The electric field in a region is given by $\vec{E}=\left(\frac{3}{5} E_{0} \hat{i}+\frac{4}{5} E_{0} \hat{j}\right)^{\frac{N}{c}}$. The ratio of flux of reported field
through the rectangular surface of area $0.2 \mathrm{~m}^{2}$ (parallel to $\mathrm{y}-\mathrm{z}$ plane ) to that of the surface of area
$0.3 \mathrm{~m}^{2}$ (parallel to $\mathrm{x}-\mathrm{z}$ plane ) is a : $\mathrm{b}$, where $\mathrm{a}=$ (round off to nearest integer)
[Here $\hat{\mathrm{i}}, \hat{\mathrm{j}}$ and $\hat{\mathrm{k}}$ are unit vectors along $\mathrm{x}, \mathrm{y}$ and z-axes respectively]
(1)
$\phi=\vec{E}_{1} \vec{A}$
$\vec{A}_{a}=0.2 \hat{i}$
$\vec{A}_{b}=0.3 \hat{j}$
$\phi_{a}=\left(\frac{3}{5} E_{0} \hat{i}+\frac{4}{5} E_{0} \hat{j}\right) \cdot 0.2 \hat{i}$
$\phi_{a}=\frac{3}{5} E_{0} \times 0.2$
$\phi_{a}=\left(\frac{3}{5} E_{0} \hat{i}+\frac{4}{5} E_{0} \hat{j}\right) \cdot 0.3 \hat{j}$
$\phi_{b}=\frac{4}{5} E_{0} \times 0.3$
$\frac{a}{b}=\frac{\phi_{a}}{\phi_{b}}=\frac{\frac{3}{5} E_{0} \times 0.2}{\frac{4}{5} E_{0} \times 0.3}=\frac{6}{12}=0.5$
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