A rectangular area of sides 4.0 cm and 5.0 cm is placed in an electric field E = \(\rm \left(4.0\times 10^2 \frac{N}{c}\right)\) i such that the normal unit vector for the area is \(\rm \left[\left(\frac{1}{2}\right)i+\left(\frac{\sqrt2}{2}\right)j\right]\). The electric flux through the rectangle is:

Concept:

•  The electric flux is defined as the total number of electric field lines crossing a given area.
• \(ϕ = \oint \overrightarrow E . \overrightarrow {d S}\)
• Where, E is electric field lines, S is the area.
• Flux is a scalar quantity 
• Dot product of i and j is equal to zero

Calculation:

Given, \(\overrightarrow E\) = \(\rm \left(4.0× 10^2 \frac{N}{c}\right)\)i, area of rectangular plate = 0.04 × 0.05 = 2 × 10^-3 m², normal unit vector for the area is \(\rm \left[\left(\frac{1}{2}\right)i+\left(\frac{\sqrt2}{2}\right)j\right]\)

So, \(\overrightarrow S \) = 2 × 10^-3× \(\rm \left[\left(\frac{1}{2}\right)i+\left(\frac{\sqrt2}{2}\right)j\right]\)

Now flux \(ϕ = \oint \overrightarrow E . \overrightarrow {d S}\)

\(⇒ ϕ = \overrightarrow E . \overrightarrow {S}\)

⇒ ϕ = \(\rm \left(4.0× 10^2 \frac{N}{c}\right)\)i . 2 × 10^-3 × \(\rm \left[\left(\frac{1}{2}\right)i+\left(\frac{\sqrt2}{2}\right)j\right]\) = 0.40 N m²​/C