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CAF123

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## Homework Statement

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##

## Homework 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##

## The Attempt at a Solution

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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!