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A charge \[q\] is placed at a point exactly above the centre of the square (of side \[a\]) at a distance \[\dfrac{a}{2}\]. Thus the flux passing through square will be:

(A) \[\dfrac{q}{{6{\varepsilon _0}}}\]
(B) \[\dfrac{q}{{12{\varepsilon _0}}}\].
(C) \[\dfrac{q}{{{\varepsilon _0}}}\].
(D) \[\dfrac{q}{{3{\varepsilon _0}}}\].

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Last updated date: 20th Apr 2024
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Answer
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Hint In this question, consider a charge placed at a point above the centre of the enclosed sin a cube and apply the Gauss law that is total electric flux of a enclosed surface is equal to the charge enclosed in the surface divided by the permittivity.

Complete step by step answer
 We are given that a charge \[q\] is placed at a point above the centre of the square (of side \[a\]) at a distance \[\dfrac{a}{2}\].
Let us consider a cube that is enclosed and the square having a point charge at the Centre. A cube has six surfaces or sides. therefore, flux is passing through the six sides or surfaces.
As we know that the Gauss law says that total electric flux of an enclosed surface is equal to the charge enclosed in the surface divided by the permittivity.
By using the Gauss law, we get,
\[ \Rightarrow \phi = \dfrac{{{Q_{in}}}}{{{\varepsilon _0}}}\]
Where, \[\phi \] is the electric flux, \[Q\] is the total charge enclosed, \[{\varepsilon _0}\] represents the permittivity
When a charge is placed at Centre of the cube, So, the flux can be written as,
\[ \Rightarrow \phi = \dfrac{q}{{{\varepsilon _0}}}\]
As the flux is link to each surface of the cube is equal
Thus, we can write the flux through each surface of the square as,
\[ \Rightarrow \phi = \dfrac{q}{{{\varepsilon _0}}} \times \dfrac{1}{6}\]
After simplification we get,
\[\therefore \phi = \dfrac{q}{{6{\varepsilon _0}}}\]

Therefore, the flux when the charge \[q\] is placed at a point exactly above the centre of the square passing through square will be \[\phi = \dfrac{q}{{6{\varepsilon _0}}}\].

Note
As we know that the surface area of each plane of the cube is the same because the sides of the cube are equal, the flux through each surface will be the same as the charge is placed at the center of the cube.