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_(3) || 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)`