A parallel plate capacitor has plate area 40 cm ^ 2 and plates...

With insertion of dielectric, the system can be considered as combination of two capacitors in series.

Thus, equivalent capacitance of the system is

$\dfrac{1}{{C}_{eq}} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}}$

$= \dfrac{1}{\dfrac{K{ε}_{0}A}{t}} + \dfrac{1}{\dfrac{{ε}_{0}A}{d - t}}$

$= \dfrac{t}{K{ε}_{0}A} + \dfrac{d - t}{{ε}_{0}A}$

$= \dfrac{1 \times {10}^{- 3}}{5{ε}_{0} \times 40 \times {10}^{- 4}} + \dfrac{1 \times {10}^{- 3}}{{ε}_{0} \times 40 \times {10}^{- 4}}$

$⇒ \dfrac{1}{{C}_{eq}} = \dfrac{1}{20{ε}_{0}} + \dfrac{1}{4{ε}_{0}}$

$⇒ {C}_{eq} = \dfrac{20 \times 4{ε}_{0}}{24} = \dfrac{10{ε}_{0}}{3}F$.