Q. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure.
The effective dielectric constant K will be :

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Solution:

$C_{12} = \frac{C_{1} C_{2}}{C_{1 }+ C_{2}} = \frac{\frac{k_{1} \in_{0} \frac{L}{2} \times L }{d/2} . \frac{k_{2} \left[\in_{0} \frac{L}{2} \times L\right]}{d/2}}{\left(k_{1} +k_{2}\right) \left[\frac{\in_{0} . \frac{L}{2} \times L}{d/2}\right]} $
$ C_{12} = \frac{k_{1}k_{2}}{k_{ 1} +k_{2}} \frac{\in_{0} L^{2}}{d} $
in the same way we get, $ C_{34} = \frac{k_{3}k_{4} }{k_{3} +k_{4}} \frac{ \in_{0} L^{2}}{d} $
$ \therefore C_{eq} = C_{12} + C_{34} = \left[\frac{k_{1}k_{2}}{k_{1} + k_{2}} + \frac{k_{3}k_{4}}{k_{3} + k_{4}} \right] \frac{\in_{0} L^{2}}{d} $ ....(i)
Now if $ k_{eq} = k , C_{eq} = \frac{k \in_{0} L^{2}}{d} $ .....(ii)
on comparing equation (i) to equation (ii), we get
$ k_{eq} = \frac{k_{1}k_{2} \left(k_{3} +k_{4}\right)+k_{3} k_{4} \left(k_{1 }+k_{2}\right)}{\left(k_{1} +k_{2} \right)\left(k_{3}+ k_{4}\right)} $
This does not match with any of the options so probably they have assumed the wrong combination
$ C_{13} = \frac{k_{1} \in_{0} L \frac{L}{2}}{d/2} + k_{3} \in_{0} \frac{L. \frac{L}{2}}{d/2} $
$ = \left(k_{1} + k_{3}\right) \frac{\in_{0} L^{2}}{d} $
$ C_{24} = \left(k_{2 } + k_{4}\right) \frac{\in_{0} L^{2}}{d} $
$ C_{eq} = \frac{C_{13} C_{24}}{C_{13} C_{24}} = \frac{\left(k_{1} + k_{3} \right)\left(k_{2} +k_{4}\right) }{\left(k_{1} + k_{2} + k_{3} + k_{4}\right)} \frac{\in _{0} L^{2}}{d} $
$ = \frac{k \in_{0}L^{2}}{d} $
$ k = \frac{\left(k_{1} +k_{3}\right)\left(k_{2} +k_{4}\right)}{\left(k_{1} +k_{2 }+ k_{3} + k_{4}\right) }$
However this is one of the four options. It must be a "Bonus" logically but of the given options probably they might go with (4)

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