If the area of each plate is A and then successive separations are `d, 2d` and `3d`, then find the equivalent capacitance across A and B.
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Answer

`(epsilon_(0)A)/(4d)`

Explanation

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Plates can be rearranged as shown in
plates `1` and `2` and plates `3` and `4` form two capacitors, which are in series between `A` and `B` plates `2` and `3` do not form any capacitor as they at the same potential.
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`C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2))=((epsilon_(0)A/d)xx((epsi_(0)A)/3d))/((epsi_(0)A)/(d)+(epsi_(0)A)/(3d))=(epsi_(0))/(4d)`