What does $E$ field look like around an inductor? What does $B$ field look like outside of a capacitor? For whatever reason I could not find a picture of $E$ field around inductor, and also couldn't find a picture of $B$ field outside of a capacitor.
The reason I think is because these quantities are zero, and due to some obvious geometric intuition, it is not addressed in most texts. It has been years since I touched upon this topic and what may have once been obvious is now confounding for me. 
Can someone provide an intuitive answer as to why these quantities are zero?
Thanks
 A: Well, obviously it depends on the geometry.
The capacitor is a little easier: if the charge on the capacitor is fixed, there aren't any currents and the magnetic field is zero everywhere. However if the capacitor is charging or discharging there will be a magnetic field circulating around the leads and around the E-field volume, obeying
$$
\vec\nabla\times\vec B - \mu_0\epsilon_0 \frac{\partial \vec E}{\partial t} = \mu_0 \vec J
$$
as famously shown by Maxwell. (See Figure 7.42 in Griffiths, for example.)
For an inductor I'll assume you mean a simple cylindrical solenoid with one lead at each end, operating at constant current. If it's a superconducting solenoid, there'll be no potential difference between one end and the other, and so no electric field.  However if it's a resistive solenoid, there'll be a small voltage drop across the solenoid, which will give the solenoid an electric dipole moment and an electric field analogous to the magnetic field around a bar magnet.
If you wanted to produce a resistive solenoid without an electric dipole moment, you could put the high- and low-potential leads at the same end of the cylinder. You'd do this by winding the wire around the form like thread around a spool: from the bottom of the cylinder up to the length of the magnet, then in a second layer back to the bottom. Between the two layers of coils there would be an electric field like that in a cylindrical capacitor, though with more E-field at the end near the leads than at the end where the wire turns back on itself. This probably as complicated as we can get, or perhaps too complicated, without illustrations.
A circuit that's dissipating power must always have both electric and magnetic fields, because the flow of energy is governed by Poynting's theorem and vanishes unless the electric and magnetic fields overlap. For a simple DC circuit consider these illustrations, which are based on an argument by Purcell.
