Gfyuki

Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.

If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by

$$F_e=frac{e^2a}{c^2d}$$

where we introduce $e^2 equiv q^2/4piepsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5).

Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.

Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration

$$2mg+F_e=2m a$$

Therefore the acceleration $a$ of the dipole is given by

$$a=glarge(1-frac{e^2}{2mc^2d}large)^{-1}$$

Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.

Surely this contradicts the equivalence principle?

The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.

However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!

From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.

In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.

It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.

This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.

But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula

$$
E = M_Gc^2
$$

where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain

$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.