$A$ parallel plate capacitor having air as the dielectric medium is charged by a potential difference of $V$ volt. After disconnecting the battery,the distance between the plates of the capacitor is increased using an insulated handle. As a result,the potential difference between the plates . . . . . . .

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 capacitor is filled with two dielectrics as shown in the figure. The area of each plate is $A \; m^2$ and the separation between the plates is $t \; m$. The dielectric constants are $k_1$ and $k_2$ respectively. Its capacitance in farad will be:

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is:

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

$A$ parallel plate capacitor with plate separation $5 \text{ mm}$ is charged by a battery. On introducing a mica sheet of $2 \text{ mm}$ thickness and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is . . . . . . .