# Why can't the flux be an integral for electric field passing over an area?

I need to know why we can't use the $$\int\vec{E}.d\vec{S}$$ as the electrical flux. But we try to define electrical flux density vector $$\vec{D}=\epsilon_{0}\vec{E}$$ and then try to use $$\int\vec{D}.d\vec{S}$$. In Sadiku Elements of electromagnetic he has just mentioned

The flux due to the electric field E can be calculated by using the general definition of flux in $$\int\vec{E}.d\vec{S}$$. For practical reasons, however, this quantity is not usually considered to be the most useful flux in electrostatics.

• Your title lacks clarity and precision. Nov 22 '20 at 16:09
• Even your quoted passage says that it can and it is defined as the integral of E over the surface.
– nasu
Nov 22 '20 at 16:35
• The total flux of $D$ over a closed surface equals the enclosed free charge. In general, there is no such interpretation, or any useful physical quantity for that matter, if you calculate the flux of the $E$-field over a closed surface, unless the matter enclosed by the surface has a constant scalar permittivity. Nov 22 '20 at 16:38

There is nothing wrong with defining the flux of the field with $$\int \mathbf{E}\cdot d\mathbf{S}$$. For any general field $$\mathbf{F}$$, flux $$\Phi$$ is defined as : $$\Phi=\int\mathbf{F}\cdot d\mathbf{S}$$
Now if we replace $$\mathbf{E}$$ to $$\mathbf{D}$$, it will become flux for a different field.
The $$\mathbf{D}$$ is what is known as the electric displacement. Defined as $$\mathbf{D}\equiv \epsilon_0 \mathbf{E}+\mathbf{P}$$ Gauss's law reads $$\nabla\cdot \mathbf{D}=\rho_f$$ This is a particularly a useful way to express Gauss’s law, in the context of dielectrics, because it makes reference only to free charges, and the free charge is the stuff we control. The bound charge comes along for the ride. when we put the free charge in place, a certain polarization automatically ensues and this polarization produces the bound charge. In a typical problem, therefore, we know $$ρ_f$$ , but we do not (initially) know $$ρ_b$$; Eq. lets us go right to work with the information at hand. In particular, whenever the requisite symmetry is present, we can immediately calculate $$\mathbf{D}$$ by the standard Gauss’s law methods.