JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the side of the cube, s=0.5m
Electric field, E=150y2

^

j

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

^

j

=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=

Qin

ε0

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

8.85×10−12

Qin=8.3×10−11C
The charge inside the cube is 8.3×10−11C.