A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,

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

A frictionless dielectric plate $S$ is kept on a frictionless table $T$. A charged parallel plate capacitance $C$ (of which the plates are frictionless) is kept near it. The plate $S$ is between the plates. When the plate $S$ is left between the plates

A medium having dielectric constant $K>1$ fills the space between the plates of a parallel plate capacitor. The plates have 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, both the plates are moved by a distance of $\frac{d}{2}$ from their original positions, 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?

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections, the field between the plates shall be