Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)

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

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants ${k_1},{k_2}$ and ${k_3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

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

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