Electromagnetic Dipole

This example can be computed by using the following source term for the purely hyperbolic Maxwell equations:
<br /> \mathbf{g}= -\cfrac{1}{\epsilon_0} (0,\, 0,\, Q\omega d\cos(\omega t),\, 0, \,0,\, 0,\,0,\,0)^T.<br />

An analytic solution for this example is given by
<br /> E_r = \cfrac{2\mathrm{cos}(\phi)Qd}{4\pi\epsilon_0} \left[\cfrac{1}{r^3}\mathrm{sin}\left(\omega t-\cfrac{\omega r}{c}\right)+\frac{\omega}{cr^2}\mathrm{cos}\left(\omega t-\cfrac{\omega r}{c}\right)\right], \\</p> <p>E_{\phi} = \cfrac{\mathrm{sin}(\phi)Qd}{4\pi\epsilon_0} \left[\left(\cfrac{1}{r^3}-\cfrac{\omega^2}{c^2r}\right)\mathrm{sin}\left(\omega t-\cfrac{\omega r}{c}\right)+\cfrac{\omega}{cr^2}\mathrm{cos}\left(\omega t-\cfrac{\omega r}{c}\right)\right], \\</p> <p>B_{\theta} = \mu_0 \omega\cfrac{\mathrm{sin}(\phi)Qd}{4\pi} \left[-\cfrac{\omega}{cr}\mathrm{sin}\left(\omega t-\cfrac{\omega r}{c}\right)+\cfrac{1}{r^2}\mathrm{cos}\left(\omega t-\cfrac{\omega r}{c}\right)\right].</p> <p>