A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$
$2$
$3$
$5$
$10$
Due to which the surface charge density arises on the surface of a dielectric slab, when it is placed in a uniform electric field ?
The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used, is $C$. If the dielectric is replaced by another with dielectric constant $20$, the capacity will become
A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$
If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as
Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant $\mathrm{K}$.