The plates of a parallel plate capacitor are charged upto $200 \mathrm{V}$. A dielectric slab of thickness 4 mm is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{mm}$. The dielectric constant of the dielectric slab is

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has the capacitance $C_1$ in air. When two dielectrics of different relative permittivity $\left({\epsilon }_1=2 \text{and} {\epsilon }_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is, 

Find the amount by which the total energy stored in the capacitor (in $\mu \mathrm{J}$) will increase in the circuit shown in the figure after the switch $K$ is closed. Consider all capacitors are of capacitance $3 \mathrm{\mu F}$ and battery voltage is $10 \mathrm{V}$.