A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of $1 \mu \mathrm{~F}$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13} \Omega \mathrm{~m}$ and resistance $17.7 \times 10^{14} \Omega$. Relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is $\_\_\_\_$ .

(Take permittivity of free space $=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$ )

The stored charge in the capacitor in steady state of the following circuit is $\_\_\_\_$ $\mu \mathrm{C}$.

The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1=2, K_2=3$ and $K_3=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .

A capacitor $P$ with capacitance $10 \times 10^{-6} \mathrm{~F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \mathrm{~F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} C$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is $\_\_\_\_$ .