QJEEM20Ph3d71: JEE Main 2020 Physics - Class 12 JEEM20Ph3d71 Solution: An electric field N/C passes through the box shown in figure.…

An electric field $\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$ N/C
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as ${\phi _I}$ and ${\phi _{II}}$
respectively. The difference between $\left( {{\phi _I} - {\phi _{II}}} \right)$ is (in Nm²/C) _______.