In a parallel plate capacitor set up, the plate area of capacitor is $2 \,m ^{2}$ and the plates are separated by $1\, m$. If the space between the plates are filled with a dielectric material of thickness $0.5\, m$ and area $2\, m ^{2}$ (see $fig.$) the capacitance of the set-up will be $.........\, \varepsilon_{0}$

(Dielectric constant of the material $=3.2$ ) and (Round off to the Nearest Integer)

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

A capacitor has some dielectric between its plates and the capacitor is connected to a $\mathrm{D.C.}$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$

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