Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.


Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite [displacement current][1] (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

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Firstly, the parallel between charge and mass comes from _gravitational mass_. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like [gravitomagnetism][2], though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.


However, these similarities don't extend to the _inertial_ property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore and explain this link)


  [1]: http://en.wikipedia.org/wiki/Displacement_current
  [2]: http://en.wikipedia.org/wiki/Gravitoelectromagnetism