$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates,then the bound charge $q_{b}$ can be expressed as:

Two point charges are kept in air with a separation $r$ between them. The force between them is $F_1$. If half of the space between the charges is filled with a dielectric of dielectric constant $K=4$,the force between them becomes $F_2$. If $1/3$ rd of the space between the charges is filled with a dielectric of dielectric constant $K=9$,then the ratio $F_1/F_2$ is:

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

$A$ parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. $A$ dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes: