$A$ parallel plate capacitor has plates of area $0.4 \pi \,m^2$ and spacing of $0.5 \,mm$. If a slab of thickness $0.5 \,mm$ and dielectric constant $4.5$ is introduced in between the plates of the capacitor, then the capacitance of the capacitor is

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if a dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where $d$ is the distance of separation between the plates. The dielectric constant of the slab is

$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?