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Electrostatics question

2020 · 9 Jan · Shift 2 · Q42
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Electrostatics question

2020 · 9 Jan · Shift 2 · Q42

JEE MainPhysicsElectrostaticsNumerical+4 / −1
An electric field E→=4xi^−(y2+1)j^\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat jE=4xi−(y2+1)j​ N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI{\phi _I}ϕI​ and ϕII{\phi _{II}}ϕII​ respectively. The difference between (ϕI−ϕII)\left( {{\phi _I} - {\phi _{II}}} \right)(ϕI​−ϕII​) is (in Nm2/C) ‾\underline{\hspace{2cm}}​. JEE Main 2020 (Online) 9th January Evening Slot Physics - Electrostatics Question 176 English
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