$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \, pF$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $\frac{x}{4} \, pF$. The value of $x$ is $............$

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \ F$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \ V$. The dielectric slab is then removed. The charge passing through the wire is:

The function of a dielectric in a capacitor is

The plates of a parallel plate capacitor are separated by a distance $d$ with air as the medium between them. $A$ dielectric slab of dielectric constant $K = 3$ is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is