Electric Dipole Radiation
#Physics
$\displaystyle V(r,\theta,t)=- \frac{p_{0}\omega}{4\pi {\varepsilon}_{0}c}\left( \frac{\cos \theta}{r} \right)\sin[\omega(t-r/ c)]$
- Voltage due to a dipole along the $\displaystyle z$ axis a distance $\displaystyle r$ from the center of the dipole and angle $\displaystyle \theta$ that is between the plane intersecting the center of and perpendicular to the dipole at a time $\displaystyle t$
- $\displaystyle p_{0}=q_{0}d$, where $\displaystyle d$ is the distance between the charge ends
- $\displaystyle \omega$ is the oscillation frequency of the charge $\displaystyle q_{0}$
$\displaystyle \vec{A}(r,\theta,t)=- \frac{{\mu}{0}p{0}\omega}{4\pi r}\sin[\omega(t-r/c)]\hat{z}$
$\displaystyle \vec{E}=-\frac{{\mu}{0}p{0}\omega ^{2}}{4\pi}\left( \frac{\sin \theta}{r} \right)\cos[\omega(t-r /c)]\hat{\theta}$
$\displaystyle \vec{B}=- \frac{{\mu}{0}p{0}\omega ^{2}}{4\pi c}\left( \frac{\sin \theta}{r} \right)\cos[\omega(t-r /c)]\hat{\phi}$
$\displaystyle {\left\langle{\vec{S}}\right\rangle}=\left( \frac{{\mu}{0}p{0}^{2}\omega^{4}}{32\pi ^{2}c} \right) \frac{\sin ^{2}\theta}{r^{2}}\hat{r}$
$\displaystyle {\left\langle{P}\right\rangle}=\frac{{\mu}{0}p{0}^{2}\omega^{4}}{12\pi c}$
- This contributes to Why The Sky is Blue
