$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

$A$ 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 the dielectric constant of the material is. $\{ \text{Use} : \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2} \}$

$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)