Find the equivalent capacitance between `A` and `B`.`A`: area of each plate, `d`: separation between the plates.

Explanation

(a) Capacitor `①,C_(1)=(K_(1)(A)/(2)in_(0))/(d)=(K_(1)Ain_(0))/(2d)`
Capacitor `②,C_(2)=(K_(2)(A)/(2)in_(0))/(d)=(K_(2)Ain_(0))/(2d)`
Now connect a battery between `A` and `B`, as shown. The positive plates of two capacitors are connected to `A` and negative plates to `B`, hence `C_(1)` and `C_(2)` are in parallel. here, we are deciding series/parallel with the help of polarity.
`C_(eq)=C_(1)+C_(2)=((K_(1)+K_(2))/(2))(Ain_(0))/(d)`
(b) `C_(1)=(K_(1)Ain_(0))/(d//2)=(2K_(1)Ain_(0))/(d)`
`C_(2)=(K_(2)Ain_(0))/(d//2)=(2K_(2)Ain_(0))/(d)`
Now, check polarity by connecting a battery between `A` and `B`. As shown in diagram, between `A` and `B`, we have, `+,-,+,-i.e., C_(1)` and `C_(2)` are in series.
`C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2))=((2K_(1)K_(2))/(K_(1)+K_(2)))(Ain_(0))/(d)`
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