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.......$\mu J$

The potential gradient at which the dielectric of a condenser just gets punctured is called

How does the polarised dielectric modify the original external field inside it ?

A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .

[ $\epsilon_0$ is the permittivity of free space]

The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be

The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is