A capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when medium between plates is air. The charge on the plates become $100\,\mu C$ when space between plates is filled with oil. The dielectric constant of oil is

While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)

A parallel plate capacitor is formed by two plates each of area $30 \pi\, cm ^{2}$ separated by $1\, mm$. A material of dielectric strength $3.6 \times 10^{7} \,Vm ^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\, C$, the value of dielectric constant of the material is

$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$

A parallel plate capacitor is made of two circular plates separated by a distance $5\ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4$ $ Vm^{-1}$ the charge density of the positive plate will be close to

In a parallel plate capacitor the separation between the plates is $3\,mm$ with air between them. Now a $1\,mm$ thick layer of a material of dielectric constant $2$ is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made......$mm$