An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now,the space between the plates is filled with a liquid of dielectric constant $K = 5$. The charge that flows from the battery to the capacitor is......$\mu C$.

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in the figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q'_1$ and $Q'_2$. Then:

$A$ 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:

$A$ parallel plate capacitor is filled with $3$ dielectric materials of the same thickness,as shown in the sketch. The dielectric constants are such that $k_3 > k_2 > k_1$. Let the magnitudes of the electric field in and potential drops across each dielectric be $E_3, E_2, E_1$ and $\Delta V_3, \Delta V_2, \Delta V_1$,respectively. Which one of the following statements is true?

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.