A parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric  of dielectric constant $5$, the percentage increase in the capacitance will be......$\%$

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in

A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

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 ....... $\times 10^{-4}\, m ^{2}$

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