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Physics Question on electrostatic potential and capacitance

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1,K2,K3,K4K_1, K_2, K_3, K_4 arranged as shown in the figure. The effective dielectric constant K will be :

A

K=(K1+K2)(K3+K4)2(K1+K2+K3+K4)K = \frac{\left(K_{1} +K_{2} \right)\left(K_{3}+ K_{4}\right)}{2\left(K_{1} + K_{2}+K_{3} +K_{4}\right)}

B

K=(K1+K2)(K3+K4)(K1+K2+K3+K4)K = \frac{\left(K_{1} +K_{2} \right)\left(K_{3}+ K_{4}\right)}{\left(K_{1} + K_{2}+K_{3} +K_{4}\right)}

C

K=(K1+K4)(K2+K3)2(K1+K2+K3+K4)K = \frac{\left(K_{1} +K_{4} \right)\left(K_{2}+ K_{3}\right)}{2\left(K_{1} + K_{2}+K_{3} +K_{4}\right)}

D

K=(K1+K3)(K2+K4)K1+K2+K3+K4K = \frac{\left(K_{1} +K_{3} \right)\left(K_{2}+ K_{4}\right)}{K_{1} + K_{2}+K_{3} +K_{4}}

Answer

K=(K1+K3)(K2+K4)K1+K2+K3+K4K = \frac{\left(K_{1} +K_{3} \right)\left(K_{2}+ K_{4}\right)}{K_{1} + K_{2}+K_{3} +K_{4}}

Explanation

Solution

C12=C1C2C1+C2=k1∈0L2×Ld/2.k2[∈0L2×L]d/2(k1+k2)[∈0.L2×Ld/2]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]}
C12=k1k2k1+k2∈0L2dC_{12} = \frac{k_{1}k_{2}}{k_{ 1} +k_{2}} \frac{\in_{0} L^{2}}{d}
in the same way we get, C34=k3k4k3+k4∈0L2dC_{34} = \frac{k_{3}k_{4} }{k_{3} +k_{4}} \frac{ \in_{0} L^{2}}{d}
∴Ceq=C12+C34=[k1k2k1+k2+k3k4k3+k4]∈0L2d\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 keq=k,Ceq=k∈0L2dk_{eq} = k , C_{eq} = \frac{k \in_{0} L^{2}}{d} .....(ii)
on comparing equation (i) to equation (ii), we get
keq=k1k2(k3+k4)+k3k4(k1+k2)(k1+k2)(k3+k4)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
C13=k1∈0LL2d/2+k3∈0L.L2d/2C_{13} = \frac{k_{1} \in_{0} L \frac{L}{2}}{d/2} + k_{3} \in_{0} \frac{L. \frac{L}{2}}{d/2}
=(k1+k3)∈0L2d= \left(k_{1} + k_{3}\right) \frac{\in_{0} L^{2}}{d}
C24=(k2+k4)∈0L2dC_{24} = \left(k_{2 } + k_{4}\right) \frac{\in_{0} L^{2}}{d}
Ceq=C13C24C13C24=(k1+k3)(k2+k4)(k1+k2+k3+k4)∈0L2dC_{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}
=k∈0L2d= \frac{k \in_{0}L^{2}}{d}
k=(k1+k3)(k2+k4)(k1+k2+k3+k4)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)