A parallel plate capacitor with plate area A and plate separation d...

This parallel plate capacitor can be divided into two capacitors. One with dielectric $\left({C}_{1}\right)$ and other without dielectric\left({C}_{2}\right)$. The two capacitors will be in series.</p><p>Initially$C = \dfrac{{ε}_{0}A}{d} = 4\mu F$</p><p>Now,${C}_{1} = \dfrac{k{ε}_{0}A}{\dfrac{d}{2}} = \dfrac{2 \times 3 \times {ε}_{0}A}{d} = 24\mu F$and${C}_{2} = \dfrac{{ε}_{0}A}{\dfrac{d}{2}} = \dfrac{2 \times {ε}_{0}A}{d} = 8\mu F$</p><p>Finally${C}^{'} = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}} = \dfrac{24 \times 8}{24 + 8} = 6\mu F