In this topic, we will discuss the important concept of electric flux, its computation. Also, an analogy between the flux of an electric field and that of water will be explained. Electric flux is an important property of an electric field. It can be considered as the number of forces that are intersecting a given area. Electric field lines are generally considered to start on positive electric charges and to finish on negative charges. Field lines directed into the closed surface are negative and those directed out of a closed surface are positive. In this article, the student will learn the Electric flux Formula with examples. Let us learn the concept!
Source:en.wikipedia.org
Electric Flux formula
What is electric Flux?
Consider the flow of water with some velocity v in a pipe in a certain fixed direction, suppose towards the right. We will take the crosssectional plane of the pipe. Let us consider a small unit area of the plane given as dA. Then the volumetric flow of the liquid crossing that plane normal to the flow will be given as v × dA.
When the plane is not exactly normal to the flow of the fluid then it will be inclined at an angle \(\theta\). Then the total volume of liquid crossing through the plane per unit time is given as \(v ×dA \; cos \theta\). Here, \(dA cos\theta\) is the projected area of the plane towards the perpendicular direction of the flow of the liquid.
The electric field is behaving in a similar way. Therefore it is analogous to the flow of liquid as discussed above. If there is no given net charge within some closed surface then every field line directed into the given surface will continue through the interior. The negative flux is just equal to the magnitude of the positive flux. So, the net or total, the electric flux will be zero.
If a net charge is contained within a closed surface, then the total flux through the surface will be proportional to the enclosed charge, i.e. positive if it is positive, negative if it is negative.
The Formula for Electric flux:
The total number of electric field lines passing through a given area in a unit time is the electric flux.
Thus,
Similar to the above example, if the plane is normal to the flow of the electric field, the total flux is given as:
\(\phi_p = E \times A\)
Also, if the same plane is inclined at an angle \theta, the projected area can be given as \(Acos\theta\) and the total flux through this surface will be,
\(\phi = E A \;cos\theta\)
Where,
\(\phi\)  Electric flux 
E  The magnitude of the electric field 
A  Area of the surface 
\(\theta\)  The angle between the plane and the axis parallel to the direction of flow of the electric field.

Solved Examples
Q.1: A planar surface has an area of 1 square meter. If an electric field crosses with an angle of \(60^{\circ}\) to it and has E= 2 Volte per meter. What will be the electric flux?
Solution:
Using the formula of the electric flux,
\(\phi = E A \;cos\theta\)
= \(2\times 1 \times cos 60^{\circ} \)
= \(2\times 1 \times \frac{1}{2} \)
= 1 Volte Meter.
Therefore electric flux will be a 1 Volt Meter.
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
M=f/g
Interesting studies
It is already correct f= ma by second newton formula…