Electric Flux & Gauss's Law question from JEE Main 2025 (4 Apr, Shift 1)

Answer: (c)

1. Each sheet gives $\frac{\sigma}{2\varepsilon_0}$ pointing away from it. Between the two identical sheets these cancel; outside they add to $\frac{\sigma}{\varepsilon_0}$.
2. At A and B only the sphere's field remains, $E \propto \frac{1}{d^2}$. A is closer to the sphere, so $E_A > E_B$.
3. At C the total field points left, at D it points right, and D is much closer to the sphere than C, so $\vec{E}_C \neq \vec{E}_D$ (different direction and magnitude).
4. So the answer is (c) $\vec{E}_C \neq \vec{E}_D;\ E_A > E_B$.