Answer 7.4

_psets/5.md

@@ -78,8 +78,41 @@ so we can write the total power as $$IV$$.
 
 {:.question}
 Find the characteristic impedance and signal velocity for a transmission line consisting of two
-parallel strips with a width $$w$$ and a separation $$h$$ (Figure 7.4). You can ignore fringing
-fields by assuming that they are sections of conductors infinitely wide.
+parallel strips with a width $$w$$ and a separation $$h$$. You can ignore fringing fields by
+assuming that they are sections of conductors infinitely wide.
+
+Take the plates to have length one. Suppose the bottom plate has a charge $$Q$$, and the top plate a
+charge of $$-Q$$. Then as we found last week the field between the plates will have magnitude
+$$Q/(\epsilon w)$$ (since the charge density is the total charge divided by area $$1 \cdot w$$), and
+point from the bottom plate toward the top plate. Integrating from the bottom plate to the top one,
+we find that the voltage difference is $$Qh/(\epsilon w)$$. Thus the capacitance (charge per
+voltage) per length is
+
+$$
+C = \frac{\epsilon w}{h}
+$$
+
+By symmetry the magnetic field must point perpendicular to the direction of current travel and
+parallel to the surfaces of the plates. Consider a square perpendicular to the current, with a width
+of $$w$$, bisected by the lower plate. Ampère's Law tells us that the integral of the
+magnetic field along the boundary of this square is $$I$$. So we deduce that the magnetic field has
+magnitude $$I/w$$. Integrating over the surface perpendicular to the field between the plates, we
+find that the flux is $$\mu_0 I h / w$$. Thus the inductance (flux per current) per length is
+
+$$
+L = \frac{\mu_0 h}{w}
+$$
+
+From here we can use the results derived in the chapter.
+
+$$
+v = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{\mu_0 \epsilon}}
+$$
+
+$$
+Z = \sqrt{\frac{L}{C}} = \frac{\mu_0 h^2}{\epsilon w^2}
+$$
+
 
 ## (7.5)