diff --git a/_psets/2.md b/_psets/2.md
index fb35dc55be9b2122aac2e293bfa551b5ad324ae2..16b433cab4b1705b9a12ba5b8e2114f05f0d99ad 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