In the figure a capacitor is filled with dielectrics. The resultant capacitance is

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The given capacitor can be considered as a combination of three capacitors $C_{1}$ , $C_{2}$ , and $C_{3}$ .

The left half consists of two capacitors $C_{1}$ and $C_{2}$ in series, each with plate area $A /2$ and plate separation $d /2$ .

$C_{1} = \frac{K _{1} ε _{0} ( A /2 )}{d /2} = \frac{K _{1} ε _{0} A}{d}$

$C_{2} = \frac{K _{2} ε _{0} ( A /2 )}{d /2} = \frac{K _{2} ε _{0} A}{d}$

The equivalent capacitance of this series combination is $C_{12}$ :

$C_{12} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{( \frac{K _{1} ε _{0} A}{d} ) ( \frac{K _{2} ε _{0} A}{d} )}{\frac{K _{1} ε _{0} A}{d} + \frac{K _{2} ε _{0} A}{d}} = \frac{ε _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

The right half is a single capacitor $C_{3}$ with plate area $A /2$ and plate separation $d$ .

$C_{3} = \frac{K _{3} ε _{0} ( A /2 )}{d} = \frac{K _{3} ε _{0} A}{2 d}$

The combination $C_{12}$ is in parallel with $C_{3}$ . The total equivalent capacitance $C_{e q}$ is:

$C_{e q} = C_{12} + C_{3} = \frac{ε _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}}) + \frac{K _{3} ε _{0} A}{2 d}$

$C_{e q} = (\frac{K _{3}}{2} + \frac{K _{1} K _{2}}{K _{1} + K _{2}}) \frac{ε _{0} A}{d}$

Answer: $(\frac{K _{3}}{2} + \frac{K _{1} K _{2}}{K _{1} + K _{2}}) \cdot \frac{ε _{0} A}{d}$

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