Electric field in a region is given by overrightarrow{ E }=frac{2}{5} E _{0} hat{ i }+frac{3}{

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1. Electric Charges and Fields

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The electric field in a region is given by $\overrightarrow{ E }=\frac{2}{5} E _{0} \hat{ i }+\frac{3}{5} E _{0} \hat{ j }$ with $E _{0}=4.0 \times 10^{3}\, \frac{ N }{ C } .$ The flux of this field through a rectangular surface area $0.4 \,m ^{2}$ parallel to the $Y - Z$ plane is ....... $Nm ^{2} C ^{-1}$

A

$624$

B

$661$

C

$620$

D

$640$

(JEE MAIN-2021)

Solution

$\phi= E _{ x } A \Rightarrow \frac{2}{5} \times 4 \times 10^{3} \times 0.4=640$

Standard 12

Physics

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