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 charge on the 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 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 $\mathrm{K}=6$.  If the battery remains connected to the capacitor, then what will be the answers of the above parts.

The capacitance of a variable radio capacitor can be changed from$50 \mathrm{pF}$ to $950 \mathrm{pF}$ By tuning the dial from $0$ to $180°.$ With the dial set at$180°$, the capacitor is connected to a $400 \mathrm{V}$ Battery. After charging, the capacitor is disconnected from the battery and the dial is turned to$0.$ What is the potential difference across the capacitor when the dial reads $0$?