moment tensor components is
\begin{eqnarray}
Q_{xx} &=&\frac{1}{2} \sum_\alpha Q_\alpha (3x^2_\alpha - r^2_\alpha)\\
Q_{yy} &=& \frac{1}{2} \sum_\alpha Q_\alpha (3y^2_\alpha - r^2_\alpha)\\
Q_{zz}&=& \frac{1}{2} \sum_\alpha Q_\alpha (3z^2_\alpha - r^2_\alpha)
\end{eqnarray}
In this case \(Q_{xz}=Q_{yz}=0\) because all charges are in \(XY\) plane
\((z=0)\). Also \(Q_{xy}=0\) due to symmetry reflection symmetry in the \(X,
Y\) axes.
For all the charges \(z=0, r=a\). Hence
\begin{equation*}
\sum_{\alpha=1}^6 z^2=0 \qquad \sum_{\alpha=1}^6 r^2= 6a^2
\end{equation*}
For four charges \(x^2= \frac{3a^2}{4}\) and for two charges \(x=0\). Therefore
\begin{equation*}
\sum_{\alpha=1}^6 3 x^2 = 9a^2
\end{equation*}
Thus we get
\begin{eqnarray}
Q_{xx} &=& \frac{1}{2} \sum_\alpha Q_\alpha (3x^2_\alpha - r^2_\alpha)=
Q (9a^2 - 6a^2) = 3Qa^2 \\
Q_{zz} &=& \frac{1}{2} \sum_\alpha Q_\alpha (3z^2_\alpha - r^2_\alpha)= - 6Qa^2
\end{eqnarray}
and the final answer is
\[Q= 2(-6Qa^2- 3Qa^2) = -18 Qa^2.\]

The Maxwell's equations for electrostatic are derived from Coulomb's law which has been formulated based on experiments. This provides initial experimental evidence for the Maxwell's equations. We discuss two applications of Maxwell's equations. The first result is that  the electric field inside an empty cavity in conductors is proved to be zero. The second result is  an  expression for electric stress tensor is derived. The surface integral of the electric tensor gives the force on charge distribution.