Question

As per Gauss law
 $\int {E.dS = \dfrac{{{q_{in}}}}{{{\varepsilon _0}}}} $
Which of the following is true about this?
A.This is valid for symmetrical surfaces only
B.$E$ is the electric field to the charge inside the surface
C.Electric flux on closed surfaces due to outside charge is always zero.
D.None of the above

Answer 1

Hint: Recall that Gauss law is also known as Gauss’s flux theorem. According to Gauss law the electric flux through any closed surface is directly proportional to the total charge enclosed by the surface. It determines the electric field generated by electric charges.

Complete answer:
Step I:
The electric field has both magnitude and direction. So it is a vector quantity . The electric field is given by the flux through that area. The number of electric field lines depends on the strength of the field, the surface area.
According to Gauss Law:
$\int {E.da = \dfrac{Q}{{{\varepsilon _0}}}} $---(i)
Because of the symmetrical surface the charge will be the same throughout the surface. But if the surface is asymmetrical then given relation in equation (i) will be wrong. This is because in an asymmetrical surface the distribution of charge throughout the surface will not be even.
Step II:
Also it is clear from equation (i) that the electric field E will depend upon the charge enclosed by the surface.
Step III:
Therefore the Gauss law is valid for symmetrical surfaces only. Also electric flux through a closed surface depends on charge enclosed between that surface

Hence, Option C is the correct option.

Note:
It is to be noted that electric flux is defined as the electric field passing through a given area in a plane perpendicular to the direction of the field. The electric field intensity is the measure of the strength of the electric field at a point. The electric flux can have both positive and negative values.