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Say there is a polarized sphere with polarization density $\vec{P} = \alpha \hat{r}$. How can I tell if the electric field outside of the sphere will also be radial? I see in many places that it is taken as obvious, but why is it?

*Edit: rephrase

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2 Answers 2

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Pay attention to the fact that the sphere you are considering is not uniformely polarized: only the magnitude of the polarization density is uniform, while its direction changes at each point. In particular the direction of $\vec{P}$ is radial. Therefore the system has a symmetry with respect to the center of the sphere and it is possible to say that the electric field outside the sphere has to have the same symmetry. This is why it is radial.

Note that if $\vec{P}$ is taken "really" uniform on the sphere (i.e. it has same magnitude and direction at each point on the sphere) the external electric field outside is not radial: it is a dipole field.

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  • $\begingroup$ You're right, I should have phrased that differently. So the explanation is symmetry? Does it have anything to do with $\vec{P} = \varepsilon_0 \chi_e \vec{E}$ ? $\endgroup$
    – Darkenin
    Commented May 13, 2020 at 18:52
  • $\begingroup$ The equation that you wrote in the comment is not connected to the expression of $\vec{E}$ outside the sphere. Indeed the relation $\vec{P}=\epsilon_{0}\chi_{e}\vec{E}$ sets a relation between the polarization and the electric field at each point of the polarized dielectric. This relation is not valid outside the sphere. Or, if you want, you can apply it also outside the sphere but it is of no help because in empty space the coefficient $\chi_{e}$ is zero. The explaination to your question is just symmetry as stated in my answer and in the answer that you approved. $\endgroup$ Commented May 13, 2020 at 19:10
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It's because the whole system (including the polarization density) has spherical symmetry.

Think of it this way, if I rotate the sphere by an arbitrary angle around an axis passing its origin, the sphere, and the associated polarization density $\mathbf{P}(\mathbf{x}) = \alpha \hat{\mathbf{r}}$ are both going to coincide with the non-rotated case. So the same has to be true for the electric field distribution; i.e. not only is the electric field outside the sphere radial, but so is the field inside.

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