JEE Mains · Physics · STD 12 - 1. Electric charges and fields
The electric field in a region is given \(\overrightarrow{ E }=\left(\frac{3}{5} E _{0} \hat{ i }+\frac{4}{5} E _{0} \hat{ j }\right) \frac{ N }{ C } .\) The ratio of flux of reported field through the rectangular surface of area \(0.2\, m ^{2}\) (parallel to \(y - z\) plane) to that of the surface of area \(0.3\, m ^{2}\) (parallel to \(x - z\) plane \()\) is \(a : b ,\) where \(a =\) ............. [Here \(\hat{ i }, \hat{ j }\) and \(\hat{ k }\) are unit vectors along \(x , y\) and \(z-\)axes respectively]
- A \(2\)
- B \(3\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right) \frac{ N }{ C }\) \(A_{1}=0.2 m ^{2}\) [parallel to \(y - z\) plane \(]\) \(=\overrightarrow{ A }_{1}=0.2 m ^{2} \hat{ i }\) \(A_{2}=0.3 m ^{2}\) [parallel to \(x - z\) plane \(]\)…
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