A frictionless dielectric plate $S$ is kept on a frictionless table $T$. A charged parallel plate capacitance $C$ (of which the plates are frictionless) is kept near it. The plate $S$ is between the plates. When the plate $S$ is left between the plates

If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become

Two capacitors of capacities $2 {C}$ 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

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.

The respective radii of the two spheres of a spherical condenser are $12\;cm$ and $9\;cm$. The dielectric constant of the medium between them is $ 6$. The capacity of the condenser will be