$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

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).

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is

Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?