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JEE Main 2020 (Online) 9th January Evening Slot
Numerical
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-0
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) _______. JEE Main 2020 (Online) 9th January Evening Slot Physics - Electrostatics Question 125 English
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