Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 35C / Solution

For a localized charge distribution and a field point $r$ outside it, expand the Coulomb kernel:Integrating against the charge density gives the electrostatic multipole expansionwhere $Q=\int \rho d^{3}x$, $p_i=\int \rho x_i{d}^{3}x$ is the electric dipole moment, and the symmetric trace-free electric quadrupole moment is $Q_{ij}=\int \rho (3x_i{x}_j-|x{|}^2{\delta }_{ij})d^{3}x$. This states the tensor normalization explicitly.

For the tetrahedral point charges, the total charge is zero and the three other vertex vectors sum to minus $(1,1,1)$. HenceEvery vertex has $|x{|}^2=3$ and each $x_i^2=1$, so each diagonal quadrupole entry vanishes. For each distinct coordinate pair the charge-weighted sum of $x_i{x}_j$ is $4q$, yieldingThe tensor is trace-free as required. If the alternative raw second-moment convention is used, its off-diagonal entries are $4q$ and the kernel contraction has the corresponding factor of three; the physical potential is unchanged.