### Pset 6, Abubakar Abid (It may take a second for the LaTeX-javascript library to load...)

8.2*) At a distance of 1 km, an isotropic antenna that is radiating 1 kW of power will have a local power density given by:
$$ P_d = P/(4 \pi r^2) = 1 \textrm{ kW} / (4 \pi * 1000 \textrm{ m})^2$$
This gives a value of 7.958*10^-5 W/m^2.
(assuming it is entirely radiative fields and not reactive fields)
The Poynting vector is the cross-product of the electric and magnetic fields, so at sufficiently far away distances, we can say that the power is given by:
$$ P = E \cross H = \sqrt{\frac{\epsilon_0}{\mu_0}} E^2 $$
This gives a value of about for 0.17 V/m for \(E\)
8.3) Let \(S\) be the source resistance and \(L\) the load resistance. The total power delivered to the load is:
$$ P = (\frac{V}{S+L})^2 L$$
Taking the derivative of this with respect to \(L\) gives
$$ \frac{dP}{dL} = \frac{V^2 (S-L)}{(S+L)}$$
Setting that to zero, we see the max occurs at
$$S = L$$
*Late