$A$ parallel plate capacitor has a plate separation of $d$. $A$ metal plate of thickness $d/2$ and the same area as the capacitor plates is inserted between them without touching the plates. The capacitance of the resulting capacitor becomes:

Between the plates of a parallel plate capacitor,a plate of thickness $t_1$ and dielectric constant $k_1$ is placed. In the rest of the space,there is another plate of thickness $t_2$ and dielectric constant $k_2$. The potential difference across the capacitor will be

$A$ parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$. What will be the capacitance if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ capacitor has capacitance $C_0$ when there is no dielectric between its plates. Two slabs of dielectric constant $K_1$ and $K_2$ respectively,with area equal to the area of the plates but thickness half of the distance between the plates,are placed in between the plates. Then the new capacitance is

The figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plates of both capacitors is filled with a dielectric of dielectric constant $K = 3$. What will be the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric?

$A$ parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor,which of the following does not change?