
$Q_{e q}=C_{e q} E=C E / 2$
As battery is connected to capacitor so $E$ remains constant.
After insert dielectric $, C_{e q}^{\prime}=\frac{K C C}{K C+C}=\frac{K}{K+1} C$
$Q_{e q}^{\prime}=C_{e q}^{\prime} E=\frac{K}{K+1} C E$
Charge flow through the battery is $\Delta Q=Q_{e q}^{\prime}-Q_{e q}=\frac{K}{K+1} C E-\frac{C E}{2}=\frac{(K-1) C E}{2(K+1)}$
As $\mathrm{E}$ is constant so for inserting dielectric the charge on the capacitor will increase and it will flow from capacitance to battery to maintain constant $E$



