 Power delivered to the second dipole

Integrating the ${\sin }^2\theta$ radiation pattern uses $\int {\sin }^2\theta d\Omega =8\pi /3$, so its lossless gain isThe receiver lies on the $x$ axis, broadside to both $z$-directed dipoles. Thus $G_t=G_r=3/2$ and their linear polarizations match. At $f=100\mathrm{MHz}$, $\lambda =c/f\simeq 3.00\mathrm{m}$, givingAt separation $r=100\mathrm{m}$, the incident flux is $P_0{G}_t/(4\pi r^2)$. The Friis transmission equation therefore gives the matched-load powerThe far-field and weak-coupling approximations are well satisfied at a distance of about $33$ wavelengths for an electrically short dipole. Matching refers to the available received power, not interception of all power in a geometric area equal to the wire's physical size.