An electric field $\vec{E} = 4 x \hat{i} - (y^{2} + 1) \hat{j} N / C$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $B C G F$ are marked as $ϕ_{1} a n d ϕ_{2}$ respectively .The difference between $(|ϕ_{1} - ϕ_{2}|)$ is $({in Nm}^{2} / C)$ ____

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Detailed Solution

$\bar{E} = 4 x \hat{i} - (y^{2} + 1) \hat{j} . N / c$

Area of ABCD is $d s_{A B C D} = l b \hat{k} = 3 \times 2 \hat{k} = 6 \hat{k}$
$ϕ_{1} = \bar{E} . d {\bar{s}}_{A B C D}$ =0

Area of BCGF $d s_{B C G F} = l b \hat{i} = 2 \times 2 \hat{i} = 4 \hat{i}$

$ϕ_{2} = \bar{E} . d {\bar{s}}_{B C G F} = [4 (3) \hat{i} - (y^{2} + 1) \hat{j}] .4 \hat{i}$
$ϕ_{2} = 12 \times 4 = 48 N m^{2} / C \Rightarrow ϕ_{1} - ϕ_{2} = - 48.00 N m^{2} / C$

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