JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- Numerical
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]
Pick an option to check your answer.
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 \(]\)
\(\overrightarrow{ A }_{2}=0.3 m ^{2} \hat{ j }\)
Now \(\phi_{a}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.2 \hat{i}]=\frac{3 \times 0.2}{5} E _{0}\)
\(\phi_{b}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.3 \hat{j}]=\frac{4 \times 0.3}{5} E _{0}\)
\(\operatorname{Now} \frac{\phi_{ a }}{\phi_{ b }}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{ a }{ b }\)
\(\Rightarrow a: b=1: 2\)
\(\Rightarrow a=1\)
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