$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$

The capacitance of a parallel plate capacitor is $2.5 \mu F$. When it is half filled with a dielectric as shown in the figure,its capacitance becomes $5 \mu F$. The dielectric constant of the dielectric is

Between the plates of a parallel plate capacitor,a plate of thickness $t_1$ and dielectric constant $k_1$ is placed. In the rest of the space,there is another plate of thickness $t_2$ and dielectric constant $k_2$. The potential difference across the capacitor will be

$A$ medium having dielectric constant $K > 1$ fills the space between the plates of a parallel plate capacitor. The plates have a large area,and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$,as shown in Figure $(a)$. Now,the plates are moved such that the distance between them becomes $2d$,with the dielectric slab of thickness $d$ remaining in between,as shown in Figure $(b)$. In the process of going from the configuration depicted in Figure $(a)$ to that in Figure $(b)$,which of the following statement$(s)$ is(are) correct?

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference $V$ and then the battery is disconnected. $A$ slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then:

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: