From 803819b60464771d997a6aa7763a111c2e239dd7 Mon Sep 17 00:00:00 2001 From: Erik Strand <erik.strand@cba.mit.edu> Date: Wed, 13 Mar 2019 23:53:50 -0400 Subject: [PATCH] Answer 7.4 --- _psets/5.md | 37 +++++++++++++++++++++++++++++++++++-- 1 file changed, 35 insertions(+), 2 deletions(-) diff --git a/_psets/5.md b/_psets/5.md index 290c523..ec5476d 100644 --- a/_psets/5.md +++ b/_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) -- GitLab