JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Medium
- Numerical
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 ....... \(Nm ^{2} C ^{-1}\)
Pick an option to check your answer.
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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