A uniform electric field \[E = 2 \times {10^3}N{C^{ - 1}}\] is acting along the positive x-axis. The flux of this field through a square of $10cm$ on a side whose plane is parallel to the yz plane is:
A) $20N{C^{ - 1}}{m^2}$
B) $30N{C^{ - 1}}{m^2}$
C) $10N{C^{ - 1}}{m^2}$
D) $40N{C^{ - 1}}{m^2}$

Hint: Recall that the electric field is defined as the area around an electric charge or charged object in which it has influence on the other objects placed. The electric flux is defined as the total number of electric field lines passing through a given area.

Complete step by step solution:
Given that the electric field is $E = 2 \times {10^3}N{C^{ - 1}}$
The electric field is acting along positive x-axis
Also the side of square is $a = 10cm$
Changing cm into meters, the side of the square becomes
$ \Rightarrow a = \dfrac{{10}}{{100}} = 10 \times {10^{ - 2}}m$
$\therefore $the area of square is given by using the formula,
$A = {a^2}$
$ \Rightarrow A = {(10 \times {10^{ - 2}})^2}{m^2}$
If the electric field is uniform, then the electric flux passing through a surface of area is given by the formula,
$\phi = EA\cos \theta $---(i)
Where E is the electric field
A is the area
$\cos \theta $is the angle between the electric field lines and the normal. In this case, the field lines and the normal are in the same direction, so the angle between them will be ${0^ \circ }$.
$\therefore $equation (i) becomes,
$ \Rightarrow \phi = EA\cos {0^ \circ }$
$ \Rightarrow \phi = EA$---(ii)
Substituting the values in the equation (ii) and solving for the value of electric flux,
$ \Rightarrow \phi = 2 \times {10^3} \times {(10 \times {10^{ - 2}})^2}{m^2}$
$ \Rightarrow \phi = 20N{C^{ - 1}}{m^2}$
$\therefore $the flux of this field is $\phi = 20N{C^{ - 1}}{m^{^2}}$

Option A is the right answer.

Note: It is important to remember that the electric flux is a property of the electric field. The electric field lines start from positive electric charge and end on negative electric charge. The electric field lines that are in a direction out of the surface are taken as positive while those in an inward direction are negative.