How can electric displacement vector field is equivalent to flux density on free space Wikipedia states that 

"In free space, the electric displacement field is equivalent to flux density, a concept that lends understanding to Gauss's law".  

Doesn't that meant that at free space $ \vec D $ is equal to $ \vec D = {\text{Total Flux} \over \text{Area} } = \frac{ \oint_s \vec E \cdot \hat n dS}{\oint_s dS} = \epsilon_0 \vec E$
 A: Normally we think of the electric field as being the physical entity, and flux is just its integral. But you can also think of the electric flux as a thing in its own right, and its area density then corresponds to what we call electric field. 
Basically, a charge produces a certain amount of electric flux, spread out over all possible directions going away from (or toward) that charge, just like a star produces a certain amount of luminous flux (light), spread out over all possible directions going away from the star. The density of the flux is the electric field in the case of the charge, and is the luminous intensity in the case of the star. As you move further away from the star, the luminous intensity decreases because the same amount of flux gets spread over a larger sphere, leading to the inverse-square law $I = P_s/4\pi r^2$, and the same happens with the electric flux, leading to the inverse-square law $\epsilon E = Q_\text{enc}/4\pi r^2$. This can be rearranged to a special case of Gauss's law for a spherical surface concentric with the charge.
