Given, the side of the cube,
s=0.5m Electric field,
E=150y2 The direction of electric field is as shown in the below figure,
At bottom surface,
y=0 As we know that, the expression of electric flux,
ϕ=E.Acosθ Here, E is the electric field passing through the cube and A is the surface area of the cube.
Substituting the values in the above equations, we get
ϕ=150y2.(0.5×0.5)×cos180° =150(0)2.(0.25)×(−1)=0 Hence, the electric flux is zero at the bottom surface.
At the top surface,
y=0.5m Electric field,
E=150y2=150(0.5)2=37.5N∕C Electric flux at the top surface,
ϕ=E.Acosθ =(37.5).(0.5×0.5)cos0° =9.375N∕C−m2 By using the Gauss's law,
ϕ= Here,
Qin= net charge enclosed in the cube
and
ε0= permittivity of the free space.
Substituting the values in the above equation, we get
9.375= Qin=8.3×10−11C The charge inside the cube is
8.3×10−11C.