When the space between the plates of a parallel plate condenser is completely filled with two slabs of dielectric constants `K_(1)` and `K_(2)` and each slab having area `(A)/(2)` and thickness equal to distance of seperation `d` as shown in the figure.

Fig. the equivalent circuit is as shown.
.

Explanation

(a) Capacity of the left half `C_(1)=K_(1)(epsilon_(0)A)/(2d)`
(b) Capacity of the right half `C_(2)=K_(2) (epsilon_(0)A)/(2d)`
( c) `C_(1)` and `C_(2)` may be supposed to ve connected in parallel then effective capacity
`C=C_(1)+C_(2)=(epsilon_(0)A)/(d)((K_(1)+K_(2))/(2))`
`C=C_(0)((K_(1)+K_(2))/(2))` where `C_(0)` is capacity of capacitor without dielectric.
(d) Effective dielectric constant `K=(K_(1)+K_(2))/(2)`.