$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

In a parallel plate capacitor,two dielectric slabs of thicknesses $t_1$ and $t_2$ and dielectric constants $K_1$ and $K_2$ respectively are inserted. What is the capacitance of this capacitor?

$A$ parallel-plate capacitor of plate area $10 \text{ cm}^2$ and plate separation $3 \text{ mm}$ is charged to a potential difference $12 \text{ V}$ and then the battery is disconnected. $A$ slab of dielectric constant $3$ is then inserted between the plates. The work done on the system in the process of inserting the slab is $\alpha \varepsilon_0$. The value of $\alpha$ is (Take $\varepsilon_0$ as the permittivity of free space).

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:

Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distances as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k = 2$. The capacitance of the system between $1$ and $3$ and between $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{C_1}{C_2}$ is