Fig shows a parallel plate capacitor of plate area `A` and plate separation `d`. Its entire space is filled with three different dielectric slabs of same thickness. Find the equivalent capacitance of the arrangment.

Answer

`C = (3 in_(0) A K_(1) K_(2) K_(3))/(d (K_(1) K_(2) + K_(2) K_(3) + K_(3) K_(1)))`

Explanation

This arrangement in Fig, is equivalent to three capacitors connected in series
`C_(1) = (K_(1) in_(0) A)/(d//3) = (3 K_(1) in_(0) A)/(d) , C_(2) = (3 K_(2) A)/(d)`,
`C_(3) = (3 K_(3) in_(0) A)/(d)`
`(1)/(C ) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (d)/(3 in_(0) A) ((1)/(K_(1)) + (1)/(K_(2)) + (1)/(K_(3)))`
`C = (3 in_(0) A K_(1) K_(2) K_(3))/(d (K_(1) K_(2) + K_(2) K_(3) + K_(3) K_(1)))`