Given,
The electric field in the region,
E=E0+E0 Here,
E0=4×103 Area of the rectangular surface,
A=0.4m2 The direction of electric field vector and area vector is same, so the angle between the electric field vector and area vector is 0 .
As we know the expression of electric flux,
φ=E⋅Acosθ...(i)
Here,
E is the electric field vector, and
A is the surface area of the surfaces.
Consider the surface parallel to the
Y−Z plane, so the area vector,
A=0.4 i {m}^{2}
Substituting the values in Eq. (i), we get
φ=E⋅Acos0∘⇒φ=E0(0.4) ⇒φ=(4×103)(0.4)=640Nm2C−1 Hence, the electric flux of the surface parallel to the
Y−Z plane is
640Nm2C−1.