Electric Flux & Gauss's Law question from JEE Main 2021 (1 Sep, Shift 2)

Answer: (b)

1. $\vec E = 150y^2\hat j$ has flux only through the faces $y = 0$ and $y = 0.5$ m.
2. At $y = 0$: $E = 0$. At $y = 0.5$: $\phi = 150(0.5)^2(0.5)^2 = 9.375$ N m$^2$/C.
3. $q = \varepsilon_0\phi = 8.85\times10^{-12}\times9.375 \approx 8.3\times10^{-11}$ C
4. So the answer is (b).