The electric field in a region is given by \overrightarrow E = 2 \over 5E_0\widehat i + 3 \over 5E_0\widehat j with E_0 = 4.0 \times 10^3N \over C . The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y-Z plane is \underline\hspace2cm Nm2C - 1.

Electrostatics PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q65 with Solution

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Correct answer: 640

\vec E = \frac{2}{5}E_0\,\hat i + \frac{3}{5}E_0\,\hat j, \qquad E_0 = 4.0\times 10^3\ \text{N/C}

So the components are:

E_x = \frac{2}{5}E_0, \qquad E_y = \frac{3}{5}E_0

The rectangular surface is parallel to the Y-Z plane.

Hence its area vector is along the X-axis:

\vec A = A\,\hat i

with

A = 0.4\ \text{m}^2

Electric flux through a surface is

\Phi = \vec E \cdot \vec A

Since only the component of electric field normal to the surface contributes, only the $x$ -component matters:

\Phi = E_x A = \left(\frac{2}{5}E_0\right)(0.4)

E_x = \frac{2}{5}\times 4.0\times 10^3 = 1.6\times 10^3\ \text{N/C}

Therefore,

\Phi = (1.6\times 10^3)(0.4)

\Phi = 640\ \text{N m}^2\text{/C}

\boxed{640}