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 $0.2m^2$ (parallel to $y-z$ plane ) to that of the surface of area $0.3m^2$ (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

Answer: 1

$\vec{E}=\left(\frac{3E_0}{5}\hat{i}+\frac{4E_0}{5}\hat{j}\right)\frac{N}{C}$
$A_1=0.2m^2$ [parallel to $y-z$ plane $]$
$={\vec{A}}_1=0.2m^2\hat{i}$
$A_2=0.3m^2[$ parallel to $x-z$ plane $]$
${\vec{A}}_2=0.3m^2\hat{j}$
Now ${\varphi }_a=\left[\frac{3E_0}{5}\hat{i}+\frac{4E_0}{5}\hat{j}\right]\cdot [0.2\hat{i}]=\frac{3\times 0.2}{5}{E}_0$
${\varphi }_b=\left[\frac{3E_0}{5}\hat{i}+\frac{4E_0}{5}\hat{j}\right]\cdot [0.3\hat{j}]=\frac{4\times 0.3}{5}{E}_0$
Now $\frac{{\varphi }_a}{{\varphi }_b}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{a}{b}$
$\Rightarrow a:b=1:2$
$\Rightarrow a=1$