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

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then:

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be: