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)epsi_(0))/(d), C_(2)=(A_(2)K_(2)epsi_(0))/(d)`
`E_(1)=V/d=(sigma_(1))/(K_(1)epsi_(0)), E_(2)=V/d=(sigma_(2))/(K_(2)epsi_(0))`

`sigma_(1)=(K_(1)epsi_(0)V)/(d), sigma_(2)=(K_(2)epsi_(0)V)/(d)`
`C=(Q_(1)+Q_(2))/(V)=(sigma_(1)A_(1)+sigma_(2)A_(2))/(V)=(K_(1)epsi_(0)A_(1))/(d)+(K_(2)epsi_(0)A_(2))/(d)`

The combination is equivalent to :
`therefore C=C_(1)+C_(2)`