The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant $8$,and a copper plate of thickness $d/2$ is introduced between the plates of $N$ ($d$ is the distance between the plates). Then the potential differences across $M$ and $N$ are,respectively,in the ratio:

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

Two parallel plate air capacitors of the same capacity $C$ are connected in series to a battery of emf $E$. Then one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacitance of the series combination is

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 capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be: