$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

$A$ 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.....$\%$

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?