$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled as shown in the figure with a dielectric constant $K = 5$, the percentage increase in the capacitance is . . . . . . .

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge:

Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

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:

Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

The voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of $10^6\,V/m$. The plate area is $10^{-4}\,m^2$. What is the dielectric constant if the capacitance is $15\,pF$? (Given $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$)