JEE Main 17 March 2021 Shift 2 Solved Paper

Given,
The electric field in the region, E=‌

2

5

E0

∧

i

+‌

3

5

E0

∧

j

Here, E0=4×103‌

N

C

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=‌

2

5

E0(0.4)
⇒‌‌φ=‌

2

5

(4×103)(0.4)=640Nm2C−1
Hence, the electric flux of the surface parallel to the Y−Z plane is 640Nm2C−1.