Question

The electric field in a region is given $\vec{ E }=\left(\frac{3}{5} E _{0} \hat{ i }+\frac{4}{5} E _{0} \hat{ j }\right) \frac{ N }{ C } .$ 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 $]$

Answer 1

$\vec{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right) \frac{ N }{ C }$
$A _{1}=0.2 m ^{2}$ [parallel to $y - z$ plane $]$
$=\vec{ A }_{1}=0.2 m ^{2} \hat{ i }$
$A _{2}=0.3 m ^{2}[$ parallel to $x - z$ plane $]$
$\vec{ A }_{2}=0.3 m ^{2} \hat{ j }$
Now $\phi_{a}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E _{0}}{5} \hat{ j }\right] \cdot[0.2 \hat{ i }]=\frac{3 \times 0.2}{5} E _{0}$
$\phi_{ b }=\left[\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right] \cdot[0.3 \hat{ j }]=\frac{4 \times 0.3}{5} E _{0}$
Now $\frac{\phi_{ a }}{\phi_{ b }}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{ a }{ b }$
$\Rightarrow a : b =1: 2$
$\Rightarrow a =1$