Question

how does \( \vec D = \epsilon_0 \vec E \) give flux density?

Answer 1

I am not sure to understand the question.

Flux of \(\vec D\) equals charge enclosed in the surface.

Answer 2

Mass density(or just density for most people) is mass per volume. The per clues you
that you need to multiply by volume to get the mass.

D is the flux per area through a surface. Again the per is the clue that you have to multiply
by an area to get the flux. Here things are a little more complex because the answer depends on the orientation of the surface, but the basic idea is the same.

Answer 3

Total flux = \( Q \over \epsilon_0\)
\[ \oint_s \vec E \cdot \hat n dS = \iiint _v \nabla \cdot \vec E dV = \iiint_v {\rho \over \epsilon_0} dV \\
\implies \nabla \cdot \vec E = { \rho \over \epsilon _0} ,\\ \implies \nabla \cdot \epsilon _0\vec E = { \rho } \\
\implies \nabla \cdot \vec D = \rho\]

Wikipedia says "In free space, the electric displacement field is equivalent to flux density"
http://en.wikipedia.org/wiki/Electric_displacement_field
how can \( \vec D \) be flux density?