$A$ 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 ....... $\times 10^{-4} \, m^{2}$.

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

While a capacitor remains connected to a battery and a dielectric slab is inserted between the plates,then:

$A$ parallel plate capacitor having plates of area $S$ and plate separation $d$,has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $(\varepsilon_1=2$ and $\varepsilon_2=4)$ are introduced between the two plates as shown in the figure,the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

Two point charges are placed at a distance $r$ in air,experiencing a force $F$. When they are placed in a medium with a dielectric constant $K$,at what distance will the force between them remain the same?