# The electric field of a polarized sphere

Assuming we have a uniformly polarized sphere. Why does it happen that it exerts an electric field outside, but with a sphere polarized with some radial polarization the electric field outside is 0? Is it because in the second instance one can use Gauss law because of the radial symmetry, while in the first instance we have no symmetry?

1) a sphere with a uniform polarization in a fixed direction, e.g. in the $$\hat z$$ direction, versus
2) a sphere with a uniform radial polarization, i.e. in the $$\pm \hat r$$ direction (which is a direction that depends upon position, unlike $$\hat z$$).
If that's what you mean, then yes, in case (2), symmetry allows you to reduce the E-field to be pointing in either the $$+ \hat r$$ or $$- \hat r$$ direction, which allows you to use Gauss's Law to evaluate it, which, when you do the volume integral for the total contained charge, gives you zero. Whereas, there is no spherical symmetry in case (1).
It's worth pointing out though that, if we are talking about external fields (i.e. some small or not-so-small distance removed from the surface of the sphere), case (1) is equivalent to taking a uniformly charged sphere of charge $$+Q$$ and displacing it an infinitesimal distance in the $$+\hat z$$ direction, then flipping the sign of the charge on the sphere to $$-Q$$ and displacing it the same infinitesimal distance in the $$-\hat z$$ direction, and then summing ("superposing") the two resultant fields together. So there is still a connection to the potential around a spherically-symmetric charged sphere, it's just a little more nuanced.