Answer 7.3

_psets/5.md

@@ -41,6 +41,33 @@ $$
 Integrate Poynting’s vector $$P = E \times H$$ to find the power flowing across a cross-sectional
 slice of a coaxial cable, and relate the answer to the current and voltage in the cable.
 
+We saw that the magnetic field has a magnitude of $$I/(2 \pi r)$$ and wraps around the inner
+conductor according to the right hand rule. The electric field has magnitude $$Q/(2 \pi \epsilon
+r)$$ and points radially from the inner conductor toward the outer. So $$P = E \times H$$ points
+along the direction of current in the inner conductor, and has magnitude
+
+$$
+| P | = \frac{I Q}{(2 \pi r)^2 \epsilon}
+$$
+
+Integrating over the area between the conductors, we find that the cross-sectional power transfer is
+
+$$
+\begin{align*}
+\int_0^{2 \pi} \int_{r_i}^{r_o} \frac{I Q}{(2 \pi r)^2 \epsilon} r \mathrm{d} r \mathrm{d} \theta
+&= \frac{I Q}{2 \pi \epsilon} \int_{r_i}^{r_o} r^{-1} \mathrm{d} r \mathrm{d} \theta \\
+&= \frac{I Q}{2 \pi \epsilon} \ln \left( \frac{r_o}{r_i} \right)
+\end{align*}
+$$
+
+But the charge is
+
+$$
+Q = CV = \frac{2 \pi \epsilon}{\ln (r_0 / r_i)}
+$$
+
+so we can write the total power as $$IV$$.
+
 ## (7.4)
 
 Find the characteristic impedance and signal velocity for a transmission line consisting of two