It has been experimentally observed that the electric field in a large region of earth's atmosphere is directed vertically down. At an altitude of 300 m, the electric field is 60  Vm−1. At  an altitude of 200 m, the field is 100  Vm−1. The net amount of charge contained in the cube of 100 m edge, located between 200 m and 300 m altitude is found to be n×105  ε0C. Find the value of n.

According to Gauss'theorem, electric flux: ϕE=qε0=∫E→.dS→The surface integral in the above equation contains six terms-the surface integral over the bottom surface, the surface integral over the top surface and surface integral over the four vertical faces.For the bottom surface, both the vectors E→  and  dS→ are in the same direction. For the top surface, they act in opposite directions while for the vertical faces, they are perpendicular to each other.Hence, ϕE=∫bottomE→1.  dS→+∫topE→2.dS→+4 ∫facesE→.dS→∫bottomE1.dScos00+∫topE2.dScos1800+4  ∫facesE dScos900∫bottomE1.dS−∫topE2.  dS=E1S−E2S=(E1−E2)SAlso, ϕE=qenclosedε0q=ε0(E1−E2)S=ε0(100   Vm−1−60  Vm−1)(100   m)2=4×105ε0C