MHTCET 16 Apr 2026 Shift 1 Paper

Concept:Electric flux is the dot product of the electric field vector and the area vector of the surface. The area vector is always perpendicular to the surface plane.Explanation:The square has side length $l$, so its area is $l^2$. Since the square is parallel to the $Y$-$Z$ plane, its area vector points along the $X$-axis. Therefore, $\vec{A} = l^2 \hat{i}$. The given electric field is $\vec{E} = a \hat{i} + b \hat{j}$. Electric flux is calculated as $\phi = \vec{E} \cdot \vec{A}$. Substituting the vectors, $\phi = (a \hat{i} + b \hat{j}) \cdot (l^2 \hat{i})$. Only the $X$-component of the field contributes because $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$. So, $\phi = a l^2$. The $b \hat{j}$ component lies parallel to the square surface and does not cross it, so it contributes zero flux.Answer:The net electric flux is $a l^2$, which corresponds to option D.