$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then

Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

$A$ 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:

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?

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