I read this derivation in Griffiths that if we obtain dipole moment vector $\vec{p}$ , then dipole term in the potential (due to general distribution far away at $\vec{r}$ ) can be written as $$\frac{\vec{p}.\hat{r}}{4\pi\epsilon_or^2}$$ I was wondering if quadrupole contribution can be written like this in terms of the quadrupole moment matrix. I mean of course, it wont be a dot product but maybe some other way.
(Couldn't find it on net!)

• You tagged this question with "multipole-expansion"; didn't you google for that? For me at least, the first link is this one, which explains that not only the monopole, dipole, and quadrupole can be written in this way, but you can keep going to arbitrarily high multipoles.
– Mike
Mar 13 '18 at 17:28

Hey the potential term you seek is the following (for gravitational potential. The functional form is exactly the same in the electric potential case.): $$\phi_{quad}(\vec r)=-G\frac{Q_{ij}\hat r_i\hat r_j}{r^3},$$ $$Q_{ij}=\int\frac{3x_ix_j-\delta_{ij}r^2}{2}_{}\rho(\vec r)d^3\vec r.$$ The quantity $\rho(\vec r)d^3\vec r$ is just the element $dm(\vec r)$ of the mass distribution generating the gravitational potential. This should correspond to the electric charge density generating the electric potential, in your case.
• +1 but I just want to point out that I've (almost?) never seen the factor of $1/2$ in the definition of $Q_{ij}$ — it's usually put in the equation before it. You may argue that this is just a matter of convention, but $Q_{ij}$ is such a common thing that conventions matter.
• Shouldn't $\phi_{quad}(\vec{r})$ have a summation? Mar 13 '18 at 18:10