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JEE Main 2021 (17 Mar, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2021 (17 Mar, Shift 2)

Q79JEE Main 2021 · 17 MAR · SHIFT 2Easy

Electric Flux & Gauss's Law

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 $E_0 = 4.0 \times 10^{3}\ \frac{\text{N}}{\text{C}}$ . The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y-Z plane is __________ Nm2C ${}^{-1}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 640

Surface parallel to the Y-Z plane: area vector along $\hat i$ , so $\phi = E_x A$ .
$\phi = \dfrac{2}{5}(4.0\times10^3)(0.4)$
$\phi = 1600\times0.4 = 640$ N m $^2$ C $^{-1}$

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