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:
Answer: (A) $\frac{\varepsilon_0A}{d}\left(\frac{1}{2} + \frac{K_1K_2}{K_1+K_2}\right)$
Half the plate area has air across the full gap $d$: $C_0 = \dfrac{\varepsilon_0(A/2)}{d} = \dfrac{\varepsilon_0A}{2d}$.
The other half has the two slabs in series, each of thickness $d/2$:
$$C_1 = \frac{K_1\varepsilon_0A}{d},\quad C_2 = \frac{K_2\varepsilon_0A}{d},\quad C_s = \frac{\varepsilon_0A}{d}\cdot\frac{K_1K_2}{K_1+K_2}$$
These two halves are in parallel:
$$C = \frac{\varepsilon_0A}{d}\left(\frac{1}{2} + \frac{K_1K_2}{K_1 + K_2}\right)$$
Solution by Sreeraj P, M.Sc Physics