Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now 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

$A$ parallel plate capacitor has charge $5 \times 10^{-6} \ C$. $A$ dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \ C$, then the dielectric constant of the slab is . . . . . . .

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil 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 . . . . . . $V$.