[2019EM/QUIZ-04#Solu]

 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
\[Q= 2(-6Qa^2- 3Qa^2) = -18 Qa^2.\]