# How can electric displacement vector field is equivalent to flux density on free space

"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 $\vec D = {\text{Total Flux} \over \text{Area} } = \frac{ \oint_s \vec E \cdot \hat n dS}{\oint_s dS} = \epsilon_0 \vec E$

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That doesn't look like the continuity equation to me. I'm pretty sure the continuity equation involves the current density J, not the flux density D. Besides that, isn't it div D = rho? –  Alfred Centauri Jul 6 '12 at 11:02
sorry ... i made mistake. but can you explain how $\vec D$ give flux density?? –  hasExams Jul 6 '12 at 11:03
I'm sorry, I'm just not sure what you you're asking. Perhaps you could provide a little more context? –  Alfred Centauri Jul 6 '12 at 11:16
I've never heard $\vec{D}$ called flux density before. It is usually called electric displacement. –  Siyuan Ren Jul 6 '12 at 13:19
@AlfredCentauri please check edit above –  hasExams Jul 6 '12 at 14:12

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.
I thin it makes quite clear "The density of the flux is the electric field in the case of the charge" ... now my problem reduces to unit. what makes $\vec D$ different from $\vec E$? Why would anyone formulate $\vec D$ when it can be trivially expressed with $\epsilon_0 \vec E$ or $\epsilon_0 \vec E + \vec P$ in dielectric. –  hasExams Jul 6 '12 at 18:11
$\vec{D}$ can be used to formulate Maxwell's equations in a medium. It'd be pretty awkward to always have to write out things like $\vec\nabla\times(\epsilon_0\vec{E} + \vec{P})$. Perhaps more importantly, it's just nice to not have to think about the polarization. –  David Z Jul 6 '12 at 19:42