$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

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

For changing the capacitance of a given parallel plate capacitor,a dielectric material of dielectric constant $K$ is used,which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$,where $d$ is the separation between the plates of the parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $(C_{0})$ is given by the following relation:

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored:

In the adjoining figure,capacitors $(1)$ and $(2)$ have a capacitance $C$ each. When a dielectric of dielectric constant $K$ is inserted between the plates of one of the capacitors,the total charge flowing through the battery is: