A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ……. $ \times 10^{-4}\; m ^{2}$

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of $\in _r = 6$ is inserted in it then energy and capacity becomes  (Assume charge on plates remains constant)

A parallel plate capacitor is of area $6\, cm^2$ and a separation $3\, mm$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be

A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of $4.5\,mm$ is introduced between them, then the capacity will become…….$times$

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