How does energy transfer between B and E in an EM standing wave? I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.
Using a standing wave as an example, I came up with the equations
$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$
$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$
I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:
$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$
Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?
 A: Ah, but there is energy stored in the field: recall that the energy density is given by
$$
u(t,x)=\frac12\left(\varepsilon_0\vec{E}\cdot\vec{E}+\frac{1}{\mu_0}\vec{B}\cdot\vec{B}\right)
$$
which is clearly non-zero at $t=0$. Surely there are zeroes in the energy at $t=0$, but for $t>0$, the energy is transferred from the nearby non-zero fields! In fact, that's exactly what the Poynting vector tells you: how much energy is transfered per unit area in a particular direction.
A: The energy conservation is written $\dfrac{\partial u}{\partial t} + div \vec S=0$, where $u$ is the energy density $\vec E^2+\vec B^2$, and $\vec S$ is the Poynting vector $\vec E \wedge \vec B$  (skipping irrelevant constant factors).
If you choose $x,t$ such as $\sin \omega t=0$ and $\cos k x=0$, both $E$, $B$, the energy density $u$ and the Poynting vector $S$ will be zero.
We have $\vec S \sim \sin 2k x \sin 2\omega t ~\hat{x}$. The divergence of the Poynting vector will be $div \vec S\sim \cos 2k x \sin 2\omega t $, so it it zero too (because of the $t$ dependence ), and so $\dfrac{\partial u}{\partial t} = 0$. However, the first time derivative of $div(\vec S)$is not zero : $\dfrac{\partial (div \vec S)}{\partial t}\sim \cos 2k x \cos 2\omega t $, so, from the energy conservation equation, the second derivative of the density energy $\dfrac{\partial^2 u}{\partial t^2} = - \dfrac{\partial (div \vec S)}{\partial t}$ is not zero.
So, you may write : 
$u(x,t+\epsilon) = \frac {\epsilon^2}{2} \dfrac{\partial^2 u}{\partial t^2} + o(\epsilon^2)$
So, at infinitesimal times after $t$, the  energy density is not zero.
A: The variation of the fields over a quarter of a period looks something like this.  

In the left diagram it is the magnetic field which is storing the energy of the system whilst in the right hand diagram it is the electric field.  
The exchange between the electric field and magnetic field follows Maxwell's equations - Faraday and Ampere.
As with all standing waves you can think of then as the superposition of two travelling waves carrying energy and in this example at a node the net transport of energy is zero for all time whereas at other positions it is the average value of the Poynting vector over half a period which is zero.
