$(A)\;\frac{3 \in _0 A}{d} \\ (B)\;\frac{\in _0}{4d} \\ (C)\; \frac{2 \in _0}{d} \\ (D)\;\frac{\in _0 A}{d} $

Given plates can be rearranged as :

$C_1= \large\frac{\in _0A}{d}$

$C_2= \large\frac{\in _0A}{2d}$

$C_3= \large\frac{\in _0A}{3d}$

$C_4= \large\frac{\in _0A}{2d}$

$C_5= \large\frac{\in _0A}{d}$

Equivalent of $C_1 \& C_2$

$\qquad= \large\frac{\Large\frac{\in_0 A}{d} \times \frac{\in _0 A}{2d}}{\large\frac{\in_0 A}{d} + \frac{\in _0 A}{2d}}$

$\qquad= \large\frac{ \in _0 A}{3d}$

Equivalent of $C_4 \& C_5$

$\qquad= \large\frac{ \in _0 A}{3d}$

Effective capacitance between A and B is

$\large\frac{ \in _0 A}{3d}+\frac{ \in _0 A}{3d}+\frac{ \in _0 A}{3d}=\large\frac{ \in _0 A}{d}$

Hence D is the correct answer.

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