The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\, \mu F$. The permittivity of the medium is……….$C ^{2} N ^{-1} m ^{-2}$

$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $………V.$ (Assuming Dielectric constant $=10$ )

Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

Between the plates of a parallel plate capacitor a dielectric plate is introduced just to fill the space between the plates. The capacitor is charged and later disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference across the plates and the length of the dielectric plate drawn out is