Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

$A$ medium having dielectric constant $K > 1$ fills the space between the plates of a parallel plate capacitor. The plates have a large area,and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$,as shown in Figure $(a)$. Now,the plates are moved such that the distance between them becomes $2d$,with the dielectric slab of thickness $d$ remaining in between,as shown in Figure $(b)$. In the process of going from the configuration depicted in Figure $(a)$ to that in Figure $(b)$,which of the following statement$(s)$ is(are) correct?

$A$ parallel plate capacitor with air between the plates has a capacitance of $4 \ pF$. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $K = 6$,then the value of capacitance will be . . . . . . . (in $pF$)

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

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: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]