Consider flow of a liquid with velocity v vector, through a small flat surface dS, in a direction normal to the surface.

The rate of flow of liquid is given by the volume crossing the area per unit time, v dS and represents the flux of liquid flowing across the plane.

If the normal to the surface is not parallel to the direction of flow of liquid, i.e., to v vector, but makes an angle θ with it, the projected area in a plane perpendicular to v vector is v dS cos θ.

Therefore, the flux going out of the surface dS is v vector ⋅ n cap dS.

For the case of the electric field, we define an analogous quantity and call it electric flux. We should, however, note that there is no flow of a physically observable quantity unlike the case of liquid flow.

In the picture of electric field lines described above, we saw that the number of field lines crossing a unit area, placed normal to the field at a point is a measure of the strength of electric field at that point. This means that if we place a small planar element of area ΔS normal to E vector at a point, the number of field lines crossing it is proportional* to E ΔS.
(* It will not be proper to say that the number of field lines is equal to EΔS. The number of field lines is after all, a matter of how many field lines we choose to draw. What is physically significant is the relative number of field lines crossing a given area at different points.)

Now suppose we tilt the area element by angle θ. Clearly, the number of field lines crossing the area element will be smaller. The projection of the area element normal to E is ΔS cos θ. Thus, the number of field lines crossing ΔS is proportional to E ΔS cosθ. When θ = 90°, field lines will be parallel to ΔS and will not cross it at all (Fig. 1.18).

FIGURE 1.18 Dependence of flux on the inclination θ between E vector and n cap .

The orientation of area element and not merely its magnitude is important in many contexts. For example, in a stream, the amount of water flowing through a ring will naturally depend on how you hold the ring. If you hold it normal to the flow, maximum water will flow through it than if you hold it with some other orientation. This shows that an area element should be treated as a vector. It has a magnitude and also a direction. How to specify the direction of a planar area? Clearly, the normal to the plane specifies the orientation of the plane. Thus the direction of a planar area vector is along its normal.

How to associate a vector to the area of a curved surface? We imagine dividing the surface into a large number of very small area elements. Each small area element may be treated as planar and a vector associated with it, as explained before.

Notice one ambiguity here. The direction of an area element is along its normal. But a normal can point in two directions. Which direction do we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given context. For the case of a closed surface, this convention is very simple. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. This is the convention used in Fig. 1.19.

FIGURE 1.19 Convention for defining normal n cap and ΔS.

Thus, the area element vector ΔS vector at a point on a closed surface equals ΔS n cap where ΔS is the magnitude of the area element and n cap is a unit vector in the direction of outward normal at that point.
We now come to the definition of electric flux. Electric flux Δφ through an area element ΔS vector is defined by
Δφ = E vector ⋅ ΔS vector = E ΔS cosθEquation (1.11)
Here, the ‘ ⋅ ’ stands for the scalar or the dot product of vectors.
which, as seen earlier, is proportional to the number of field lines cutting the area element. The angle θ here is the angle between E vector and ΔS vector. For a closed surface, with the convention stated already, θ is the angle between E vector and the outward normal to the area element. Notice we could look at the expression E ΔS cos θ in two ways: E (ΔS cosθ) i.e., E times the projection of area normal to E vector, or (E cosθ) × ΔS, i.e., component of E vector along the normal to the area element times the magnitude of the area element.

The unit of electric flux is N C^−1 m^2.

The basic definition of electric flux given by Equation (1.11) can be used, in principle, to calculate the total flux through any given surface. All we have to do is to divide the surface into small area elements, calculate the flux at each element and add them up. Thus, the total flux φ through a surface S may be written as:
φ ≅ Σ E vector ⋅ ΔS vector Equation (1.12)

The approximation sign is used because the electric field E vector is taken to be constant over the small area element. This is mathematically exact only when you take the limit ΔS → 0 and the sum in Equation (1.12) is written as an integral.
Or, Electric flux, φ = ∫ over s of E vector ⋅ ds vector.

As you can see that here we have placed an equal to sign instead of the approximately equal to sign, that clearly means that the results that we obtain using integration are the best and the accurate answers. It is therefore very important to learn integration, before you start prepairing for the ultimate subject. By Dr. T. K. Bansal.