How does the induced E field for a uniformly varying B field obey spatial symmetry?

Question

Consider this answer giving the induced E field by a uniformly varying B field. In this case, if you place a point positive charge at the origin it will experience 0 force, but if you put it at some other point (such as (1, 0)) it would experience a force (in the positive Y direction) due to the induced E field.

If the laws of physics obey translational symmetry, then the particle at (0, 0) should experience the same force as the one at (1, 0). Furthermore if the laws of physics obey rotational symmetry then both forces must be 0. However this is not what is predicted by any solution to Faraday's law, since any vector field with a constant, nonzero curl cannot be everywhere 0.

How then should Faraday's law be interpreted in the context of a uniformly varying B field? In other words, given some system (for instance, a system of 2 electric charges, one at (0, 0) and one at (1, 0)), what constraints can be deduced to choose the particular solution of Faraday's law that must be used for a given prediction (such as, "What force will be exerted on each particle?")?

Answer 1

The thing is, there is an infinite amount of solutions to this problem, i.e. if B points in the z-direction, $E(x,y,t) =E_0 (t) ((x-x_0) \hat{y} - (y-y_0) \hat{x})$ solves Faraday's law for every value of $x_0$ and $y_0$. As a consequence, the particle can accelerate in any direction!

The problem with this situation is that we are missing one important ingredient to determine the "good" field: Boundary conditions. Adding boundary conditions to this problem will force the selection of a unique solution among the infinite solution set. In essence, this is not a physical situation as in reality you always have boundary conditions.

The most usual boundary condition used, but not the only one that would produce a unique solution, is generally E->0 and B-> 0 at infinity, which in this case is obviously violated.