I just did a back-of-the-envelope calculation, which surprised me. I calculated the difference in acceleration (due to repelling like-charges) experienced by two sides of an electron the size of the classical electron radius, when placed one angstrom from another electron. I used purely classical formulas:
$$F = m_{e}\Delta a ~=~ \frac{k_e q_e^2}{r^2} - \frac{k_e q_e^2}{(r+r_e)^2},$$
Where $m_e$, $q_e$ are the mass and charge of the electron, $r_e$ is the classical electron radius, and $\Delta a$ is the difference in acceleration (the tidal effect) between the two sides of the electron.
Using $r = One\,\, Angstrom = 10^{-10}m$, I get: $$\Delta a ~=~ 1.5 \times 10^{18} m/s^2.$$
In other words, assuming I didn't make a mistake, the electromagnetic tidal effect is enormous!
This brings up some questions:
Would this effect be measurable if the electron were not point-like (or far smaller than $r_e$)?
Can we prove a particle is nearly point-like by considering electromagnetic tidal effects like the above?
Are these sorts of effects studied or considered at all in QFT?