## Show Solution

## The Electrostatic Potential

Let us write down the **Green's function** for the given system. Of course we know that for a single charge at \[\mathbf{r}^{\prime} = \left(\begin{array}{c}x^{\prime}\\y^{\prime}\\z^{\prime}\end{array}\right)\ ,\]the **mirror charge** would have to be placed at \[\mathbf{r}_{m}^{\prime} = \left(\begin{array}{c}x^{\prime}\\y^{\prime}\\{\color{red}-}z^{\prime}\end{array}\right)\ .\] Then, the Green's function is given by \[G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \frac{1}{4\pi\varepsilon_{0}}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|}\right)\ .\]To find the **electrostatic potential** for the dipole, we have to use the charge distribution of it:\[\rho_{D}\left(\mathbf{r}\right) = -\mathbf{p}\cdot\nabla\delta\left(\mathbf{r}-\mathbf{r}_{D}\right)\ .\]We obtained this formula in "The Electric Field of a Dipole" , if you want to check the expression. So, we use the solution of the **Poisson equation** to formally find\[\begin{eqnarray*} \phi\left(\mathbf{r}\right)&=&\int G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\rho\left(\mathbf{r}^{\prime}\right)dV^{\prime}\\&=&-\mathbf{p}\cdot\int G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\nabla_{\mathbf{r}^{\prime}}\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime}\\&=&\mathbf{p}\cdot\int\left[\nabla_{\mathbf{r}^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\right]\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime} \end{eqnarray*}\]by partial integration. We now have to determine the **gradient** of the Green's function with respect to its second variable! First for the \(x\) component:\[\begin{eqnarray*} \frac{\partial}{\partial x^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)&=&\frac{1}{4\pi\varepsilon_{0}}\frac{\partial}{\partial x^{\prime}}\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|}\right)\\&=&\frac{1}{4\pi\varepsilon_{0}}\left(\frac{x-x^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}-\frac{x-x^{\prime}}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|^{3}}\right) \end{eqnarray*}\]and likewise for the \(y\) component. Note that we have used \(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|=\sqrt{\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}+\left(z-z^{\prime}\right)^{2}}\) and that we have **differentiated** with respect to the **second variable** of the Green's function! Now for the \(z\) component:\[\frac{\partial}{\partial z^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right) = \frac{1}{4\pi\varepsilon_{0}}\left(\frac{z-z^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}}{\color{red}+}\frac{z-z^{\prime}}{\left|\mathbf{r}-\mathbf{r}_{m}^{\prime}\right|^{3}}\right)\ .\]We can already see the **difference** between both components of the Green's function's gradient: a **sign change** (denoted red). Now we put these components of the gradient into the solution of the Poisson equation and perform the integration:\[\begin{eqnarray*} \phi\left(\mathbf{r}\right)&=&\mathbf{p}\cdot\int\left[\nabla_{\mathbf{r}^{\prime}}G\left(\mathbf{r},\mathbf{r}^{\prime}\right)\right]\delta\left(\mathbf{r}^{\prime}-\mathbf{r}_{D}\right)dV^{\prime}\\&=&\frac{1}{4\pi\varepsilon_{0}}\mathbf{p}\cdot\left(\begin{array}{c}\left(x-x_{D}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}-\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\\\left(y-y_{D}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}-\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\\\left(z-z_{D}^{\prime}\right)\left[\left|\mathbf{r}-\mathbf{r}_{D}\right|^{-3}+\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{-3}\right]\end{array}\right)\ \text{with}\\\mathbf{r}_{mD}&=&\left(\begin{array}{c}x_{D}\\y_{D}\\{\color{red}-}z_{D}\end{array}\right)\ . \end{eqnarray*}\]This is the **solution** of the electrostatic potential of a dipole in front of a flat metallic surface. We can directly see that it corresponds to a two-dipole-solution: a mirror-dipole appears at \(\mathbf{r}_{mD}\) (see figure). Now let us try to **understand the potential** a little further

## Taylor Expansion of the Potential

From a heuristic point of view it is clear that the dipole in \(z\) direction causes a mirror dipole pointing in the same direction. The \(x\)-polarization, parallel to the surface, corresponds to dipoles with opposite direction. Hence, the \(z\)-polarization should correspond to an overall dipolar field and the \(x\) polarization to a quadrupolar one. This is what we should find out.

Far away from the sources for \(\left|\mathbf{r}_{D}\right|\ll\left|\mathbf{r}\right|\), we may **Taylor-expand** the denominators in the potential. The terms look rather similar, so we just have to calculate:\[\frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}} \approx \frac{1}{\left|\mathbf{r}\right|^{3}}+\frac{3\mathbf{r}\cdot\mathbf{r}_{D}}{\left|\mathbf{r}\right|^{5}}\] and analog for \(\mathbf{r}_{mD}\), where the \(z\) component switches the sign. This is the **important difference** and we find\[\begin{eqnarray*} \frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}}-\frac{1}{\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{3}}&\approx&\frac{6z\, z_{D}}{\left|\mathbf{r}\right|^{5}}\ \text{and}\\\frac{1}{\left|\mathbf{r}-\mathbf{r}_{D}\right|^{3}}+\frac{1}{\left|\mathbf{r}-\mathbf{r}_{mD}\right|^{3}}&\approx&\frac{2}{\left|\mathbf{r}\right|^{3}}+\frac{6\left(x\, x_{D}+y\, y_{D}\right)}{\left|\mathbf{r}\right|^{5}}\ .\end{eqnarray*}\]In this far-field approximation, the potential reads as\[\phi\left(\mathbf{r}\right) \approx \frac{1}{4\pi\varepsilon_{0}}\mathbf{p}\cdot\left(\begin{array}{c}6x\, z\, z_{D}/\left|\mathbf{r}\right|^{5}\\6y\, z\, z_{D}/\left|\mathbf{r}\right|^{5}\\2z/\left|\mathbf{r}\right|^{3}+6z\left(x\, x_{D}+y\, y_{D}\right)/\left|\mathbf{r}\right|^{5}\end{array}\right)\] and we have shown that the potential has a stronger fall-off behaviour for the \(x\) and \(y\) polarizations which goes effectively like \(\propto Q_{ij}x^{i}x^{j}/r^{5}\) whereas the \(z\) polarization behaves as \(\propto p_{z}z/r^{3}\). These potentials correspond to the quadrupolar and dipolar fields. We will find out later that oscillating dipoles radiate much faster than quadrupoles. This is the reason why a dipole polarized parallel to the surface will basically heat the metal whereas the perpendicula polarization has a chance to escape the system. The polarization has thus a **dramatic effect** on the effectivity of emitter/metal-plate-systems.

Check again the multipole moment section for more information on multipole moments.