Dipoles

An electric dipole simply consists of two opposite charges with the same absolute value $q$. The direction is defined from the negative to the positive charge. The superposition of both potentials at a given point can then simply be calculated as $$\phi_\mathrm{D}(\vec{R}) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{|\vec{R}-\frac{\vec{d}}{2}|}-\frac{q}{|\vec{R}+\frac{\vec{d}}{2}|}\right)$$ Although this formula could already be computed numerically, we want to take a look now at large distance (R >> d). In this case, the following Taylor series can be applied: $$\frac{1}{|\vec{R}\pm\frac{\vec{d}}{2}|} = \frac{1}{R}\left(1\mp \frac{1}{2}\frac{\vec{R}\cdot\vec{d}}{R^2}+\dots\right)$$ Inserting this equation into the formula for $\phi_\mathrm{D}$ leads to the following formula: $$\boxed{\phi_\mathrm{D} = \frac{q}{4\pi\varepsilon_0}\frac{\vec{d}\cdot\vec{R}}{R^3}}$$ Electric dipoles can be characterized with their dipole moment $\vec{p}$ which is defined as $$\boxed{\vec{p} = q \vec{d}}$$ With the help of this definition the formula above can be written as $$\phi_\mathrm{D} = \frac{\vec{p}\cdot\vec{R}}{4\pi\varepsilon_0 R^3}$$ In order to obtain the electric field from this equation, the gradient will be calculated according to $$\vec{\nabla}\phi_\mathrm{D} = \frac{q}{4\pi\varepsilon_0}\left[(\vec{d}\cdot\vec{R})\vec\nabla\frac{1}{R^3} + \frac{1}{R^3}\vec\nabla(\vec{d}\cdot\vec{R})\right]$$ This leads to the following formula for the electric field of an electric dipole $$\boxed{\vec{E}_D(\vec{R}) = \frac{1}{4\pi\varepsilon_0R^3}(3p\hat{\vec{R}}\cos\theta - \vec{p})}$$ where $\hat{\vec{R}}$ is the unit vector of $\vec{R}$. Because the field does not depend on the azimuthal angle $\varphi$, it is useful to use cylindrical coordinates instead. In this case, $\vec{E}$ can be written as $$\vec{E}(\vec{R}) = -\begin{pmatrix}\frac{\partial \phi_D}{\partial R}\\ \frac{1}{R}\frac{\partial \phi_D}{\partial \theta}\\ \frac{1}{R\sin\theta}\frac{\partial \phi_D}{\partial \varphi}\end{pmatrix}$$ Inserting the equation for $\phi_\mathrm{D}$ finally leads to the three components of the electric field in radial, polar, and azimuthal direction. $$E_R = \frac{2p\cos\theta}{4\pi\varepsilon_0R^3}, \quad E_\theta = \frac{p\sin\theta}{4\pi\varepsilon_0 R^3}, \quad E_\varphi = 0$$