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Six plates of same area A are arranged as shown. Equivalent capacitance of arrangement between A and B is :


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

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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.
answered Jan 13, 2014 by meena.p

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