A medium having dielectric constant K > 1 fills the space between...

Capacitance of both capacitors are ${C}_{a} = \dfrac{K{ε}_{0}A}{d}$ and ${C}_{b} = \dfrac{{ε}_{0}A}{d + \left(\dfrac{d}{K}\right)} = \dfrac{{ε}_{0}AK}{d\left(K + 1\right)}$

Now, charge stored is ${q}_{a} = \dfrac{K{ε}_{0}A}{d}V$ and ${q}_{b} = \dfrac{{ε}_{0}AKV}{d\left(K + 1\right)}$.

Electric field between each parallel plate capacitor is ${E}_{a} = \dfrac{{q}_{a}}{KA{ε}_{0}} = \dfrac{K{ε}_{0}AV}{dK{ε}_{0}A}$$= \dfrac{V}{d}$ and

${\left({E}_{b}\right)}_{dielectric} = \dfrac{{E}_{air}}{K}$$= \dfrac{{q}_{b}}{KA{ε}_{0}}$$= \dfrac{{ε}_{0}AKV}{d\left(K + 1\right)\left(KA{ε}_{0}\right)}$$= \dfrac{V}{d\left(K + 1\right)}$

So, from above relations, we can see that capacitance decrease by a factor of $\dfrac{1}{K + 1}$.

Work done in the process $W = {U}_{f} - {U}_{i}$

$= \dfrac{1}{2}\left({C}_{f} - {C}_{i}\right){V}^{2}$

$= \dfrac{1}{2}\left(\dfrac{{ε}_{0}AK}{d\left(K + 1\right)} - \dfrac{K{ε}_{0}A}{d}\right){V}^{2}$

$= \dfrac{1}{2}{V}^{2}\dfrac{{ε}_{0}AK}{d}\left(\dfrac{1}{K + 1} - 1\right)$

$= \dfrac{1}{2}\dfrac{{ε}_{0}AK{V}^{2}}{d}\dfrac{1 - K - 1}{K + 1}$

$= \dfrac{1}{2}\dfrac{{ε}_{0}A{V}^{2}}{d}\left(\dfrac{- {K}^{2}}{K + 1}\right)$

Thus, the work done in the process depends upon the dielectric material.