If a parallel plate capacitor is charged with charge Q and -Q on its plates, respectively, and then isolated, then we place a dielectric into the space between the plates, does the charge on the plates decrease, or stay the same.
The Attempt at a Solution
I saw Walter Lewin's lecture in which he says that the surface charge density is changed after you put the dielectric in place (sigma_0 becomes sigma_0 - sigma_induced) and that made sense to me. The induced polarisation of the dielectric material displaces some bound charge onto the surface of the plates, creating an induced electric field to counter the existing field.
However, in a tutorial tonight one of my tutors argued that the total charge Q stays the same after the dielectric is put in place, the capacitance goes up and the voltage goes down.
Are these two explanations compatible? How can the areal surface charge density decrease but the charge on the plates remains the same?
Thanks in advance