If inserting a dielectric has the effect of reducing the magnitude of the electric field in a capacitor (holding all other variables constant), then why is the energy stored in a capacitor directly proportional to the relative permittivity of the dielectric? This seems contradictory to me. A higher relative permittivity leads to a more reduced electric field magnitude. It seems like a decreased electric field magnitude would mean reduced energy stored in the capacitor. What am I missing here?
EDIT: Folks, let's assume that the cap is connected to a constant voltage source, so V cannot change. Also, what I mean by hold all other variables constant is to hold V, plate area, and plate distance constant.
EDIT: After stumbling across (web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide05.pdf), it appears that the answer to my specific scenario question can be found on page 5-22. With a constant voltage source in play, Q has to change and the E field appears to remain constant across the cap since E=V/d. So it's not always true that the E field strength decreases with dielectrics added, contrary to what certain sources imply (Wikipedia and parts of hyperphysics)- only under certain conditions is that the end result for a cap (i.e., if the voltage across the cap can change/is not held constant by a source).
Best Answer
It seems like a decreased electric field magnitude would mean reduced energy stored in the capacitor
If that were true, then where did the energy go when the dielectric was inserted?
The energy stored in a capacitor depends on the charge and the capacitance of the capacitor. By inserting the dielectric you changed (increased) the capacitance of the capacitor! Since the energy and charge must remain the same, the voltage must decrease.