Can one force the electric quadrupole moments of a neutral charge distribution to vanish using a suitable translation? For a system of electric charges $q_i$, at positions $\mathbf{r}_i$, with a nonzero net charge $Q=\sum_i q_i$, one can define a "centre of charge" in the obvious way as
$$
\mathbf{r}_c=\frac{1}{Q}\sum_i q_i\mathbf{r}_i.
$$
This concept is definitely not as useful as one might naively hope, but it does have a physical significance as the position of the origin that sets the system's dipole moment (and therefore the dipolar term in a multipolar expansion) to zero. That means, then, that the monopole approximation is far better in the far field than normally: the electrostatic potential goes as
$$
\Phi(\mathbf{r})=\frac{Q}{|\mathbf{r}-\mathbf{r}_c|}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_c|^{3}}\right)
$$
where the subleading term is of order $1/r^3$, instead of $1/r^2$ as usual.

For a neutral system with $Q=0$, however, the concept of centre of charge is meaningless and there is no monopole term in the expansion. The relevant concept is then the dipole moment, 
$$
\mathbf{d}=\sum_i q_i(\mathbf{r}_i-\mathbf{r}_0),
$$
which is independent of the position $\mathbf{r}_0$ of the origin. However, this means that the relative importance of the subleading term in the multipole expansion is higher than above:
$$
\Phi(\mathbf{r})=\frac{\mathbf{d}\cdot\mathbf{r}}{|\mathbf{r}-\mathbf{r}_0|^2}+O\left(\frac{1}{|\mathbf{r}-\mathbf{r}_0|^{3}}\right)
$$
My question is: for a neutral system, is it possible to find a suitable position $\mathbf{r}_0$ for the origin that will set the subleading, quadrupole term to zero? Since this entails a system of five equations (linear in $\mathbf{r}_0$ when $Q=0$), I suspect this is impossible in general geometries. If this is the case, which geometries allow for vanishing quadrupole moments and which ones don't? In the cases where one can do this, does this position have a special name? More generally, if all multipole moments up to some $l\geq0$ are zero, (when) can the subleading term be made to vanish?
 A: I think it's easier to see in Cartesian coordinates.  Define the "primitive" moments
\begin{align}
q & = \int \rho(\mathbf r)\, \text{d}^3\mathbf{r}\\
p_i & = \int r_i \rho(\mathbf r)\, \text{d}^3 \mathbf{r}\\
Q_{ij} & = \int r_i r_j \rho(\mathbf r)\, \text{d}^3 \mathbf{r}
\end{align}
Assume $q = 0$, $p_i$ not all 0.
If you displace the origin by $\mathbf{d}$ and call the new quadrupole moments $Q'_{ij}$, then
\begin{align}
Q'_{ij} & = \int (r_i - d_i)(r_j - d_j) \rho \text{d}^3 \mathbf{r}\\
        &= Q_{ij} - p_i d_j - p_j d_i + q d_i d_j.
\end{align}
If q = 0, finding a translation that makes $Q'_{ij} = 0$ is now a (overspecified) linear problem of six equations in three unknowns.  To make it even simpler, assume the dipole is aligned along the z-axis ($p_1 = p_2 = 0$).  Then it is easy to see that no translation will change $Q_{11}, Q_{12}, \text{ or } Q_{22}$, so if any of them are nonzero then translating the distribution will not make them zero.  If they are all zero, then the quadrupole moment can be entirely zeroed out by setting $d_1 = Q_{13}/p_3$, $d_2 = Q_{23}/p_3$, and $d_3 = Q_{33}/p_3$.
