diff --git a/_psets/6.md b/_psets/6.md
index 611e574f3d70881c0adfc9d216df46466a34df9e..e2a2ada2c723ed2136e35832286e7106fce04df5 100644
--- a/_psets/6.md
+++ b/_psets/6.md
@@ -7,7 +7,109 @@ title: Problem Set 6
 {:.question}
 Find the electric field for an infinitesimal dipole radiator.
 
-So much math... check back later.
+This problem is most naturally expressed in spherical coordinates. We know that the vector potential
+is
+
+$$
+\begin{align*}
+A_r &= \mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \cos \theta \\
+A_\theta &= -\mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \sin \theta \\
+A_\phi &= 0
+\end{align*}
+$$
+
+So the electric field is "simply"
+
+$$
+E = \frac{1}{i \omega \mu_0 \epsilon_0} \nabla ( \nabla \cdot A) - i \omega A
+$$
+
+First let's find $$\nabla \cdot A$$.
+
+$$
+\begin{align*}
+\nabla \cdot A
+&= \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 A_r)
+ + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} ( A_\theta \sin \theta )
+ + \frac{1}{r \sin \theta} \frac{\partial A_\phi}{\partial \phi} \\
+&= \frac{\mu_0 I_0 d}{4 \pi} \left(
+   \frac{\cos \theta}{r^2} \frac{\partial}{\partial r} \left( r e^{-ikr} \right)
+ - \frac{e^{-ikr}}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \sin^2 \theta \right) \\
+&= \frac{\mu_0 I_0 d}{4 \pi} \left( \frac{\cos \theta}{r^2} \left( e^{-ikr} - r i k e^{-ikr} \right)
+ - \frac{e^{-ikr}}{r^2 \sin \theta} \left( 2 \sin \theta \cos \theta \right) \right) \\
+&= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left(
+   \frac{1}{r^2} - \frac{ik}{r} - \frac{2}{r^2} \right) \\
+&= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{r^2} + \frac{ik}{r} \right)
+\end{align*}
+$$
+
+For any function $$f(r, \theta, \phi)$$ in spherical coordinates,
+
+$$
+\nabla f = \frac{\partial f}{\partial r} \hat{r}
+         + \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta}
+         + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \hat{\phi}
+$$
+
+So the three components of $$\nabla (\nabla \cdot A)$$ are
+
+$$
+\begin{align*}
+[\nabla (\nabla \cdot A)]_r
+&= \frac{\partial (\nabla \cdot A)}{\partial r} \\
+&= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta \left(
+     \left( \frac{-2}{r^3} - \frac{ik}{r^2} \right) e^{-ikr}
+     + \left( \frac{1}{r^2} + \frac{ik}{r} \right) (-ik) e^{-ikr} \right) \\
+&= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left(
+     \frac{-2}{r^3} - \frac{ik}{r^2} - \frac{ik}{r^2} + \frac{k^2}{r} \right) \\
+&= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left(
+   \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) \\
+
+[\nabla (\nabla \cdot A)]_\theta
+&= \frac{1}{r} \frac{\partial (\nabla \cdot A)}{\partial \theta} \\
+&= - \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right)
+     \frac{\partial}{\partial \theta} \cos \theta\\
+&= \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) \sin \theta \\
+
+[\nabla (\nabla \cdot A)]_\phi &= 0
+\end{align*}
+$$
+
+Putting these together (and recalling that $$k = \omega \sqrt{\mu_0 \epsilon_0}$$) we find that
+
+
+$$
+\begin{align*}
+E_r
+&= \frac{1}{i \omega \mu_0 \epsilon_0} \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr}
+   \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right)
+   - i \omega \frac{\mu_0 I_0 d}{4 \pi} \frac{\cos \theta}{r} e^{-ikr} \\
+&= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{i \omega \mu_0 \epsilon_0}
+    \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) - \frac{i \omega}{r} \right) \\
+&= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3}
+ + \frac{2k}{\omega \epsilon_0 r^2} - \frac{k^2}{i \omega \epsilon_0 r}
+ - \frac{i \omega \mu_o}{r} \right) \\
+&= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3}
+ + \frac{2 \omega \sqrt{\epsilon_0 \mu_0}}{\omega \epsilon_0 r^2}
+ - \frac{\omega^2 \epsilon_0 \mu_0}{i \omega \epsilon_0 r} - \frac{i \omega \mu_o}{r} \right) \\
+&= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3}
+ + \frac{2}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}}
+ + \frac{i \omega \mu_0}{r} - \frac{i \omega \mu_o}{r} \right) \\
+&= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3}
+ + \frac{2}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \right) \\
+
+E_\theta
+&= \frac{1}{i \omega \mu_0 \epsilon_0} \frac{\mu_0 I_0 d}{4 \pi} \sin \theta e^{-ikr}
+   \left( \frac{1}{r^3} + \frac{ik}{r^2} \right)
+ + i \omega \frac{\mu_0 I_0 d}{4 \pi} \frac{\sin \theta}{r} e^{-ikr} \\
+&= \frac{I_0 d}{4 \pi} \sin \theta e^{-ikr} \left( \frac{1}{i \omega \epsilon_0 r^3}
+ + \frac{k}{\omega \epsilon_0 r^2} + \frac{i \omega \mu_0}{r} \right) \\
+&= \frac{I_0 d}{4 \pi} \sin \theta e^{-ikr} \left( \frac{1}{i \omega \epsilon_0 r^3}
+ + \frac{1}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} + \frac{i \omega \mu_0}{r} \right) \\
+
+E_\phi &= 0
+\end{align*}
+$$
 
 
 ## (8.2)