Question

Total electric flux coming out of a unit positive charge put in the air is
A. ${\varepsilon _0}$
B. ${\varepsilon _0}^{ - 1}$
C. \[{\left( {4{\varepsilon _0}} \right)^{ - 1}}\]
D. \[\left( {4\pi {\varepsilon _0}} \right)\]

Answer 1

Hint: The law that states the total flux of the closed surface is equal to the charge that is divided by the permittivity is called Gauss law. The electric flux mentioned in the law is described by the product of the electric field and the area of the projected plane. Using this law, the question can be solved.

Complete step by step solution:
The total electric flux linked with the closed surface is $\dfrac{1}{{{\varepsilon _0}}}$ times of the charge enclosed by the closed surface. This law is known as Gauss law.
By using this law, the question can be solved.
 The total electric flux of a unit positive charge when put in the air is $\dfrac{q}{{{\varepsilon _0}}}$.
The charging unit is one. Substitute the value in $\dfrac{1}{{{\varepsilon _0}}}$.
Therefore, the flux is given as $\dfrac{1}{{{\varepsilon _0}}}$ or ${\varepsilon _0}^{ - 1}$.
Hence option $\left( B \right)$ is the correct answer.

Additional information:
The gauss theorem statement gives an important corollary. The electric flux from any closed surface is only due to the positive charge sources and sinks the negative charge sources of the electric fields enclosed by the surface. Any charges that are outside the surface will not contribute to the electric flux. Also, the electric charges act as a source or sink the electric field. by changing the magnetic fields, cannot act as source or sinks of the electric field.

Note:
The electric flux in an area can be defined as the electric field multiplied by the area of the plane that is projected and it is perpendicular to the field. The gauss law is only the restatement of the coulomb’s law. If the gauss law is applied to the point on the closed sphere, we will get back the coulomb’s law.