Answer 8.2

_psets/6.md

@@ -7,6 +7,9 @@ title: Problem Set 6
 {:.question}
 Find the electric field for an infinitesimal dipole radiator.
 
+So much math... check back later.
+
+
 ## (8.2)
 
 {:.question}
@@ -14,11 +17,44 @@ What is the magnitude of the Poynting vector at a distance of 1 km from an anten
 power, assuming that it is an isotropic radiator with a wavelength much less than 1 km? What is the
 peak electric field strength at that distance?
 
+Since it's an isotropic radiator, the Poynting vector must point along $$\hat{r}$$. Similarly its
+magnitude must depend only on $$r$$, so we can compute its time average value by dividing the total
+power by the total area.
+
+$$
+\begin{align*}
+\langle |P| \rangle &= \frac{10^3 \si{W}}{4 \pi \cdot 10^6 \si{m^2}} \\
+&= \num{8e-5} \si{W/m^2}
+\end{align*}
+$$
+
+Since $$\lambda << 1\si{km}$$, locally the radiation will look like a plane wave. Thus E and H are
+perpendicular, and $$|H| = \sqrt{\epsilon_0/\mu_0} |E|$$.
+
+$$
+\begin{align*}
+\langle |P| \rangle &= \langle |E| |H| \rangle \\
+&= \sqrt{\frac{\epsilon_0}{\mu_0}} \langle |E|^2 \rangle \\
+&= \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E^2_\text{max}
+\end{align*}
+$$
+
+Solving for $$E_\text{max}$$,
+
+$$
+\begin{align*}
+E_\text{max} &= \left( 2 \sqrt{\frac{\mu_0}{\epsilon_0}} \langle P \rangle \right)^\frac{1}{2} \\
+&= \sqrt{2 \cdot 377 \si{\ohm} \cdot \num{8e-5} \si{W/m^2}} \\
+&= 0.24 \si{W/m}
+\end{align*}
+$$
+
 ## (8.3)
 
 {:.question}
 For what value of $$R_\text{load}$$ is the maximum power delivered to the load in Figure 8.3?
 
+
 ## (8.4)
 
 {:.question}