How can we verify the direction of electric displacement is radial?

Question

I have two questions about electric displacement $D$.

The following is a problem in Griffiths.

A long straight wire, carrying uniform line charge $\lambda$, is surrounded by rubber insulation out to a radius $a$. Find the electric displacement.

(solution in the text) Drawing a cylindrical Gaussian surface, of radius $s$ and length $L$, applying Gauss theorem, we find $$D(2\pi s L) = \lambda L$$ $$D = \frac{\lambda}{2\pi s}\hat{s}$$

1. I think for using the symmetry, we, in advance, verify that the direction of $D$ is radial. How can we verify it?

Griffiths also wrote the following advice.

When you are asked to compute the electric displacement, first look for symmetry. If the problem exhibits spherical, cylindrical, or plane symmetry, then you can get $D$ directly from the Gauss theorem like the above problem. (Evidently in such cases $\nabla \times P$ is automatically zero, but since symmetry alone dictates the answer you're not really obliged to worry about the curl.)

2. How can we get $\nabla \times P=0$ when there is symmetry? Also does $\nabla \times P = 0$ implies that direction $D$ is radial? I want to know how to prove.

Answer 1

1. Let $$\mathbf D = D_\rho \hat{\mathbf \rho}+ D_\varphi \hat{\mathbf \varphi} +D_z \hat{\mathbf z}. $$Then because of the cylindrical symmetry $\frac{\partial}{\partial \varphi}=0$, $D_\varphi $ should not be a function of $\varphi$. When we take a circle of raidus $\rho_0$ which is parallel to the xy plane and have a point of intersection with the z axis, the line integral of $\mathbf D$ along the circle should be zero under an assumption that relative permittivity is constant over all domain. $$ \oint_C \mathbf D \cdot d\mathbf l = \oint_C \mathbf D \cdot \hat{\mathbf \varphi} \rho d\phi =2\pi\rho_0 D_\varphi = 0 $$ So we get $D_\varphi =0$. Using similar way, we also obtain $D_z=0$.

2. $\mathbf P$ is propotional to $\mathbf E$ as long as the medium is isotropic and linear. Therefore $\nabla\times \mathbf P =0$ is trivial in electrostatics.