The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

$A$ parallel plate capacitor has plates of length $l$,width $w$,and separation $d$. It is connected to a battery of emf $V$. $A$ dielectric slab of the same thickness $d$ and dielectric constant $k = 4$ is being inserted between the plates. At what length $x$ of the slab inside the plates will the energy stored in the capacitor be two times the initial energy stored?

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

$A$ 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?

An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase:

In the given figure,there are two capacitors in series. $A$ rigid conducting slab of thickness $b$ can slide vertically. Find the equivalent capacitance of the system.