$A$ parallel plate air capacitor has a capacitance of $100\,\mu F$. The plates are at a distance $d$ apart. If a slab of thickness $t$ $(t < d)$ and dielectric constant $K = 5$ is introduced between the parallel plates,then the new capacitance will be:

$A$ parallel plate capacitor has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

$A$ 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 capacitor with width $4\,cm$,length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. $A$ dielectric slab of dielectric constant $5$ having length $1\,cm$,width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of the parallel plate capacitor. The electrostatic energy of this system will be ......... $\epsilon_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is: