$A$ parallel-plate capacitor with plate area $A$ has a separation $d$ between the plates. Two dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$,each having an area of $A/2$ and thickness $d/2$,are inserted in the space between the plates as shown in the figure. The equivalent capacitance of the capacitor will be:

• A
  $\frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{K_{1} K_{2}}{K_{1} + K_{2}} \right)$
• B
  $\frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{K_{1} K_{2}}{2(K_{1} + K_{2})} \right)$
• C
  $\frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{K_{1} + K_{2}}{K_{1} K_{2}} \right)$
• D
  $\frac{\varepsilon_{0} A}{d} \left( \frac{1}{2} + \frac{2(K_{1} + K_{2})}{K_{1} K_{2}} \right)$