$A$ capacitor of capacitance $20 \ \mu F$ has a distance of $2 \ mm$ between its plates. If a dielectric slab of thickness $1 \ mm$ and dielectric constant $K = 2$ is inserted between the plates,the new capacitance will be ..... $\mu F$.

$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

There is an air-filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax,the capacitance increases to $2\,pF$. The dielectric constant of wax is

$A$ parallel plate capacitor is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored: