How big of an area can a classical charge absorb an EM wave from? In electromagnetism, accelerating charges can emit or absorb the EM radiation.
My understanding is that when a plane wave passes a stationary charge, the charge is jiggled in such a way that it emits its own EM waves but in such a way that the waves from the charge cancel out the incoming waves. The result is that we can think of the charge as absorbing part of the incoming wave.
Are there references that analyze this scenario mathematically? How can I analyze this scenario?
My main question though is, how wide an area would the electron absorb the radiation from? Does the question even make any sense to ask (now that I write this I suspect the answer is no)? 

My question comes from thinking about wire grid polarizers.
I know that a wire grid polarizer for microwaves has impressively large gaps between the metal wires, but the EM waves (if the polarization is aligned with the wires) can be completely absorbed because the electrons are moved up and down along the wires. 
This led me to ask, how much can a single electron absorb? How much larger is the area that can be absorbed compared to the electron cross-section (assuming we imagine it as a classical sphere)?
Edit: When I'll get the chance, I might give a stab at the math to show what I'm looking for. 
 A: This problem is handled by the theory of Thomson scattering,
which is the scattering of photons by a charged particle (e.g. an electron).
The differential Thomson cross-section is
$$\frac{d\sigma}{d\Omega} = \left(\frac{q^2}{4\pi\epsilon_0 mc^2}\right)^2 \frac{1+\cos^2\chi}{2}$$
where $q$ and $m$ are charge and mass of the scattering particle,
and $\chi$ is the deflection angle of the outgoing photon.
The important feature is that the cross section is independent of photon frequency.
By integrating over the solid angle, we get the total Thomson cross-section
$$\sigma = \frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0 mc^2}\right)^2$$
For an electron this gives
$$\sigma_\text{electron} = 6.65\cdot 10^{-29}\text{ m}^2$$
A: It is difficult to give the cross section for one electron, but it is interesting that graphene has a transmittance of 97.7 % up to the frequencies of visible light. This value is linked to the QED fine-structure constant $\alpha \approx 1/137$:
$$t = \frac{1}{(1 + \pi \ \alpha_{\rm QED}/2)^2}.$$
Sheehy and Schmalian, PRB (2009) on Researchgate 
