Electric Flux & Gauss's Law question from JEE Main 2020 (7 Jan, Shift 1)

Answer: (d)

1. Each sheet gives $\frac{\sigma}{2\varepsilon_0}$, perpendicular to the sheet and pointing away from it.
2. Lower (horizontal) sheet: region is above it, so $\vec{E}_1 = \frac{\sigma}{2\varepsilon_0}\hat{y}$.
3. Inclined sheet (at $30^\circ$): region is below it, so its normal away from the sheet is $-\sin 30^\circ\,\hat{x} - \cos 30^\circ\,\hat{y}$: $\vec{E}_2 = \frac{\sigma}{2\varepsilon_0}\left(-\frac{1}{2}\hat{x} - \frac{\sqrt{3}}{2}\hat{y}\right)$.
4. $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\sigma}{2\varepsilon_0}\left[\left(1 - \frac{\sqrt{3}}{2}\right)\hat{y} - \frac{\hat{x}}{2}\right]$
5. So the answer is (d).