$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity $10$ is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is

$A$ 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 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 separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?