A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant $K_{1}$ and $K_{2}$ of same area $\frac{A}{2}$ and thickness $\frac{d}{2}$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

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C eq = C 1 + C 2 C 3 C 2 + C 3 C eq = A ε 0 2 d + K 1 A ε 0 d × K 2 A ε 0 d K 1 A ε 0 d + K 2 A ε 0 d C eq = A ε 0 2 d + A ε 0 d K 1 K 2 K 1 + K 2 = A ε 0 d 1 2 + K 1 K 2 K 1 + K 2

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