Question

The electric field in a region is given by $\overrightarrow{E}=(\frac{3}{5}{E}_o\hat{i}+\frac{4}{5}{E}_o\hat{j})\text{N/C}$. The ratio of flux of reported field through the rectangular surface of area $0.2{\text{m}}^2$ $($parallel to $y-z$ plane$)$ to that with the surface of area $0.3{\text{m}}^2$ $($parallel to $x-z$ plane$)$ is $a:2$, where $a=$_______.
$[$Here $\hat{i},\hat{j}$ and $\hat{k}$ are unit vectors along $x$, $y$ and $z$ - axes respectively$]$

Answer 1

Given:
$\overrightarrow{E}=(\frac{3}{5}{E}_o\hat{i}+\frac{4}{5}{E}_o\hat{j})\text{N/C}$
$\overrightarrow{A_1}=\left(0.2\hat{i}\right){\text{m}}^2$
$\overrightarrow{A_2}=\left(0.3\hat{j}\right){\text{m}}^2$

We know that,
$\varphi =\overrightarrow{E}.\overrightarrow{A}$

So,
${\varphi }_1=(\frac{3}{5}{E}_o\hat{i}+\frac{4}{5}{E}_o\hat{j}).\left(0.2\hat{i}\right)$

$\Rightarrow {\varphi }_1=0.12E_o$

${\varphi }_2=(\frac{3}{5}{E}_o\hat{i}+\frac{4}{5}{E}_o\hat{j}).\left(0.3\hat{j}\right)$

$\Rightarrow {\varphi }_2=0.24E_o$

Further,
$\frac{{\varphi }_1}{{\varphi }_2}=\frac{0.12E_o}{0.24E_o}=\frac{1}{2}=1:2=a:2$

$\Rightarrow a=1$