$C_a=\frac{K \varepsilon_0 A}{d}$

$$ \therefore $$ $q_a=\frac{K \varepsilon_0 A}{d} V$

$ E_a =\frac{q_a}{K A \varepsilon_0}$

$=\frac{K \varepsilon_0 A V}{d K \varepsilon_0 A} $

$ =\frac{V}{d}$

$\begin{aligned} C_b & =\frac{\varepsilon_0 A}{d+\left(\frac{d}{K}\right)} \\\\ & =\frac{\varepsilon_0 A K}{d(K+1)}\end{aligned}$

$q_b=\frac{\varepsilon_0 A K V}{d(K+1)}$

$\left(E_b\right)_{\text {dielectric }} $

$=\frac{E_{\text {air }}}{K} $

$=\frac{q_b}{K A \varepsilon_0} $

$=\frac{\varepsilon_0 A K V}{d(K+1)\left(K A \varepsilon_0\right)}$

$ =\frac{V}{d(K+1)}$

$\begin{aligned} &\therefore\text { Capacitance decrease by a factor of } \frac{1}{K+1} \\\\ & \text { Work done in the process }=U_f-U_i \\\\ &=\frac{1}{2}\left(C_f-C_i\right) V^2 \\\\ &=\frac{1}{2}\left(\frac{\varepsilon_0 A K}{d(K+1)}-\frac{K \varepsilon_0 A}{d}\right) V^2 \\\\ &=\frac{1}{2} V^2 \frac{\varepsilon_0 A K}{d}\left(\frac{1}{K+1}-1\right) \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A K V^2}{d} \frac{1-K-1}{K+1} \\\\ &=\frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\left(\frac{-K^2}{K+1}\right)\end{aligned}$