A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of $4.5\,mm$ is introduced between them, then the capacity will become.......$times$

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be

A parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$, the dielectric constant of the medium is

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