The area of each plate of a parallel-plate capacitor is $100 {\mathrm{cm}}^2$ and the distance between them is$0.05 \mathrm{cm}.$ When it is filled with a dielectric, its capacitance becomes $3.54 \mathrm{x} {10}^{-10} \mathrm{F}.$Find the dielectric constant.

When a $20 \mathrm{V}$ Battery is connected to an air-capacitor, a charge of $30 \mathrm{\mu C}$ Is obtained on the capacitor. When oil is filled between the plates of the capacitor, then a charge of $75 \mathrm{\mu C}$Is obtained. Calculate the dielectric constant of the oil.

When a $20\mathrm{V}$ Battery is connected to an air-capacitor, a charge of $30\mathrm{\mu C}$ Is obtained on the capacitor. When oil is filled between the plates of the capacitor, then a charge of $75\mathrm{\mu C}$ Is obtained. Calculate the energy contained in the oil-capacitor.

A parallel-plate air capacitor has a plate area of$6 \times {10}^{-3}{\mathrm{m}}^2$ and a plate separation of $3 \mathrm{mm}$. A potential difference of $100 \mathrm{V}$ Is established between its plates by a battery. After disconnecting the battery, the space between the plates is filled by a $3 \mathrm{mm}$ thick mica sheet of dielectric constant$\mathrm{K} = 6$. Find the new$\mathrm{PD}$ Between the plates.

A parallel-plate air capacitor has a plate area of $6\times {10}^{-3}{\mathrm{m}}^2$ And a plate separation of $3 \mathrm{mm}$. A PD of $100 \mathrm{V}$ Is established between its plates by a battery. After disconnecting the battery, the space between the plates is filled by a $3 \mathrm{mm}$ thick mica sheet of dielectric constant $\mathrm{K}=6$. Find the final capacitance.