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: (1) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2}\right)$ (2) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{2(K_1 + K_2)}{K_1 K_2}\right)$ (3) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 + K_2}{K_1 K_2}\right)$ (4) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{2(K_1 + K_2)}\right)$
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:
(1) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2}\right)$
(2) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{2(K_1 + K_2)}{K_1 K_2}\right)$
(3) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 + K_2}{K_1 K_2}\right)$
(4) $\frac{\varepsilon_0 A}{d}\left(\frac{1}{2} + \frac{K_1 K_2}{2(K_1 + K_2)}\right)$