Electric Flux Density (D)| Definition| Significance| How is it different from E?

# What is Electric Flux Density (D)?

Electric flux density or D is a common term in Electromagnetics particularly in the study of Electrostatics and Gauss’ Law. This article explains its significance along with the comparison with E.

## What is Electric Field Intensity (E)?

Every charge posses their electric field and strength of the field at a particular point of interest is defined as the Electric Field Intensity or E.

Technically, it is defined as the force per unit charge when placed in the electric field. Check out the following article for detailed definition of E.

What is Electric Field and its intensity?

So from the simple definition, we know that the Electric Field because of point charge (Q) at the distance of r from it is given as,

$E=\frac1{4\pi\varepsilon}\frac Q{r^2}$

Likewise using the concepts for deriving an Electric Field Intensity for distributed charges, we can derive formulas of E for any type of charge.

Take any formula of E, you will notice that, this Electric Field Intensity E is dependent on the surrounding medium.

But for most of the practical reasons, we require the flux of the Electric field which is independent of the medium. Had we considered the E as it is for defining the flux, we would get that flux dependent on the medium. But, we need it to be independent of the medium.

So we define a new term called Electric Flux Density or D from the Electric Field Intensity (E).

## Electric Flux Density (D)

We know that there is ε i.e. permittivity of the medium present in the denominator of each expression of E. So if we define the term D as εE, then we could get rid of the dependence of the medium.

$\boxed{D=\varepsilon E}$

So, Electric Flux Density, D, is the Electric Field Intensity, E, multiplied by the permittivity of the corresponding medium.

It is another way to represent the strength of the Electric Field independent of the medium.

It is also a vector quantity exactly the same as E.

## How D is independent of the Medium?

As I explained above, intuitively we can say that multiplication of ε with E, cancels out the ε present in the denominator of E and making D medium free. Well that’s mathematical reasoning. Technically, we can prove it by going slightly deeper into the definition.

Consider the air filled parallel plate capacitor. We know that Electric Field Intensity E within its plates is given as,

$E=\frac\sigma{\varepsilon_0}$

Here, σ is the charge density on the plates of the capacitor and ε0 is the permittivity of the free space.

Now, when we insert any other medium i.e. dielectric within the plates of the capacitor then the polarization of the dielectric takes place. These polarized charges reduce the E within the plates or in other words reduce the charge density σ on the plates.

$E=\frac{\sigma-\sigma_p}{\varepsilon_0}$

Here, σp are the induced polarized charges.

$\therefore\sigma=\varepsilon_0E+P$

Here, P is the polarization of charges which are then related to E by introducing the concept of susceptibility (χ) and relative permittivity which we will discuss in future articles.

$\therefore\sigma=\varepsilon_0E+\chi_eE=\varepsilon E$

And the right-hand side of the above expression is termed as Electric Flux Density or D.

Now we can clearly observe from the above expression, the left-hand side is the charge density (σ) or charge per unit area of the plate of the capacitor which is absolutely independent of the medium. It depends on the external deposition of the charges on the plates.

Conclusion, the D is independent of the medium.

#### Unit of D

From the above explanation, it is evident that the unit of D must be equal to the surface charge density (σ) i.e. C/m2.

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