Falling electric dipole contradicts equivalence principle? Consider an electric dipole, with total mass $M$, consisting of charges $q$ and $-q$, separated by a distance $d$. The total mass $M$ includes the mass defect due to the negative electrostatic energy associated with the opposite charges.  
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 $e^2 \equiv q^2/4\pi\epsilon_0$ 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 of motion to the dipole we have: gravitational force (gravitational mass times field strength) plus electric force equals the inertial mass times acceleration
$$Mg+F_e=M a$$
where, following the equivalence principle, we assume gravitational and inertial masses are equal.
Therefore the acceleration $a$ of the dipole is given by
$$a=g\large(1-\frac{e^2}{Mc^2d}\large)^{-1}.$$
Thus the dipole accelerates faster than gravity. An observer falling with the dipole sees it move away from him whereas in deep space the observer does not see the dipole move away.
Surely this contradicts the equivalence principle?
P.S. I think that if advanced EM fields are somehow included in the calculation of the electric force $F_e$ then $F_e=0$ and the equivalence principle is obeyed. 
 A: 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 (in this case) 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 if $M_G\neq 0$, the dipole will move in accordance with the equivalence principle.
A: 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.
A: You can’t apply the gravitational force to a free falling body.
What is the difference of a free falling body near the surface of the earth (your dipole) and a body free floating in space or orbiting the earth? There is no difference, both following their geodesic path and both are not feeling any acceleration. Being you in these positions, you will feel weightless.
A: Instead of doing explicit calculations, let me point out why (naïvely) applying equivalence principle to a point charge (or a dipole with a small but fixed separation between the charges) in accelerated reference frame is a wrong idea. Equivalence principle is a statement about local physics. Expressions attempting to derive the self-consistent forces / laws of motion for a charge are non-local, because instead of treating only local entities (that would be the charge/dipole and electromagnetic field in its vicinity) those expressions remove the degrees of freedom of the EM field and present only expressions depending on the motion of the charge. This could be done, of course, but as a result we are loosing locality, since EM field is a long-range,  these expressions would carry information about the long-distance structure of space-time.
A simple example from electrostatic of point charges and extended conducting bodies. A force acting on a point charge:
$$ 
\mathbf{F}= q \,\mathbf{E}_\text{reg}(\mathbf{r}_q),
$$
(where  $\mathbf{E}_\text{reg}$ is a regularized, with divergent part removed, electric field at the position of a charge), is a local expression. One needs the value of an electric field only in the small vicinity of a charge (to perform regularization) to compute the force. The expression is applicable for a wide variety of problems, including moving charges, fields of radiation the only requirement is it must be applied in the reference frame where the charge is a (momentarily) at rest.
A force acting on a point charge at a distance $d$ from a conducting plane:
$$
 \mathbf{F} = - \frac{q^2 }{4d^2} \mathbf{n},
$$
is a non-local expression, since it encodes a structure of the boundary conditions on electric field over the whole domain of a problem. The appearance of the quantity $d$ is an indicator of non-local character of the equation. The expression however is easier to use but is applicable only to a specific problem.
Likewise, if we consider self-action of a dipole in a gravitational field, or in a constant acceleration, expressions for forces like the one in the body of a question are non-local by their origin, so the equivalence principal does not directly apply to them. The analogy  with my electrostatic example above could be made more explicit if we remember, that in a reference frame following a constantly accelerating dipole there is a Rindler horizon. This surface provides a specific boundary conditions for electrostatic field of a dipole that could be interpreted as an induced surface charge, that would provide additional force on the dipole.
