Find the capacitances of the capacitance shown In figure .The plate area is `A`and the separation between the plate is `d`. Different dielectric slabe in a particular part of the figure area of the same thickness and the entire gap between the plate is filled with the dielectric slabs.

Explanation

`C_1 = K_1 eplison_0 A / (d / 2) , C_2 = K_2 eplison_0 A / (d / 2)`
` C_2 = 2K_1 eplison_0 A / d , C_2 = 2K_2 eplison_0 A / d` (fig)
` C = C_1 xx C_2 / C_1 + C_2 `
`((2K_1 epsilon_0 A ) / (d)xx(2K_2 epsilon_0 A ) / (d))/((2K_1 epsilon_0 A ) / (d)+(2K_2 epsilon_0 A ) / (d))`
`2 K_1 K_2 epsilon_0 A / d (K_1 + K_2)
(b) Similarly as (a) but this figure three parts
` C_1 = K_1 epsilon_0 A / (d /3) = 3 K_1 epsilon_0 A / d `
(fig.) `C_2 = 3 K_2 epsilon_0 A / d `
C_3 = 3 K_3 epsilon_0 A / d `
C = C_1 xx C_2 xx C_2 / C_1 C_2 + C_2 C_3 + C_3 C_1 `
` = 3 epsilon_0 K_1 K_2 K_3 / K_1 K_2 + K_2 K_3 + K_3 K_1`
( c) C_1 = (K_2 epsilon_0 A / 2) / d = K_1 epsilon_0 A / 2d`
C_2 = K_2 epsilon_0 A / 2d `
These two part are in parallel ,
So, ` C = C_1 + C_2 `
` = epslion_0 A / 2d (K_1 + K_2)`.
,