$(A)\;\frac{acd}{2} \in_0 \\ (B)\;\frac{abc \in _o}{2} \\ (C)\;\frac{abd \in _0}{2} \\ (D)\;None $

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$\phi ABCD=-acd$

$\phi CDEF=-bcE_0$

$\phi ABEF=bcE_0 + c \int \limits _o^a (d+x) dy$

$\qquad= bcE_0 +acd+ c \int \limits_0^a x dxy$

$\qquad= bcE_0 +acd +\large\frac{ca}{b} \int\limits_0^b x dx$

$\bigg[ \therefore \large\frac{x}{b}+ \frac{y}{a} $$=1=> \frac{dx}{b} =\frac{-dy}{a}\bigg]$

$\qquad= bcE_0 +acd +\large\frac{acb}{2}$

$\phi _{net}=\large\frac{q_{in}}{\in _0}$

$q_{in}=\large\frac{abc \in _o}{2}$

hence b is the correct answer.

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