Consider a 4 nF parallel-plate capacitor with nothing between the plates that is charged by a battery at a potential difference of V. The capacitor stores an energy of 1.02 × 10^-5 J. The volume between the plates is totally filled with a sheet of dielectric while the power source is still connected. As a result, the energy-storing capacity of the capacitor is enhanced by 2.78 × 10^-5 J. i) Determine V and ii) the dielectric constant (K) of the sheet.

A student totally fills the space between parallel capacitor plates with a square piece of glass. The capacitor has a capacitance of 75 nF and is permanently connected to a 9V battery. Glass has a dielectric constant of K = 7. i) Calculate the supplementary charge that will flow onto the plates due to the insertion of the piece of glass. ii) Calculate the charge generated on each side of the glass piece, and iii) determine the impact of the glass piece on the electric field between the plates.

A camera flash unit uses a 150 µF parallel-plate capacitor without any dielectric between the plates. The two parallel plates are 3.00 mm apart, and the electric field between them is 6.00 × 10⁴ V/m. i) Determine the maximum charge Q_max on the capacitor. ii) Suppose that the volume between the capacitor plates is totally filled with teflon dielectric with K = 2.1. If the electric field between the plates remains constant at 6.00 × 10⁴ V/m, calculate the maximum charge Q_max,new on the capacitor.