The electric field in a region is given by

Question:

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]

Solution:

(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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