Two parallel plate capacitors of capacity $C$ and $3\,C$ are connected in parallel combination and charged to a potential difference $18\,V$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be $V$

The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:

$K(x) = K_0 + \lambda x$ ( $\lambda  =$ constant)

The capacitance $C,$ of the capacitor, would be related to its vacuum capacitance $C_0$ for the relation

The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out

A parallel plate capacitor with air between the plate has a capacitance of $15 pF$. The separation between the plate becomes twice 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 is of area $6\, cm^2$ and a separation $3\, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be