The plates of a parallel plate capacitor are separated by a distance 'd' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

The graph shows the variation of voltage ' V ' across the plates of two capacitors $A$ and $B$ versus increase in charge ' $Q$ ' stored in them. Then

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