# $\vec{E}$-field in toroidal polarized dielectric

In an electrostatic situation, imagine a dielectric torus that is permanently polarized with polarization given by:

$$\vec{P} = P(r, z) \hat{\phi}$$

(Here cylindrical coordinates are used with the $$z$$-axis pointing through the middle of the hole in the torus and $$z=0$$ is in the vertical-middle of the torus.)

Then, the bound surface and volume charges are given by:

$$\sigma_{b} = \hat{n}\cdot\vec{P} = 0$$

$$\rho_b = -\vec{\nabla}\cdot\vec{P} = -\frac{1}{r}\frac{\partial{P_{\phi}}}{\partial{\phi}} = 0$$

So the bound charges are zero everywhere. Assuming no free charges then the divergence of the electric field is zero everywhere so we have no sources or sinks for the field. Does this mean that we have no electrical field inside or outside the torus.

If so this seems counterintuitive to me since the dipoles in the dielectric normally produce an electric field. Intuitively I would except the field to circle around in the toroidal direction (i.e. $$\vec{E} = E(r,z)\hat{\phi}$$). However, this also seems counterintuitive since this yields an electric field with a closed loop, which is unphysical.

So both solutions I can think of seems counterintuitive!

If you assume a circularly homogeneous polarization $$\mathbf {P} = P(r, z) \hat{\mathbf {\phi}}$$ with the bound surface and volume charges $$\sigma_{b} = \hat{\mathbf n}\cdot\mathbf P= 0\\ \rho_b = -\mathbf{\nabla}\cdot\mathbf{P} = -\frac{1}{r}\frac{\partial{P_{\phi}}}{\partial{\phi}} = 0$$ then, as you noticed, since $$\nabla \times \mathbf P \ne \mathbf 0$$ this cannot be the polarization of a regular isotropic dielectric for which $$\mathbf P = \mathbf P(\mathbf E)$$ with $$\mathbf P(\mathbf 0)=\mathbf 0$$.
In other words there must be some remanent polarization at zero bias field, that is $$\mathbf P = \mathbf P(\mathbf E) = \mathbf P_0 + \chi \mathbf E + \mathbf a_2 \mathbf E\circ \mathbf E ...$$ With this assumption we can satisfy the required electrostatic equations as: $$\mathbf P = \mathbf P_0\\ \mathbf E=\mathbf 0 \\\nabla \times \mathbf E = \mathbf 0\\ \mathbf D = \epsilon_0 \mathbf P$$