$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

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

$A$ parallel plate capacitor is made of two square parallel plates of area $A$,and separated by a distance $d << \sqrt{A}$. The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. $A$ square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates,so that it is always parallel to the plates. How much work has been done on the metal slab by an external agent while it is being inserted?

There is an air-filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax,the capacitance increases to $2\,pF$. The dielectric constant of wax is