# Multipole Moments of the Electric Field

The multipole moments of a charge distribution $$\rho\left(\mathbf{r}\right)$$ arise naturally if we try to solve Poisson's equation for the electrostatic potential, $$\Delta\phi\left(\mathbf{r}\right)= -\rho\left(\mathbf{r}\right)/\varepsilon_{0}$$:
The Green's function of the Laplace operator obeying $$\Delta_{\mathbf{r}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$$, is given by $$G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = -1/4\pi\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$$, which provides us the well-known

$\phi\left(\mathbf{r}\right)=\frac{1}{4\pi\varepsilon_{0}}\int\frac{\rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV\ .$

The Green's function can now be expanded in a Taylor series around $$\mathbf{r}^{\prime}=0$$ yielding (see also in the solution of "The Electric Field of a Dipole")

$\begin{eqnarray*}\phi\left(\mathbf{r}\right) & = & \frac{1}{4\pi\varepsilon_{0}}\left\{\int\frac{\rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}\right|}dV^{\prime}+\int\frac{\mathbf{r}\cdot\mathbf{r}^{ \prime}\rho\left(\mathbf{r}^{\prime}\right)} {\left|\mathbf{r}\right|^{3}}dV^{\prime}+\dots\right\}\\&=& \frac{1}{4\pi\varepsilon_{0}}\left\{\frac{Q}{r}+\frac{\mathbf{p}\cdot\mathbf{r}}{r^{3}}+\dots\right\}\end{eqnarray*}$

with $$Q=\int\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}$$ as the total charge, the dipole moment $$\mathbf{p}=\int\mathbf{r}^{\prime}\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}$$ and $$r=\left|\mathbf{r}\right|$$. You can see that the moments we already know are just the beginning of a fundamental representation of the electrostatic potential. In the following problems we shall make ourselfs familier with the multipole moments. Enjoy!