A parallel plate capacitor of capacitance $12.5 \mathrm{pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab $\left(\epsilon_{\mathrm{r}}=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is_______.$\times 10^{-12} \mathrm{~J}$.

Between the plates of a parallel plate condenser there is $1\,mm$ thick paper of dielectric constant $4$. It is charged at $100\;volt$. The electric field in $volt/metre$ between the plates of the capacitor is

A parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric  of dielectric constant $5$, the percentage increase in the capacitance will be......$\%$

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

A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$

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)