91Ó°ÊÓ

It is given that a charge qsits at the back comer of a cube.as shown in following figure

The flux of electric field through the shaded side need to be evaluated.

If there is a surface area enclosing a volume, possessing a charge inside the volume then the electric field due to the surface or volume charge is given as$âˆ\circledR E.da=\frac{q}{{\mathrm{Î\mu }}_0}$

Here${}^{\mathbf{q}}$is the total charge inside the volume,${}^{{\mathbf{Î\mu }}_{\mathbf{0}}}$is the permittivity of free surface.

The left side expression of the gauss law,${}^{âˆ\circledR E.da=\frac{q_{en\cos ed}}{E_0}}$, that is ${}^{âˆ\circledR E.da}$is called the electric flux and it is equal to the quantity ${}^{\frac{{\mathit{q}}_{\mathbf{enclosed}}}{{\mathrm{Î\mu }}_0}}$

There are 4 cubes above the charge 2 cubes at the side of the charge, and 2 cubes below the charge. Thus there are 8 cubes in total surrounding the charge

The charge passes through each cube equally. So, the charge per face of each cube becomes ${}^{\frac{1}{6}}$.The ${}^{\frac{1}{6}\mathrm{th}}$part is quarter of the large cube as shown in following figure,

So, the total flux per face of each cube becomes ${}^{\frac{1}{6}\left(\frac{1}{4}\right)}$.that is ${}^{\frac{1}{24}}$.Thus, the flux through the shaded area is

Thus, the electric flux through the shade area is $âˆ\dots =\frac{q}{24{\mathrm{Î\mu }}_0}.$