In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. Then,

$A$ parallel plate capacitor is connected to the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of thickness $4.5\,mm$ is introduced between them,then the capacity will become how many times the original capacity?

$A$ parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$,the dielectric constant of the medium is

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

$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \ \mu C$ flows through the battery. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.

Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.