A point charge q is placed at the centre of the cubical box. If $\mathrm{q}=12000{\in }_0 C$, find the flux through shaded area of surface(in N -m²/C).
 

Total flux coming out of the cubical box is given as ${\mathrm{\varphi }}_0=\frac{\mathrm{q}}{{\mathrm{ϵ}}_0}$
This total flux is emerging equally from each of the six face of the box. Thus flux through each face can be given as $\mathrm{\varphi }=\frac{\mathrm{q}}{6{\mathrm{ϵ}}_0}$
The flux through shaded portion would be one fourth of the flux through each face as through all the four triangular portions flux would be same due to symmetry thus we have ${\mathrm{\varphi }}^{\mathrm{\prime }}=\frac{\mathrm{\varphi }}{4}=\frac{\mathrm{q}}{24{\mathrm{ϵ}}_0}=\frac{12000{\mathrm{ϵ}}_0}{24{\mathrm{ϵ}}_0}=500 \mathrm{N}-{\mathrm{m}}^2/\mathrm{C}$