Electric Dipole Radiation

$V(r,\theta ,t)=-\frac{p_0\omega }{4\pi {\epsilon }_{0}c}\left(\frac{\cos \theta }{r}\right)\sin [\omega (t-r/c)]$

• Voltage due to a dipole along the $z$ axis a distance $r$ from the center of the dipole and angle $\theta$ that is between the plane intersecting the center of and perpendicular to the dipole at a time $t$
• $p_0=q_{0}d$, where $d$ is the distance between the charge ends
• $\omega$ is the oscillation frequency of the charge $q_0$

$\vec{A}(r,\theta ,t)=-\frac{{\mu }_0{p}_0\omega }{4\pi r}\sin [\omega (t-r/c)]\hat{z}$

$\vec{E}=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi }\left(\frac{\sin \theta }{r}\right)\cos [\omega (t-r/c)]\hat{\theta }$

$\vec{B}=-\frac{{\mu }_0{p}_0{\omega }^2}{4\pi c}\left(\frac{\sin \theta }{r}\right)\cos [\omega (t-r/c)]\hat{\varphi }$

$⟨\vec{S}⟩=\left(\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\right)\frac{{\sin }^2\theta }{r^2}\hat{r}$

$⟨P⟩=\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c}$

• This contributes to Why The Sky is Blue