$A$ metal plate of thickness $2 \,mm$ and area $36 \pi \,mm^2$ is slid into a parallel plate capacitor of plate spacing $6 \,mm$ and area $36 \pi \,cm^2$. The metal plate is at a distance $3 \,mm$ from one of the plates. What is the capacitance of this arrangement (in $\,pF$)? (Let $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,N m^2 C^{-2}$)

$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

$A$ capacitor has capacitance $C_0$ when there is no dielectric between its plates. Two slabs of dielectric constant $K_1$ and $K_2$ respectively,with area equal to the area of the plates but thickness half of the distance between the plates,are placed in between the plates. Then the new capacitance is

The distance between the plates of a parallel plate capacitor is $d$. $A$ metal plate of thickness $d/2$ is introduced between the plates. The capacitance will then be

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.

The parallel combination of two air-filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage $V$. When the capacitors are fully charged,the battery is removed,and after that,a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is