ϕ=ExA⇒52×4×103×0.4=640Nm2C−1
The electric field in a region is given by E= 2 5 E_0 i+ 3 5 E_0 j with E_0=4.0× 10^3NC^-1. The flux of this field through a rectangular surface,…
JEE Main 2021 — Physics Electromagnetism
The electric field in a region is given by E=52E0i^+53E0j^ with E0=4.0×103NC−1. The flux of this field through a rectangular surface, area 0.4m2 parallel to the Y−Z plane is _______ Nm2C−1.
Held on 17 Mar 2021 · Verified 6 Jul 2026.
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