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! 

The Electric Field of a Dipole

How two charges become a dipole in the limit of vanishing separation.The electric field of a dipole can be seen as the result of two charges approaching each other. Learn in this problem how to use the Taylor expansion of \(1/\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\) to calculate this field. Find out how this series expansion yields multipole moments, a very powerful description of the electric field.

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Multipole Moments in Rotational Symmetry

a charge distribution with rotational symmetry - schematic.In rotational symmetry, the multipole moments of a charge distribution are described by just one component per order \(l\). This is a drastic difference to the usual \(2l+1\) independent components. Get to know this simplifications with two examples and derive it for any charge distribution with rotational symmetry!

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