The electric field in a region is given by E = 2/5E₀ i + 3/5E₀ j with E₀ = 4.0 × 10³N/C. The flux of this field through a rectangular surface area 0.4 m² parallel to the Y-Z plane is _ _ _ _ _ Nm²C⁻¹.

Electrostatics: JEE Main 2021 Physics question

JEE Main 2021 · 17 Mar, Evening shift · Q63

JEE Main 202117 Mar, Evening shiftPhysicsMediumElectrostatics167× since 2002

The electric field in a region is given by

\overrightarrow E = {2 \over 5}{E_0}\widehat i + {3 \over 5}{E_0}\widehat j

with

{E_0} = 4.0 \times {10^3}{N \over C}

. The flux of this field through a rectangular surface area 0.4 m² parallel to the Y-Z plane is __________ Nm²C^

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Your answerType your answer. Integers and decimals are accepted.

JEE Main 2021 Electrostatics Question with Answer (17 Mar Evening shift, Q63) | JEE Arc