The capacitance of a parallel plate capacitor with plate area `A` and separation `d` is `C`. The space between the plates in filled with two wedges of dielectric constants `K_(1)` and `K_(2)` respectively. Find the capacitance of resulting capacitor.

Answer

`[(epsilon_(0)AK_(1)K_(2))/((K_(1)-K_(2))d) "log" ((K_(1))/K_(2)))]`