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

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Solution

Given:
$\to E = (\frac{3}{5} E_{o}^i + \frac{4}{5} E_{o}^j) N/C$
$_{1} = (0.2^i) m^{2}$
$_{2} = (0.3^j) m^{2}$

We know that,
$ϕ = \to E . \to A$

So,
$ϕ_{1} = (\frac{3}{5} E_{o}^i + \frac{4}{5} E_{o}^j) . (0.2^i)$

$\Rightarrow ϕ_{1} = 0.12 E_{o}$

$ϕ_{2} = (\frac{3}{5} E_{o}^i + \frac{4}{5} E_{o}^j) . (0.3^j)$

$\Rightarrow ϕ_{2} = 0.24 E_{o}$

Further,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{0.12 E_{o}}{0.24 E_{o}} = \frac{1}{2} = 1 : 2 = a : 2$

$\Rightarrow a = 1$

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Given: →E=(35Eo^i+45Eo^j) N/C −→A1=(0.2 ^i) m2 −→A2=(0.3 ^j) m2 We know that, ϕ=→E.→A So, ϕ1=(35Eo^i+45Eo^j).(0.2 ^i) ⇒ϕ1=0.12Eo ϕ2=(35Eo^i+45Eo^j).(0.3 ^j) ⇒ϕ2=0.24Eo Further, ϕ1ϕ2=0.12Eo0.24Eo=12=1:2=a:2 ⇒a=1