Find the capacitance between A and B if two dielectric slabs (each of thickness d) of dielectric constants `K_(1)` and `K_(2)` and areas `K_(1)` and `K_(2)` and inserted between the plates of a parallel plate capacitor of plate area `A`.
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Explanation

`C_(1)=(A_(1)K_(1)in_(0))/(d),C_(2)=(A_(2)K_(2)in_(0))/(d)`
`E_(1)=(V)/(d)=(sigma_(1))/(K_(1)in_(0)),E_(2)=(V)/(d)=(sigma_(2))/(K_(2)in_(0))`
`sigma_(1)=(K_(1)in_(0)V)/(d) " " sigma_(2)=(K_(2)in_(0)V)/(d)`
`C=(Q_(1)+Q_(2))/(V)=(sigma_(1)A_(1)+sigma_(2)A_(2))/(V)=(K_(1)in_(0)A_(2))/(d)`
The combination is equivalent to : `C=C_(1)+C_(2)`