$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ has a capacitance of $4 \, \mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K = 3$ (as shown in figure) will be .........$ \mu F$.

$A$ 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 $6$?

If two charges $q_1$ and $q_2$ are separated by a distance $d$ and placed in a medium of dielectric constant $K$, what will be the equivalent distance between the charges in air for the same electrostatic force?

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:

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is

Two identical capacitors $A$ and $B$ are connected as shown in the circuit. Initially,the switch $S$ is closed. Now,the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K = 3$. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is