##### The Electric Field of a Dipole
ems.ca.me.Ed

To compute the electric field of a dipole, we will apply relation Emgp on p_di.

Let us first proceed in a slightly pedestrian manner, putting \({\bf d}\) along \(\hat{\bf z}\). Then, p_di becomes \[ \phi_d ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{p \cos \theta}{r^2} \] Taking the gradient (in spherical coordinates, c.f. sph_grad),

\begin{align*} E_r &= -\frac{\partial \phi}{\partial r} = \frac{1}{4\pi \varepsilon_0} \frac{2p\cos \theta}{r^3}, \nonumber \\ E_\theta &= -\frac{1}{r} \frac{\partial \phi}{\partial \theta} = \frac{1}{4\pi \varepsilon_0} \frac{p \sin \theta}{r^3}, \nonumber \\ E_\phi &= -\frac{1}{r \sin \theta} \frac{\partial \phi}{\partial \varphi} = 0, \end{align*}

we get

- Gr (3.103)

\[ {\bf E}_d (r, \theta) = \frac{1}{4\pi \varepsilon_0} \frac{p}{r^3} (2\cos \theta ~\hat{\bf r} + \sin \theta ~\hat{\bf \theta}) \tag{E_di_1}\label{E_di_1} \]

A smarter derivation directly applies the gradient on p_di, \[ E_d ({\bf r}) = - \frac{1}{4\pi \varepsilon_0} \nabla \frac{{\bf p} \cdot {\bf r}}{r^3} \] immediately yielding the coordinate-free expression

- Gr (3.104)

\[ {\bf E}_d ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{r^3} \left[3 ({\bf p} \cdot \hat{\bf r}) \hat{\bf r} - {\bf p}\right] \tag{E_di}\label{E_di} \]

Dipole energy (tbd)

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42