Q86 / 90 · JEE Main 2021 (17 Mar Shift 2) · Physics

The electric field in a region is given by $\vec{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} N C^{- 1} .$ The flux of this field through a rectangular surface, area $0.4 m^{2}$ parallel to the $Y - Z$ plane is _______ $N m^{2} C^{- 1}$ .

Correct answer: 640

Solution

ϕ = E_{x} A \Rightarrow \frac{2}{5} \times 4 \times 10^{3} \times 0.4 = 640 N m^{2} C^{- 1}

Hence, the correct answer is 640.

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