Between the plates of a parallel plate capacitor of plate area $A$ and capacity $0.025 \mu F$,a metal plate of area $A$ and thickness equal to $1/3$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \ V$,then the amount of work done to remove the metal plate from the capacitor is (in $\mu J$)

$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

$A$ parallel plate capacitor $C$ with plates of unit area and separation $d$ is filled with a liquid of dielectric constant $K=2$. The initial level of the liquid is $\frac{d}{3}$. Suppose the liquid level decreases at a constant speed $V$. Find the time constant $\tau$ as a function of time $t$.

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

$A$ parallel plate capacitor is formed by two plates each of area $30 \pi \, cm^{2}$ separated by $1 \, mm$. $A$ material of dielectric strength $3.6 \times 10^{7} \, Vm^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6} \, C$,the value of the dielectric constant of the material is. $\{ \text{Use} : \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2} \}$