The capacity of air filled parallel plate capacitor is $\mathrm{C}_0$. One-half of the space between the plates is filled with a dielectric constant ' K ' as shown in figure. The new capacity becomes $\mathrm{C}_{\mathrm{n}}$. The ratio $\mathrm{C}_{\mathrm{n}}$ to $\mathrm{C}_0$ is

Seven capacitors each of capacitance $2 \mu \mathrm{~F}$ are to be connected in a configuration to obtain an effective capacitance $\left(\frac{10}{11}\right) \mu \mathrm{F}$. The combination is

Two identical metal plates are given charges $q_1$ and $q_2\left(q_2 < q_1\right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' $V$ ' between the plates is

Five capacitors, each of capacity ' $C$ ' are connected as shown in the figure. The resultant capacity between A and B is $14 \mu \mathrm{~F}$. The capacity of each capacitor is