A parallel-plate capacitor with plate area A=2.0 m² and plate separation d=3.0 mm is connected to a V₀=15 V battery. With the capacitor still connected to the battery, a slab of glass with the dielectric constant K=5.0 and thickness d/2=1.5 mm is inserted between the plates of the capacitor, so that the half of the gap is filled with the dielectric as shown in the figure. Note: permittivity constant ε₀=8.85x10⁻¹² (F/m).
a) Between the plates, find the electric fields in the air and in the dielectric respectively.
b) Find the magnitudes of the (free) charge on the surface of the plates, and the (bound) induced charge on the surface of the dielectric respectively.
c) Find the capacitance of this capacitor and the energy stored in the capacitor.