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JEE Mains · Physics · STD 12 - 1. Electric charges and fields

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 ....... \(Nm ^{2} C ^{-1}\)

  1. A \(624\)
  2. B \(661\)
  3. C \(620\)
  4. D \(640\)
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Answer & Solution

Correct Answer

(D) \(640\)

Step-by-step Solution

Detailed explanation

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