In a parallel plate capacitor,the separation between plates is $3x$. This separation is filled by two layers of dielectrics,in which one layer has thickness $x$ and dielectric constant $3k$,and the other layer has thickness $2x$ and dielectric constant $5k$. If the plates of the capacitor are connected to a battery,then the ratio of the potential difference across the dielectric layers is

$A$ parallel plate air capacitor has a capacitance of $10 \ \mu F$. As shown in the figure,if this capacitor is divided into two equal parts and these parts are filled with dielectric materials of dielectric constants $K_1 = 2$ and $K_2 = 4$,then the capacitance of this arrangement is ............. $\mu F$.

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$,the percentage increase in the capacitance 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?

When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?

In a parallel plate capacitor,the separation between the plates is $3\,mm$ with air between them. Now,a $1\,mm$ thick layer of a material of dielectric constant $K = 2$ is introduced between the plates,due to which the capacity increases. In order to bring its capacity back to the original value,the separation between the plates must be increased to......$mm$. (in $.5$)