Question

A plane surface of area S is inclined at an angle @ with a uniform
field Ē as shown in the figure. Find the magnitude of flux of
Ē over S.​

Answer 1

Answer ⇒ $\phi = \vec{E}\vec{A}Cos\alpha$

Explanation ⇒ By the definition of the electric flux, it is the dot product of  Electric filed intensity and area vector.

∴ Φ = Electric field intensity.Area Vector.

∴ $\phi = \vec{E}.\vec{A}$

∴ $\phi = \vec{E}\vec{A}Cos\theta$

Now, here,

θ is the Angle between the Electric field Intensity and area vector.

Here, θ = α

Hence, magnitude of the Electric flux = Φ

$\phi = \vec{E}\vec{A}Cos\alpha$

Hope it helps.