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 dielectric slabs of dielectric constants $K_1$ and $K_2$ and of the same thickness are inserted in a parallel plate capacitor. Given $K_1 = 2K_2$. If the potential differences across the slabs are $V_1$ and $V_2$ respectively,then:

If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $4$,the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant $4$,the capacitance is $C_2$. If in both cases,the dielectric is placed parallel to the plates of the capacitor,then $C_1: C_2=$

$A$ capacitor has capacity $C$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material of dielectric constant $K$ having an area equal to that of the plates but thickness $\frac{d}{2}$ is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be:

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

The distance between the plates of a charged parallel plate capacitor is $5 \ cm$ and the electric field inside the plates is $200 \ V/cm$. An uncharged metal bar of width $2 \ cm$ is fully immersed into the capacitor. The length of the metal bar is the same as that of the plate of the capacitor. The voltage across the capacitor after the immersion of the bar is......$V$