QJEEM21Ph314e: JEE Main 2021 Physics - Class 12 JEEM21Ph314e Solution: The electric field in a region is given by with . The flux of…

Question

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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^{$-$ 1}.

Accepted Answer

\phi = \overrightarrow E \,.\,\overrightarrow A

= {{{E_0}} \over 5}\left( {2\widehat i + 3\widehat j} \right)\,.\,\left( {0.4\widehat i} \right)

= {{4000} \over 5}\left( {2 \times 0.4} \right)

= 640

Nm² C^{$-$ 1}

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