Calculate the capacitance of the parallel plate capacitor shown in the following figure between point X and Y.

Explanation

The capacitor can be broken apart into the following circuit.
(i) Series `(C_(2) "and" C_(3))`
`C_(2)= epsilon_(0)K_(2)(A)/(d) , C_(3)= epsilon_(0)k_(3)(A)/(d)`
`implies (1)/( C) = (1)/(C_(2))+(1)/(C_(3))`
(ii) Parallel `(C_(1)||C') `
`C_(1)= epsilon_(0)K_(1)(A)/(2d)`
`implies` Equivalent capacitance between X and Y is
`C= C_(1)+C'=C_(1)+(C_(2)C_(3))/(C_(2)+C_(3))`
So the capacitance is `((k_(1)k_(2)+2k_(2)k_(3)+k_(3)k_(1))/(2(k_(2)+k_(3))))(epsilon_(0)A)/(d)`