The equivalent capacitance between plates 'A' and 'B' (A-area of each plate, d-separation between plates) ( $ϵ_{0}$ - permittivity of free space) is

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The equivalent capacitance between plates 'A' and 'B' (A-area of each plate, d-separation between plates) ( ϵ 0 ​ - permittivity of free space) is

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The given capacitor can be considered as two parts connected in parallel: the left half and the right half.

For the left half, the capacitor $C_{1}$ has a plate area of $A /2$ , a separation distance of $d$ , and is filled with a dielectric of constant $K_{1} = 4$ .
Its capacitance is given by:
$C_{1} = \frac{K _{1} ϵ _{0} ( A /2 )}{d} = \frac{4 ϵ _{0} A}{2 d} = \frac{2 ϵ _{0} A}{d}$

The right half consists of two capacitors, $C_{2}$ and $C_{3}$ , connected in series.
Capacitor $C_{2}$ has an area of $A /2$ , a separation of $d /2$ , and a dielectric constant $K_{2} = 3$ .
$C_{2} = \frac{K _{2} ϵ _{0} ( A /2 )}{d /2} = \frac{3 ϵ _{0} A}{d}$

Capacitor $C_{3}$ has an area of $A /2$ , a separation of $d /2$ , and a dielectric constant $K_{3} = 6$ .
$C_{3} = \frac{K _{3} ϵ _{0} ( A /2 )}{d /2} = \frac{6 ϵ _{0} A}{d}$

The equivalent capacitance of the right half ( $C_{r i g h t}$ ) is the series combination of $C_{2}$ and $C_{3}$ :
$C_{r i g h t} = \frac{C _{2} C _{3}}{C _{2} + C _{3}} = \frac{( \frac{3 ϵ _{0} A}{d} ) ( \frac{6 ϵ _{0} A}{d} )}{\frac{3 ϵ _{0} A}{d} + \frac{6 ϵ _{0} A}{d}} = \frac{18}{9} \frac{ϵ _{0} A}{d} = \frac{2 ϵ _{0} A}{d}$

The total equivalent capacitance $C_{e q}$ between plates A and B is the parallel combination of the left and right halves:
$C_{e q} = C_{1} + C_{r i g h t} = \frac{2 ϵ _{0} A}{d} + \frac{2 ϵ _{0} A}{d} = \frac{4 ϵ _{0} A}{d}$

Answer: $\frac{4 A \in _{0}}{d}$

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