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An electric field E=4x i-(y^2+1) jN/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are…

JEE Main 2020 — Physics Electromagnetism

2020integereasy

An electric field E⃗=4xi^−(y2+1)j^N/C\vec{E}=4x\hat{i}-({y}^{2}+1)\hat{j}N/CE=4xi^−(y2+1)j^​N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCDABCDABCD and BCGFBCGFBCGF are marked as ϕI{\phi }_{\text{I}}ϕI​ and ϕII{\phi }_{\text{II}}ϕII​ respectively. The difference between (ϕI−ϕII)({\phi }_{I}-{\phi }_{II})(ϕI​−ϕII​) is (in Nm2/CN{m}^{2}/CNm2/C ) ____________.

JEE Main 2020 Physics, Electrostatics — question figure

Official previous-year question

Held on 9 Jan 2020 · Verified 6 Jul 2026.

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