From 454d31af091c30068b4655766c4fd94f9aa86012 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Wed, 20 Feb 2019 16:47:03 -0500
Subject: [PATCH] Finish writing answers to 3.3

---
 _psets/2.md | 25 +++++++++++++++++++++++++
 1 file changed, 25 insertions(+)

diff --git a/_psets/2.md b/_psets/2.md
index fb35dc5..16b433c 100644
--- a/_psets/2.md
+++ b/_psets/2.md
@@ -172,12 +172,37 @@ volts.
 {:.question}
 What size capacitor has voltage fluctuations that match the magnitude of this Johnson noise?
 
+Since a farad is equal to a second per ohm, purely dimensional considerations imply that the noise
+energy of the capacitor is approximately $$kT/C$$. Indeed this turns out to be exactly correct. The
+energy in a capacitor is $$C V^2 / 2$$, so by the equipartition theorem, $$kT/2 = \langle C V^2 / 2
+\rangle$$. This implies that $$\langle V^2 \rangle = k T / C$$. Thus the capacitor that matches the
+voltage fluctuations found in part (b) at room temperature has capacitance
+
+$$
+\begin{align*}
+C &= \frac{k T}{\left \langle V^2 \right \rangle} \\
+&= \frac{\num{1.38e-23} \si{J/K} \cdot 300 \si{K}}{\num{3.3e-12} \si{V^2}} \\
+&= \num{1.25e-9} \si{F}
+\end{align*}
+$$
+
 ### (c)
 
 {:.question}
 If it is driven by a current source, how large must it be for the RMS shot noise to be equal to 1%
 of that current?
 
+RMS shot noise is the square root of $$2 q I \Delta f$$. So we want $$\sqrt{2 q \Delta f I} = 0.01
+I$$, which is solved by
+
+$$
+\begin{align*}
+I &= \frac{2q \Delta f}{0.01^2} \\
+&= \frac{2 \cdot \num{1.6e-19} \si{C} \cdot 20 \si{kHz}}{10^{-4}} \\
+&= \num{6.4e-11} A
+\end{align*}
+$$
+
 
 ## 3.4
 
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