Question

A point charge of $+6\mu C$ is placed at a distance $20cm$ directly above the centre of a square of side $40cm$. The magnitude of flux through the square is

& \left( 1 \right){{\epsilon }_{0}} \\\ & \left( 2 \right)\dfrac{1}{{{\epsilon }_{0}}} \\\ & \left( 3 \right){{\epsilon }_{0}}\times {{10}^{-6}} \\\ & \left( 4 \right)\dfrac{1}{{{\epsilon }_{0}}}\times {{10}^{-6}} \\\ \end{aligned}$$

Answer 1

First consider a Gaussian surface including the side of the square. So the Gaussian surface will be a cube with a side $40cm$. As the charge is located at the centre so first calculate the flux linked with the whole cube which can be calculated from the Gauss law. So the flux through the side of the cube will be $\dfrac{1}{6}$ times the total flux.

Formulas used:
From Gauss law the flux linked with a closed surface containing a charge $q$ is given by
$\phi =\dfrac{q}{{{\epsilon }_{0}}}$
Where ${{\epsilon }_{0}}$ is the permittivity of free space and its value is ${{\epsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

Complete step by step answer:
The gauss law states that the total electric flux linked with a close surface containing a charge $q$ is
$\phi =\dfrac{q}{{{\epsilon }_{0}}}$
Given that a point charge of $+6\mu C$ is placed at a distance $20cm$ directly above the centre of a square of side $40cm$ . As the square is not close we have to consider a Gaussian surface .
Consider an Gaussian surface as a cube with one side as the square of side $40cm$.
So the total electric flux through the Gaussian surface i.e. the cube is
$\phi =\dfrac{q}{{{\epsilon }_{0}}}=\dfrac{6\mu C}{{{\epsilon }_{0}}}$
This flux is the total flux through the entire cube, i.e. this flux is through the six surfaces of the cube. . So the flux through its any face will be $\dfrac{1}{6}$ times the total flux. i.e.
$\phi '=\dfrac{\phi }{6}=\dfrac{1}{6}\times \dfrac{6\mu C}{{{\epsilon }_{0}}}=\dfrac{1\mu C}{{{\epsilon }_{0}}}=\dfrac{1}{{{\epsilon }_{0}}}\times {{10}^{-6}}N{{m}^{2}}{{C}^{-1}}$
So the correct option is $\left( 4 \right)\dfrac{1}{{{\epsilon }_{0}}}\times {{10}^{-6}}$

Note:
Gauss law can be applied to any close surface and the electric flux can be calculated. By using gauss law the electric field also can be calculated. Gauss law is an easier method to calculate the electric field due to a continuous charge distribution. Also Gauss law only can be applicable to closed surfaces. So if a surface is not close then consider a Gaussian surface and Gauss law can be applied to that surface.