A parallel plate capacitor has plate area $100\, m ^{2}$ and plate separation of $10\, m$. The space between the plates is filled up to a thickness $5\, m$ with a material of dielectric constant of $10 .$ The resultant capacitance of the system is $'x'$ $pF$. The value of $\varepsilon_{0}=8.85 \times 10^{-12} F \cdot m ^{-1}$ The value of $'x'$ to the nearest integer is…………

Explain the effect of dielectric on capacitance of parallel plate capacitor and obtain the formula of dielectric constant.

A parallel plate capacitor with air between the plates has a capacitance of $9\,pF$. The separation between its plates is $'d'$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac {d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $\frac {2d}{3}$ . Capacitance of the capacitor is now……..$pF$

A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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 $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$