# Homework Help: Total power radiated from two dipoles (electrodynamics)

1. May 18, 2015

### CAF123

1. The problem statement, all variables and given/known data
The retarded vector potential $\mathbf{A}(\mathbf{r}, t)$ in Lorenz gauge due to a current density $\mathbf{J}(\mathbf{r}, t)$ contained entirely within a bounded region of size d is $$\mathbf{A}(\mathbf{r},t) = \frac{1}{4\pi c}\int_V' \frac{\mathbf{J}(\mathbf r', t')}{|r-r'|} dV',$$ where $t' = t-|r-r'|/c$ If this current density is due to a charge density oscillating harmonically with frequency $\omega \ll c/d,$ then at distances $r \gg d,$ $\mathbf{A}(\mathbf{r}, t)$ the current density is proportional to the real part of $i\omega \mathbf{p}e^{i\omega(t−r/c)}$ , where $\mathbf p \cos \omega t$ is the dipole moment of the charge distribution.

Now consider two oscillating electric dipoles of equal moment $\mathbf p$ that are placed next to each other on the z axis, a small distance d apart. The two dipoles have the same angular frequency $\omega$, but are out of phase with each other by $\pi$. Show that the total power radiated by this system is smaller than that of the single dipole by a factor $\omega^2 d^2/5c^2$

2. Relevant equations
E and B field formulae and the Poynting vector. Total power radiated given by $\langle P \rangle = \int \langle S \rangle \cdot d \mathbf A$

3. The attempt at a solution

I have solved the time averaged power for the single dipole by computing the far field E and B from the given A and then computing the poynting vector before the total time averaged power. Since this second dipole is out of phase by pi relative to the other, then can write $\mathbf p_1(t) = \mathbf p \cos \omega t$ for the first one and $\mathbf p_2(t) = -\mathbf p \cos \omega t$ for the second. I am just a bit unsure of what to do next. I guess I am to superimpose the E and B fields from the two dipoles and then compute P but how does the superposition work? Thanks!

2. May 23, 2015