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## Main Question or Discussion Point

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^2/4\pi\epsilon_0## for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5) (http://inspirehep.net/record/206900/files/slac-pub-3529.pdf).

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

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

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?

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) (http://inspirehep.net/record/206900/files/slac-pub-3529.pdf).

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

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

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?

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