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The electric field produced inside a uniformly polarized sphere of radius R is equal to: $$ \bf E = - {\bf P \over \rm 3 \epsilon_0}$$

It doesn't depend on the sphere radius. Does that mean... as long as I have uniformly polarized spheres made of the same dielectric their size won't matter?

But then, if I have two identical dielectrics in front of me, except one of them has a spherical cavity inside (its volume is very small compared to the dielectric but big enough to contain a statistically significant number of elementary dipoles), I could tell the difference! For the one intact I'd measure $\bf E $ and in the other $\bf E + {\bf P \over \rm 3 \epsilon_0}$. But that's ridiculous! How can a small hole make such a macroscopic difference?

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The key here is that this expression for the electric field is valid only inside the sphere. The field on the exterior of the sphere depends on the radius of the sphere. So based on the exterior fields, the spheres could be distinguished.

As for a sphere of uniformly polarized material with a hole punched in the center: this is no longer a sphere of uniformly polarized material and hence the expression $\boldsymbol{E}=-\frac{\boldsymbol{P}}{3\epsilon_0}$ would no longer be correct. The boundary conditions will have changed and hence the solution will generally have changed as well, meaning Maxwell's equations must be resolved for this new problem.

If the cavity is very small, it might be reasonable to say that the expression for a uniformly polarized sphere might be a good approximation to the true field, but this would only be an approximation.

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