Question

What is the relation between electric flux and electric field intensity

Answer 1

"Gauss's Law. The total of the electric flux out of a closedsurface is equal to the charge enclosed divided by thepermittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surfaceprojected in a planeperpendicular to the field."

Answer 2

Answer:

The relationship is not likely what you imagine.

There is this all too common equation…

∫SE⋅dA=Qϵ0

Where the LHS defines the flux and  E  is the electric field intensity, defined as either the gradient of the electric potential or force per unit charge.

But this equation contains a bit of deception, a little trickery concealing the underlying meaning of the fields. The proper equation is

∫SD⋅dA=Q

Where  D  is the displacement field or electric flux density and operationally defined as the charge density that manifests on a test capacitor.

The equations look the same, sans a constant. What is being hidden is that  E  and  D  are completely distinct but complementary aspects of the electromagnetic field.

In general, there is no specific relationship between  E  and  D .

In free-space the the two fields can be made numerically equal and in a way which is quite simple  D=ϵ0E . So the equation above is written for free-space as:

∫SE⋅dA=∫S(ϵ0D)⋅dA

and this substitution obscures the fact that the fields are different.

The relationship between  E  and  D  are called the “constitutive relations” are determined empirically for particular materials and contexts.

Why the constitutive relations work is that for any configuration of charges you will have both aspects of the electric field present at each point, and if you know one field it is often possible to calculate the other field.

Explanation: