Air capacitor has capacitance ' $\mathrm{C}_1$ '. The space between two plates of capacitor is filled with two dielectrics as shown in figure. The new capacitance of the capacitor is ' $\mathrm{C}_2$ '. The ratio $\frac{C_1}{C_2}$ is $(d=$ distance between two plates of capacitor, $\mathrm{K}_1$ and $\mathrm{K}_2$ are dielectric constants of two dielectrics respectively)

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $\mathrm{K} . \mathrm{Q}_{\mathrm{A}}$ and $\mathrm{Q}_{\mathrm{B}}$ are the charges stored in A and B . When the dielectric slab is removed, the corresponding charges are $\mathrm{Q}_{\mathrm{A}}^{\prime}$ and $\mathrm{Q}_{\mathrm{B}}^{\prime}$. Then

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference 6 V . Another parallel combination of $\mathrm{n}_2$ capacitors, each of value $\mathrm{C}_2$ is charged by a source of potential difference 2 V . Total energy of both the combinations is same. The value of $\mathrm{C}_2$ in terms of $\mathrm{C}_1$ is

In the circuit shown in the following figure, the potential difference cross $3 \mu \mathrm{~F}$ capacitor is