When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

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

Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

Two square-shaped metal plates of side $1 \,m$, kept $0.01 \,m$ apart in air, form a parallel plate capacitor. It is connected to a battery of $500 \,V$. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of $0.001 \,m/s$. If the dielectric constant of the oil is $11$, then the current drawn from the battery during this process is:

$A$ parallel plate capacitor has plate area $40 \ cm^2$ and plate separation $2 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and dielectric constant $5$. The capacitance of the system is ($\varepsilon_0 =$ permittivity of vacuum)

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity