If a dielectric is inserted into a charged capacitor (battery removed),then the quantity that remains constant is

$A$ parallel plate capacitor is connected to a battery. If a dielectric slab is inserted between the plates,which of the following quantities increases?

$A$ parallel plate capacitor of capacitance $12.5 \ pF$ is charged by a battery connected between its plates to a potential difference of $12.0 \ V$. The battery is now disconnected and a dielectric slab $(\epsilon_{r}=6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is . . . . . . . $\times 10^{-12} \ J$.

$A$ parallel plate capacitor has a capacitance of $C_0$. If a dielectric slab of relative permittivity $\varepsilon_r$ and thickness equal to one-fourth of the distance between the plates is inserted,the new capacitance is $C$. Then,the ratio $\frac{C}{C_0}$ is:

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

$A$ 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: