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:

A

$$\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2}\right)$$

B

$$\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{2(K_1 + K_2)}{K_1 K_2}\right)$$

C

$$\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 + K_2}{K_1 K_2}\right)$$

D

$$\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{2(K_1 + K_2)}\right)$$