$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?

$A$ simple pendulum of mass $m$, length $l$, and charge $+q$ is suspended in the electric field produced by two conducting parallel plates as shown. The value of the deflection of the pendulum in the equilibrium position will be -

$A$ 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 . . . . . . .

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)