Pre-Quantum Electrodynamics

Put \({\bf d}\) along \(\hat{\bf z}\). Then, (\ref{eq:electric_dipole}) becomes \[ V_{\mbox{\tiny di}}({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{p \cos \theta}{r^2} \label{Gr(3.102)} \] Taking the gradient, \[ E_r = -\frac{\partial V}{\partial r} = \frac{1}{4\pi \varepsilon_0} \frac{2p\cos \theta}{r^3}, \hspace{1cm} E_\theta = -\frac{1}{r} \frac{\partial V}{\partial \theta} = \frac{1}{4\pi \varepsilon_0} \frac{p \sin \theta}{r^3}, \hspace{1cm} E_\phi = -\frac{1}{r \sin \theta} \frac{\partial V}{\partial \phi} = 0, \] we get \[ {\bf E}_{\mbox{\tiny di}} (r, \theta) = \frac{1}{4\pi \varepsilon_0} \frac{p}{r^3} (2\cos \theta \hat{\bf r} + \sin \theta \hat{\bf \theta}) \label{Gr(3.103)} \] or in a better coordinate-free form \[ {\bf E}_{\mbox{\tiny di}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{r^3} \left[3 ({\bf p} \cdot \hat{\bf r}) \hat{\bf r} - {\bf p}\right] \label{eq:dipole_field} \]

Dipole energy