$A$ parallel plate capacitor with air between the plates has a capacitance of $1.0 \text{ pF}$. If the distance between the plates is doubled and the space between them is filled with a dielectric substance,the capacitance becomes $2.0 \text{ pF}$. Then the value of the dielectric constant of the dielectric substance is . . . . . . .

$A$ capacitor of capacitance $15 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.5$,dielectric strength $30 \, MV/m$,and potential difference $V = 30 \, V$. The area of the plate is ....... $\times 10^{-4} \, m^{2}$.

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

$A$ capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=2$ is inserted between the plates. The potential difference across its terminals becomes

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

$A$ 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: