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

Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is $2 \times {10^5}\,V/m$. When the space is filled with dielectric, the electric field becomes $1 \times {10^5}\,V/m$. The dielectric constant of the dielectric material

A parallel plate capacitor with plate separation $5$ $\mathrm{mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is_____.

Due to which the surface charge density arises on the surface of a dielectric slab, when it is placed in a uniform electric field ?

A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$

A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$