The potential gradient at which the dielectric of a condenser just gets punctured is called
Dielectric constant
Dielectric strength
Dielectric resistance
Dielectric number
When air in a capacitor is replaced by a medium of dielectric constant $K$, the capacity
Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be.....$m$
Two dielectric slab of dielectric constant $K_1$ and $K_2$ and of same thickness is inserted in parallel plats capacitor and $K_1 = 2K_2$ . Potential difference across slabs are $V_1$ and $V_2$ respectively then
The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively ${C_o}$ and ${W_o}$. If the air is replaced by glass (dielectric constant $= 5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be
Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is