In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

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 the bound charge $q_{b}$ can be expressed as:

$A$ parallel plate capacitor has a capacitance $C$ and the distance between the plates is $d$. If the space between the plates is filled with a dielectric material of dielectric constant $K$ as shown in the figure,find the new capacitance of the capacitor.

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

$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

$A$ parallel plate capacitor whose capacitance $C$ is $14 \, pF$ is charged by a battery to a potential difference $V = 12 \, V$ between its plates. The charging battery is now disconnected and a porcelain plate with dielectric constant $k = 7$ is inserted between the plates. The plate would oscillate back and forth between the plates with a constant mechanical energy of $.......... pJ$. (Assume no friction)