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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]
$2$
$3$
$4$
$1$
Solution
$\overrightarrow{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right) \frac{ N }{ C }$
$A_{1}=0.2 m ^{2}$ [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 $\phi_{a}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.2 \hat{i}]=\frac{3 \times 0.2}{5} E _{0}$
$\phi_{b}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}\right] \cdot[0.3 \hat{j}]=\frac{4 \times 0.3}{5} E _{0}$
$\operatorname{Now} \frac{\phi_{ a }}{\phi_{ b }}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{ a }{ b }$
$\Rightarrow a: b=1: 2$
$\Rightarrow a=1$
