A capacitor has air as dielectric medium and two conducting plates of area $$12 \mathrm{~cm}^2$$ and they are $$0.6 \mathrm{~cm}$$ apart. When a slab of dielectric having area $$12 \mathrm{~cm}^2$$ and $$0.6 \mathrm{~cm}$$ thickness is inserted between the plates, one of the conducting plates has to be moved by $$0.2 \mathrm{~cm}$$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $$\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$)

In the network shown below, the charge accumulated in the capacitor in steady state will be: