Meir Achuz
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I got to this thread late, but just noticed your "paradox".Dadface said:So basically you have increased the capacitance by increasing the plate area.It is true that if you take a large and a small capacitor and charge them to the same voltage then the large one will store more energy.With this paradox we can have different situations:
1.BATTERY PERMANENTLY CONNECTED.In this case V will equalise when the capacitance is varied but Q will vary.If C is increased, more charge will flow from the battery and the energy stored will increase.
2.CAPACITOR CHARGED AND BATTERY DISCONNECTED.In this case Q will be constant but V will vary.If C is increased, V will decrease and the energy stored will decrease.
In my post above I was referring to situation two and although both situations give opposite results the paradox still exists.
This is treated in most EM textbooks. A battery connected to keep the capacitor at constant voltage does twice as much work as the negative energy difference at constant Q. This work done by the battery effectively changes the sign. The arithmetic is -1+2=+1.