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
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
where, following the equivalence principle, we assume gravitational and inertial masses are equal.
Therefore the acceleration $a$ of the dipole is given by
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.