If two plates of a capacitor with constant charge have their separation doubled, the energy stored also doubles. I'm confused as to why this happens, since I thought E-stored was related to C, and C decreased as distance of separation increased. I'm looking at the equation Energy Stored = 0.5CV^2.
You are right that the energy stored in a capacitor if related to C, the capacitance. However, you have to consider what is being held constant as the separation doubles. If you look at the equation: U=(1/2)CV2, you will notice that U is proportional to C, however you have a factor of V2 that changes in a way you can't determine directly. If instead, you re-write the equation substituting in V=Q/C, so that you have: U=(1/2)Q2/C, then you know that Q is a constant, so we can talk about U being proportional to 1/C. Thus, as C decreases, U increases.
It might help you to look at the example I posted on this if you haven't already: Chapter 17, #50. Note also that since the amount of energy stored in the capacitor is increased, work must be done to increase the separation between the plates.
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