$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is:

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:

$A$ parallel plate capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be:

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become: